Chapter+1

= = =Chapter 1= toc

__Section 1__
9/7 __What Do You See?__ The blue car is about to crash into the yellow and orange car, which have both already crashed into each other. The Blue car is trying to slow down so he wouldn't crash into the cars. The person driving the Blue car is panicking and stepping on the brakes, which show the smoke from the brakes and the three lines above his head, indicating that he's panicking. The driver's hair is pointing backwards, indicating that his hair and hat are being blown back by the wind and the acceleration of the car. In the picture, the blue car is being shown stopping its front wheels, making the back of the car lift. On the side of the picture, there is a bunny sitting on the side, which tells the reader that the bunny had caused the crash between the yellow and orange cars. __What Do You Think?__ What factors affect the time you need to react to an emergency situation while driving? Tires, speed, distance, size and color of the car, bunny crossing the road, size of the person sitting next to you, the landscape, and the way the road is going affect the reaction time. If the road is going downhill, the driver may have a shorter reaction time because when they see the car and try to stop, they realize they would have to press on the brakes harder in order to stop the car before crashing. If the bunny catches your attention, you would have to react fast so you wouldn't hit the bunny or swerve off the road. __Lab:__

Group Reaction:

Investigation GROUP 6 Foot pedal- kim: 39 steps [2.6 seconds] nicky: 42 steps [2.4 seconds]

Method A. Starting and stopping (seconds)- alex: (s)8.63 | 12.36 | (s)4.84 nicky: 8.89 | (s)12.15 | 5.11 average- .56 units kim: (s)5.51 | 5.84 | (s)7.53 jyll: 5.83 | (s)5.68 | 7.63 average- .193 units

Method B. Dropping stick- jyll: 24 cm [.22 seconds] mel: 25 cm [.23 seconds] kim: 24 cm [.22 seconds] Class Investigation Chart __**Class' Reaction Times:**__
 * || Reaction Distance || Reaction Time || Average Reaction Distance || Average Reaction Time ||
 * Jyll || 24 cm || .22 Seconds || 24.3 cm || .22 seconds ||
 * Mel || 25 cm || .23 Seconds || 24.3 cm || .22 seconds ||
 * Kim || 24 cm || .22 Seconds || 24.3 cm || .22 seconds ||
 * || Time to Move Your Foot From Gas to Break || Reaction Time measured with **Starting and Stopping Stopwatches** || Reaction Time measured with **Catching Ruler** || Reaction with **Decisions** added. || Reaction Time with **Distractions and Decisions.** ||
 * Group1 || .527 seconds || .173 seconds || .18 seconds || .268 seconds || .19 seconds ||
 * Group2 || .84 seconds || .517 seconds || .23 seconds || .273 seconds || .23 seconds ||
 * Group3 || .37 seconds || .25 seconds || .15 seconds || .14 seconds || .12 seconds ||
 * Group4 || .71 seconds || .32 seconds || .15 seconds || .17 seconds || .19 seconds ||
 * Group5 ||  ||   ||   ||   ||   ||
 * Group6 || .36 seconds || .19 seconds || .14 seconds || .15 seconds || .14 seconds ||

__Gas to Brake Pedal Time:__ Fastest: .36 seconds Slowest: .84 seconds Average: .5614 seconds __Reaction Time: Starting and Stopping Stopwatches__ Fastest: .173 seconds Slowest: .517 seconds Average: .29 seconds __Reaction Time: Catching the Ruler__ Fastest: .14 seconds Slowest: .23 seconds Average: .17 seconds __Reaction with Decisions__ Fastest: .14 seconds Slowest: .273 seconds Average: .2 seconds __Reaction Time: Distractions and Decisions__ Fastest: .12 seconds Slowest: .23 seconds Average: .174 seconds

What factors may affect reaction time for people of the same age, like your classmates?
 * People of the same age have similar reaction times. Older classmates would have different reaction times or also almost the same reaction times.

__Method B. Dropping red & green-Reaction Time with Decisions__ kim: 15, 18, 10, 13, 10 jyll: 22, 17, 21, 11, 15 nicky: 24, 21, 16 , 19 , 13 alex: 6, 18, 20, 18, 16, mel: 25, 12, 22, 19, 20 Average time: .16 seconds

__How does your reaction time with needing to make a decision compare with your reaction time without needing to make a decision?__
 * My reaction time with needing to make decisions is slower than without needing to make decisions. When I was catching the ruler on green or red, I had to make a very quick decision because one mistake would cost the previous recorded reaction times. Most of the time my mind was focused on green, and when it was on red, I would catch it because I thought it was on green. My reaction time with needing to make decisions is not very different from without, but it is slower.

__What does this difference in reaction time when making a decision apply to your ability to avoid road hazards?__
 * Since the difference between making a decision and not making one is not very great, my ability to avoid road hazards are not very good. I would not react fast enough to avoid any hazards on the road.

__Method C. Dropping & texting-Reaction Time with Decisions and Distractions__ kim: 21 [.21 seconds] nicky: 25 [.23 seconds] jyll: 22 [.229 seconds]

__How does your reaction time with needing to make a decision while distracted with texting compare with your reaction time without the distraction of texting?__
 * When I'm distracted, I'm slower in reacting. My reaction time was slower than without the distraction of texting.

__What does this difference in reaction time when distracted apply to your ability to avoid road hazards while texting?__
 * Although I would not text while driving, this difference in reaction time would be okay to my ability to avoid road hazards while texting. I would not react fast, but I will react before I crash into another oncoming car.

1. How do distractions affect reaction time?
 * __Checking Up Questions 9/8__**
 * Distractions slow down reaction time and increase the chances of being involved in car crashes.

2. Why is driving under the influence of alcohol or drugs illegal?
 * It affects your reaction time by slowing it down.

3. Name three factors in addition to distractions and drugs or alcohol that can affect reaction time.
 * Sneezing, eating, and changing a radio station or a CD all affect a driver's reaction time. Sneezing cannot be avoided, but changing a radio station or CD can be done when your car is stopped, and the same goes for eating also.

9/9 1. How long does it take a fastball to reach home plate? 2. Why if your reaction time is under .5 seconds do you still sometimes not hit the ball? 3. Explain what else might take some time before you hit a pitch besides simply reacting to it? Likewise when you are driving what else do you need to do besides react to an obstacle before you come to a stop? 4. Can you really hit a 90 mph fastball? What factors are __not__ included in the program that decide whether you could hit the baseball in real life? 5. What strategy might you use while driving to reduce reaction time once you are alert to a possible obstruction? In the same way what might a batter do to reduce reaction time once they are alert to an incoming pitch? 6. Why might someone yell from the dugout at a batter, "hey batter batter swing batter batter!", when the batter is trying to hit the ball? How is this similar to texting while driving?
 * __Reaction Time: Science of the Fastball__**
 * About half a second.
 * Yes, if your reaction time is about a half under .5 seconds, you may not hit the ball.
 * Waiting for the pitcher to pitch and readying yourself to react. Also you would need to make sure no one is in the way of you hitting the fast ball. When you are driving and you see an obstacle, you would have to react fast by moving your foot to the brake and gas pedals.
 * Yes, but in real life it is different. In the program, it does not show you if the ball is coming or not. It simply shows 'Swing Batter' and once you see it, you have to click. In real life, you have to react when the ball is coming towards you.
 * Slow down and put your foot over the brake. A batter would stay in one stance to ready themselves for the ball to come.
 * It is a distraction to the batter, like texting is a distraction to driving.

__**Active Physics Plus 9/9**__

d=1/2at^2
 * d = distance an object is dropped (centimeters)
 * t = Time it takes the object to drop 'd' (seconds)
 * a = acceleration(speed) of the object. (980cm/s^2)

d = 1/2at^2 --> (cm) = 1/2 ((cm)/(s^2)) (s)^2 --> (cm) = 1/2(cm)


 * Distance || Time ||
 * 4.9 cm || .1 sec. ||
 * 19.6 cm || .2 sec. ||
 * When time is doubled, the distance is being quadrupled.

d = 1/2at^2 .3 cm=1/2(980 cm/s^2) (.025^2)
 * __Reaction Time Ruler 9/9__**


 * Distance || Time ||
 * .3 || .025 sec ||
 * 1.225 || .05 sec ||
 * 2.756 || .075 sec ||
 * 4.9 || .1 sec ||
 * 7.656 || .125 sec ||
 * 11.025 || .15 sec ||
 * 15.006 || .175 sec ||
 * 19.6 || .2 sec ||
 * 24.806 || .225 sec ||

__**9/12 Do Now**__

Does a race car driver need a faster reaction time than someone driving in a school zone? Explain your answer, giving examples of the dangers each driver encounters.
 * Yes, because on the race track, all drivers are driving much faster than people who drive in school zones. If there is an accident on the race track, a driver who is about to drive into the two crashed cars would have to react much faster. If they act too slow or brake too quickly, they could crash into the wall. In school zones, you must watch out for students crossing the street, keep an eye on the speed and watch for stop signs.



Dollar Bill: d = 15.4 cm Reaction time needed to catch the dollar bill: .177 seconds
 * __Can You Catch A Dollar Bill?__**

d = 1/2at^2 15.4 cm = 1/2 (980 cm/s^2) **t^2** 15.4 cm = (490 cm/s^2) **t^2** 15.4 cm/(490cm/s^2) = **t^2** 15.4 / 480 s^2 = **t^2** Square root both sides(.0314 s^2 = **t^2**)
 * .177s = t**


 * Parent's Average Reaction Time Data:**


 * Distance(cm) || Time(seconds) ||
 * 6.4 || .219 ||
 * 17.2 || .16 ||
 * 12 || .19 ||
 * 11.1 || .2 ||
 * 10.3 || .2 ||
 * 11 || .149 ||
 * 5 || .226 ||
 * 5.2 || .226 ||
 * 1.5 || .241 ||

9/14 __**Physics To Go**__ 6. What are the consequences of driving if one's reaction time is slow rather than quick? 7. Even though teenagers often have good reaction times, why is auto insurance more expensive for teenage drivers than it is for older, more experienced drivers?
 * If a driver has a slow reaction time, that will risk the chances of being safe from car accidents. They wouldn't react fast enough to drive out of the way of another car or an obstacle and would drive and crash into them. If they have a long reaction time, they would have a longer reaction distance.
 * Teenagers have more expensive auto insurance because that is the age where they are first starting out as drivers. They get their license at an early age, so the risk of being in a car crash would likely, because they are still getting used to driving. They get more distracted from phones and other distractions.

1. What are the top two causes of accidents on the road? 2. What is rubbernecking? Does rubbernecking constitute a decision or distraction?
 * __Reflecting on the Chapter Challenge__**
 * Phone activity and drunk driving are the top two causes. Phone activities like texting or calling can distract a driver greatly. They're doing two things at once, texting/talking and driving at the same time. They could easily get into a car accident being distracted by that. Also, drunk drivers cause the most accidents, because they never know what exactly they're doing. Their eyesight becomes different and they wouldn't concentrate on anything on the road.
 * Rubbernecking is act of gawking at something of interest. This constitutes a distraction, in contrast to making a decision. It distracts the driver because they slow down to watch or look at whatever is happening and they wouldn't keep their eyes on the road.

Notes: standard form of a parabola: ax^2+bx+c=0 d = (1/2) a*t^2 (x,y) (t,d) (0.02, 0.196) .196 = (1/2) (980 cm/s^2) (.02s)^2 .196 = (490)(.0004) .196 = .196

Section 1 Quiz: -Know how to measure reaction time: -Distractions -Reaction Time Ruler
 * -to move your foot
 * -with stopwatches
 * -with dropping ruler
 * -dropping a ruler with distractions/decisions
 * Alcohol/Drugs, Driver fatigue, Rubbernecking, Talking on cellphones or texting, and speed
 * - d = (1/2)at^2
 * Distance(cm)
 * Acceleration: 980 (cm/s^2)
 * Time: Seconds

Example(s): Ruler drops 15 cm (distance) T = reaction time 15 = (1/2) (980 cm/s^2) t^2 15 = 490t^2 15/490 = t^2 (.03 = t^2) <--square root .17 seconds = t

__Section 2__
9/14 Grade: __What do you see?__ - In the picture, there is a little girl and a teenage boy walking next to a measuring tape ruler. The older boy is walking farther than the little girl is, because her legs are short, and his legs are long. Their strides are different. On the other side of the ruler, there is another boy who looks like he is recording how long their stride is or how many steps it takes for each of them to get to the other side. He seems like he is recording how long it takes for each of them to get there and the difference between the both of them. The teenage boy seems like he is going to get to the end of the ruler first. __What do you think?__ Two students measure the length of the same object. One reports a length of 3 m, the other reports a length of 10 m. Has one of them made a mistake? If the students reported measurements of 3 m and 3.01 m, do you think one of them has made a mistake? __Learning Outcomes__ In this section, you will
 * One of the students had made a mistake in the first question. The lengths seem too long for one object. It is not possible or logical for an object to be a length of 3 meters and 10 meters.
 * None of the students seemed to have made a mistake, because the lengths seem very close together for one object.
 * __Calibrate__ the length of a stride
 * __Measure__ a distance by pacing it off and by using a meter stick.
 * __Identify__ sources of error in measurement
 * __Evaluate__ estimates of measurements as reasonable or unreasonable.

9/15/11

__**Section 2 Investigate: Measurement: Errors, Accuracy, and Precision**__ __**Group member information:**__ Nicole: 21 strides Stride length: 74 centimeters/.74 meters Distance: 1554 centimeters Meterstick measurement: 13 meters/1300 centimeters
 * 1554 cm * (.01 meters) = 15.54 meters

__Classroom Group information__


 * Group || Strides || Metersticks || Group || Strides || Metersticks ||
 * 1. || (13 strides)*(.93m)=12.09m || 13.16 m || 4. || (22 strides)*(.5m)=11m || 14m ||
 * 2. || (20 strides)*(.54m) = 10.8m || 13.2m || 5. || (21 strides)*(.74m) = 15.54m || 13m ||
 * 3. || (18 strides)*(.74m) = 13.34m || 13.31m || 6. || (18.3 strides)*(.56m) = 10.25m || 13.5m ||

__Average Stride Measurement__ 12.7 m __Average Meterstick Measurement__ 13.33 m

1. Do the measurements listed on your class wiki agree? 2. By how many meters do the results vary? 3. Why are there differences in the measurements made by different groups? List as many reasons as you can think of. 4. Suggest a method of making the class' measurements more precise. If all groups use your measurements how would this reduce the range of measurements collected? 5. What do you think would happen if each group were given a really long tape measure? Do you think each group would get the exact same value? Why or why not? 6. Can you develop a system that will produce measurements that would agree exactly or will there always be differences in measurement? Justify your answer. 7. Read #8 on p. 23-24 in your book then answer letters a) and b) in your wiki. (When they say 'using each technique in letter "b" they mean 'strides or meterstick?' technique) 8. If you did not have any systematic errors then name 3 ways a systematic error may have been introduced. __**Physics Vocabulary:**__ __Random Error:__ An error that cannot be corrected by calculation. __Systematic Error__: An error produced by using the wrong tool or using the tool incorrectly for measurement and can be correct by calculation. __Accuracy:__ An indication of how close a series of measurements are to an accepted value. __Precision:__ An indication of the frequency with which a measurement produces the same results.
 * The measurements on the wiki agree.
 * For the strides, the results vary from 10.25 to 15.54 with 15.54 being the highest.
 * The measurements for each group's group member varies because of their height. If one member is taller, their strides are longer, and therefore their measurements are bigger than the other members. A short member would have a smaller stride and also a smaller measurement. Everybody has a different stride length. One persons stride isn't always the same length each time they take a step. Their strides could vary from big or small lengths.
 * If everybody started on the same place and were of similar height, they could have had close and precise measurements.
 * Some people might start on different places on the tape measure. Also their strides would be different depending on the person's stride.
 * There would always be some differences in the measurements because every person's stride length is different and unpredictable.
 * 1) Our group did not have any systematic errors.
 * A systematic error would have happened if someone used the meterstick and calculated in yards, but then changed it to meters.
 * Another error would have happened if the meterstick was moved from its original position, which could have changed the correct information.
 * If someone had calculated the information incorrectly.
 * __9/16 Homework:__**

__**Checking Up:**__ 1. Explain the difference between systematic and random errors. Systematic errors can be corrected by calculations while random errors cannot. 2. Explain why there will always be uncertainty in measurement. The measurements that a person gets depends on whether or not they have used the right tools or calculated it correctly. 3. What would the positions of arrows on a target need to be to illustrate measurements that are neither accurate nor precise? The arrows would have to be put randomly.

__**DeMo: How long is the tube? /4 sided ruler**__


 * Interval || 1 meter || .1 m || .01 m || .001m ||
 * || .75 || .84 || .81 || .815 ||
 * || .82 || .85 || .81 || .814 ||
 * || .75 || .85 || .81 || .815 ||
 * || .72 || .85 || .81 || .8145 ||
 * || .71 || .82 || .81 || .813 ||
 * || .78 || .82 || .81 || .815 ||
 * || .76 || .84 || .8 || .814 ||
 * || .73 || .82 || .82 || .812 ||
 * || .79 || .81 || .81 || .812 ||
 * || .8 || .83 || .815 || .813 ||
 * || .86 || .82 || .81 || .812 ||
 * || .83 ||  ||   ||   ||
 * || .8 ||  ||   ||   ||


 * __9/19 SI system__**
 * Quantity || Base Unit || Symbol ||
 * Distance || Meters || m ||
 * Mass || Gram || g ||
 * Time || Seconds || s ||

1 m = .001 km || 1 m = 100 cm || 1 m = 1000 m ||
 * Prefix || Symbol || Multiple of 10 || Exp ||
 * Kilo || K || *10^3 (*1000) || 1 km = 1000 m
 * Centi || C || *10^-2 (*.01) || 1 cm = .01 m
 * Milli || m || *10^-3 (*.001) || 1 mm = .001 m

Groups 1, 3 and 5 all had a different meterstick and started at 2 cm of the stick, making the length of the copper tube longer than it actually is. This was a systematic error. The measurement is supposed to be 64 cm.
 * Groups || Measurement of Copper Tube ||
 * Group 1 || 66 cm ||
 * Group 2 || 64.15 cm ||
 * Group 3 || 66.1 cm ||
 * Group 4 || 64 cm ||
 * Group 5 || 66 cm ||
 * Group 6 || 64.1 cm ||

__**Active Physics Plus 9/20**__ 1. What is the range of lengths for 50 m pools that have an uncertainty of +/- 10 cm? +/- 1cm? +/- 1mm? (For example, if the uncertainty of the pool were +/- 1 m the range of lengths would be 49-51 m.) 10cm 10(.01)m +/-.1 m = 50.1 m - 49.9 m +/- .01 m = 50.01 m - 49.99 m +/- .001(1 millimeter) m = 50.001 m - 49.999 m

2. How much extra time does it take to swim 50.01 m than 49.99 m (a difference of 2 cm)? Assume a good swimmer can swim 50 m in 25 sec. +/- 1 cm 50.01 m - 49.99 m (50m)/(25sec) = (___m)/(1 sec) (50m)/(25sec) = (2m)/(1 sec) (2m)/(1sec) = (.02 m)/( t sec) t = Time to swim extra distance caused by the random error (2t = .02)/2 t = .01 sec

3. Estimate how long it takes to swim 60 cm. Assume a good time for the 1500 m race is 15 min. 30 laps * .02 m = .6 m(60 cm) 1500 m/ 15 min = 1500 m/ 15(60 sec) = 1500 m/ 900 sec = .6 m/ t sec t = time takes to swim extra distance. 1500t = .6 m(900 sec) ((.6 m)(900 sec))/1500t = .36 sec

4. In watching the Olympic Games, you hear that someone just broke the record for the 1500 m swim by 1/1000 of a second. 15.36 sec -- 50.01 m 15.35 sec -- 49.99 m
 * 50.01*30laps = 1500.3 m
 * 1599.3 m/15.36 sec = 97.68 m/ 1 sec
 * 49.99*30laps = 1499.7 m
 * 1499.7 m/ 15.35 sec = 97.7 m/ 1 sec
 * This swimmer was faster although the pool was shorter.

__**9/20 Physics to Go Homework**__ p.32/33 #3,4,6,7,8,9, Inquiring Further #1 3. Give an estimated value of something that you and your friend would agree on. Then, give an estimated value of something that you and your friend would not agree on. 4. An oil tanker is said to hold five million barrels of oil. In your estimate, how accurate is the measurement? Suppose each barrel of oil is worth $100. What is the possible uncertainty in value of the oil tanker's oil? 6. Are the following estimates reasonable? Explain your answers. a) A 2-L bottle of soft drink is enough to serve 12 people at a meeting. b) A mid-sized automobile with a full tank of gas can travel from Boston to New York City without having to refuel. 7. If you are off by 1 m in measuring the width of a room, is that as much as an error as being off by 1 m in measuring the distance between your home and your school? 8. You are driving on a highway that posts a 65 mi/h (105 km/h) speed limit. The speedometer is accurate within 5 mi/h (8 km/h). a) What speed should you drive as shown on the speedometer to guarantee that you will not exceed the speed limit? b) What could a passenger in the vehicle do while you are driving to estimate how accurate the speedometer is? (Hint: The road has mile markers, and the passenger has a wristwatch that shows seconds.) 9) Many accidents are caused by speeding. To limit the number of collisions, police officers give speeding tickets to drivers. If the speed limit were 30 mi/h (50 km/h) in a residential neighborhood, a person may get a ticket for driving at 40 mi/h (65 km/h). Legally, they could also get a ticket for traveling at 31 mi/h (51 km/h). Given the uncertainties in measurements (the driver has to keep the gas pedal “just right”), you may wish to mention how these uncertainties are a part of safe driving. You may wish to explain why driving 31 mi/h in a 30 mi/h zone does or does not warrant a ticket. If you do not think that 31 mi/h deserves a ticket, you will need to explain what speed should get a ticket and why. Inquiring Further: I would expect the vegetable weight to be precise, because the tools and weights that NIST has created today may give precise and accurate information.
 * My friend and I would agree that a car ride from my house to school would be about 5 minutes, but we would not agree that the amount of time to walk from my house to school would be about 20 minutes.
 * The amount of oil in the tanker may not be exactly five million, but it could be close. The value of the oil depends on how much is in the tank, which would not be five million barrels.
 * A 2-L bottle of soft drink would not be enough to serve 12 people. It can be enough for a few people.
 * A 2-L bottle of soft drink is about 67 fl oz. 1 cup is 8 fl oz. 67/8 = about 8 cups. There wouldn't be enough for 12.
 * The distance from Boston to New York City is very far, so a full tank of gas would not be able to cover that distance.
 * The distance from Boston to NYC is 225 miles and there are 34 miles for 1 gallon of gas.
 * A room is much smaller than the distance from home and school, so the error for a room is much more drastic than measuring the distance between home and school. The measurement of a room should be more precise because it is small, and the measurement of the distance between home and school can be estimated since it is a large distance.
 * I would drive 60-63 mph so I wouldn't exceed the limit.
 * The passenger can time how long it takes to get to each marker and estimate the accuracy of the speedometer.
 * The driver must be concentrated on the roads at all times so they would decrease the chances of getting a speeding ticket. Getting a speeding ticket for 31 mph would not be right, because the difference of speed between 30 and 31 mph is not that great. 31 mph is not a lot faster than 30 mph, and it would not cause an accident. A speed of about 38 mph or more than that, the chances of being in an accident is very likely if other cars on the road are driving slower than you.

__**9/21 Do Now Section 2 pg 31**__

__What does it mean?__ Suppose your friend mistakes a yardstick for a meterstick and measures the length of an intersection in your neighborhood. Is this error random or systematic? Which of these types of errors affect precision or accuracy? __Why do you believe?__ Suppose you want to buy some gold jewelry. The jeweler tells you that the jewelry contains exactly 1 oz of gold. How do you know that the jeweler cannot be sure that it is exactly 1 oz? __Reflecting on the section and the challenge?__ __Answer question:__ Do you think it is reasonable to tell a police officer who pulls you over for going 75 mph in a 30 mph zone that your speed is due to the uncertainty present in her radar gun? __**Page 24: a-i**__ a.) A college football player has a mass of 100 kg.(220 lb) b) A high school basketball player is 4 m(13 ft) tall. c) Your teacher works 1440 minutes every day. d) A poodle has a mass of 60 kg (about 132 lb). e) Your classroom has a volume of 150 m^3 (about 5300 ft^3) f) The distance across the school grounds is 1 km(about 0.6 miles) g) On a rural road, while driving 50 mi/h (about 80 km/h), you encounter a tractor moving very slowly. You are about 1/4 miles (0.4 km) away when you see that another automobile is coming toward you, travelling at 50 mi/h. Is it safe to pass the tractor? h) While driving your pickup truck on a rural road, you approach a narrow bridge and see you will reach it at the same time as a dump truck that is coming from the opposite direction. What must you estimate in order to decide whether to stop and wait for the dump truck to cross the bridge first, or to go ahead and squeeze by the dump truck while on the bridge?
 * This is a systematic error because they can convert the yards to meters or recalculate it. Systematic errors affect accuracy and random errors affect precision.
 * The weight could be another number that is extremely close to the actual weight of the jewelry.
 * No, because the radar gun rounds up to the nearest number, so if you were going 73, but it turned out to be 75 mph, there wouldn't be much of a difference because you were going over 30 mph in a 30 mph zone.
 * This is true because a football player needs a lot of muscle to survive on the football field.
 * This is not true because it is not possible to be 13 ft tall.
 * There is no way that a teacher works 1440 minutes per day. 1440 minutes equal 24 hours. They would not go through an entire day of teaching without rest.
 * This is unreasonable because the poodle would most likely die or kill its owner with its tremendous weight. The heaviest poodle is about 95 pounds.
 * This is reasonable because a classroom is usually 10 by 15 feet.
 * This is reasonable because the school grounds are usually very big, covering a few big sports fields and a large track and parking lot.
 * Yes it is safer because there would be a car accident involving the tractor and your car.
 * No, I would not risk being in a car accident with the dump truck by squeezing through the bridge. The dump truck is big enough for the narrow bridge, so it is better to wait for the truck to pass through

__Section 3__
__**Average Speed: Following distance and Models of Motion**__
 * ** Section3 ** || **Points** ||
 * WDYSee/Think: || /10 ||
 * Investigate: || /20 ||
 * PhysicsTalk: || /20 ||
 * PhysicsPlus: || /20 ||
 * PhysicsToGo: || /20 ||
 * Wiki || /10 ||
 * **TOTAL POINTS** || **100** ||

__What Do You See?__ __What Do You Think?__ What is a safe following distance between your automobile and the vehicle in front of you? How do you decide what a safe following distance is?
 * There are three cars on one side of the road. The first one is being rear-ended by the second car, who is being rear ended by the 3rd car. There is a bunny on the side of the road where the 3 cars are, watching the scene. The lady in the first car seems to be distracted by the bunny because it's looking at the bunny. On the other side, there is a lake/ocean and 2 cars that are driving unharmed. The bunny seems to have created the rear-end collision by distracting the lady in the blue car with its cuteness.
 * Being about 5 feet behind the vehicle is a safe distance, because when you have to brake, the car takes time to stop.
 * The speed of your car and the car in front of you. The speeds should be about the same, because if your car is going faster, the chances of crashing into them when stopping is likely.

__**Learning Outcomes**__ In this section, you will
 * **Define** and contrast average speed and instantaneous speed.
 * **Use** strobe photos, graphs, and an equation to describe speed.
 * **Use** a motion detector to measure speed.
 * **Construct** graphs of your motion.
 * **Interpret** distance-time graphs.
 * **Calculate** speed, distance and time using the equation for average speed.

__**Investigation**__

__Strobe Photo:__ A combined photo of pictures taken at equal intervals of time. Strobe photo, combined photo of pictures taken at equal intervals of time 1. Hook up motion detector to usb port 2. Plug in usb 3. Open data studio 4. Click "create experiment" 5. Screenshot the graphs we make into the wiki 6. Use scrap paper, draw axes graph 2b) Is the automobile the same distance apart between successive photos? Were your images farther apart or closer together than they were at 30 mi/h (50 km/h)? How far does each car go in one minute? The 30 mph car goes 0.5 mph while the 45 mph car goes 0.75mph. 3a) In which diagram is the automobile traveling the slowest? In which diagram is the automobile traveling the fastest? Explain how you made your choice. The car is traveling the slowest in diagram C because the photos are closer together but the car is traveling faster in diagram A because the photos are farther apart. 3b) Is each automobile traveling at a constant speed? How can you tell? Yes, because they are the same distance apart. 4a) Graph of a person walking towards the detector at a normal steady speed. b) A graph of a person walking away form the detector at a normal steady speed. c) This is a graph of a person walking away from the detector at a slow speed.

d) A graph of a person walking in both directions at a fast speed.

e) Describe the similarities and differences among the graphs. Explain how the direction and speed that the person walked contributed to these similarities and differences. When walking towards the detector the graph goes down because you're getting closer and when you walk away you're getting farther away so the line goes up. When you go faster the line goes down or up more quickly and when you move slowly the line takes a lot longer to go down or up.

5. Predicted graph of walking towards the detector slowly and walking away quickly. 6a) Graph of a person walking slowly away

6b) Graph of a person walking quickly away 6b) The graph that looks like it's more stretched out is most likely the slower one and the one that takes quicker to go up is the faster one. 7a) From your graph, determine the total distance you walked in the most recent trial. 7b) How long did it take you to walk each distance? 7c) Divide the distance you walked (your change in position) (d) by the time it took for the most recent trial (t). 7d) How could you go about predicting your position after walking for twice the time in trial 2? When you extrapolate data, you make an assumption about the walker. What is the assumption? (Extrapolate means to estimate a value outside the known data points.) 8. a) If the reaction time is 0.5 seconds, how far does the automobile travel in this time? b) How much farther will the automobile travel if the driver is distracted by talking on a cell phone or unwrapping a sandwich, so that the reaction time increases to 1.5 s? c) Answer the questions in Steps 8a and 8b for an automobile moving at 50 ft / s( about 35 mi/h or 56 km/h) d) Repeat the calculation for Step 8c) for 70 ft / s (about 48 mi/h or 77 km/h). e) Imagine a driver in an automobile in traffic moving at 40 ft/s (about 28 mi/h or 45 km/h). The driver ahead has collided with another vehicle and has stopped suddenly. How far behind the preceding automobile should a driver be to avoid hitting it, if the reaction time is 0.5 s? f) An automobile is traveling at 60 ft/s (about 40 mi/h or 65 km/h). How many automobile lengths does it travel per second? A typical automobile is 15 ft (about 5 m long).
 * 2.3 meters
 * 3.4 seconds
 * 0.676 d/t
 * Since I am walking for twice the time in trial 2, I would double the time and assume that the walker is at a consistent speed.
 * 60 ft / 1 s = x / 0.5 s
 * x = 30 ft
 * 60 ft / 1 s = x / 1.5 s
 * x = 90 ft
 * 8a) 0.5s = 25ft
 * 1.5s = 75 feet
 * 40 ft/s *0.5 s = 20ft
 * (60 ft/s) / (15 ft) = 4 automobile lengths

__**9/24/11**__ __Speed:__ The distance traveled per unit time; speed is a scalar quantity, it has no direction. __Constant Speed:__ Speed that does not change over a period of time. __Average Speed:__ The total distance traveled divided by the time it took to travel that distance. __Instantaneous Speed:__ The Speed at a given moment __Velocity__: The speed in a given direction __Doppler Effect:__ The change in the pitch, or frequency of a sound (or the frequency of a wave) for an observer that is moving relative to the source of the sound (or the source of the wave) __Reaction Distance:__ The distance that a vehicle travels in the time it takes the driver to react.
 * Physics Words:**

1. Explain how the average speed of a vehicle is different from instantaneous speed. The average speed of a vehicle is average, but instantaneous speed is the speed that people get when they drive. For instantaneous speed, you do not find the average of it. 2. How are the speed and velocity of an object different? Speed has no direction, but velocity does. 3. If the distance-time graph shows a straight, inclined line, what does the line represent? The person driving is traveling at a fast speed. As time increases, there is a greater change in position. 4. How does reaction time affect reaction distance? If a driver reacts slowly, their distance would be long because they took too long to notice an obstacle. If they reacted fast, their distance would be small because they reacted fast to avoid an obstacle.
 * Checking Up Questions:**

1. Is the Doppler effect used for anything else other than to measure speeds of galaxies? 2. Is the only way to measure the speed of galaxies to use the Doppler effect?
 * Self-Made Checking up Questions**


 * Notes:**

Equation using Quantity symbols: V(av)= ∆ D/ ∆ T

V(av) is Average Speed ∆ D is change in position or total distance traveled ∆ T is change in time or elapsed time. ∆ means a change in.

If you drive a distance of 400 mi (about 640 km) in 8 h, What is your average speed? __Strategy:__ You can use the equation for average speed. __Given:__ ∆ D = 400 mi Delta T = 8 h __Solution:__ V = d/t V = 400 mi/ 8 h V= 50 mi/h
 * Sample Problem:**

Average Speed is 50 mi/h (80 km/h) Average Speed = Distance Traveled / time elapsed

Or Time = distance/average speed
 * Using Algebra:** Distance = Average Speed * time

Quantity Symbol form: V = d/t D = v*t T = D/V

__**9/27/11 Do Now**__ Show Calculations in your wiki __1. An automobile is traveling at 90 ft/s (60 mph). If the driver's reaction time is .6 s, how far does the automobile travel during this time?__ v = ∆d / ∆t 90 ft/s = d / .6 s (90 ft/s)*(.6 s) / (.6 s)*(d / .6 s) __54 ft = d__

__2. How much further will the car travel if the driver is distracted by texting, so that reaction time is increased to 1.5 s?__ v = ∆d / ∆t 90 ft/s = d / 1.5 s (1.5 s)*(90 ft/s) = (1.5 s)*( d / 1.5 s) __135 ft = d__

d = 200m t = 1:27 -1:11 = 16 sec v = 200 m/ 16 s = 12.5 m/s = 25 mph.
 * Example:**

__**Physics To Go page 49-51 Questions: #1-11**__ 1. Describe the motion of each automobile below. The diagrams of strobe photos were taken every 3 s (seconds). 2. Sketch diagrams of strobe photos of the following: 3. A race car driver travels at 350 ft/s (that’s almost 250 mi/h) for 20 s. How far has the driver traveled during this time? 4. A salesperson drives the 215 mi from New York City to Washington, DC, in 4.5 h. 5. If you planned to bike to a park that was five miles away, what average speed would you have to maintain to arrive in about 15 min? (Hint: To compute your speed in miles per hour, consider this: What fraction of an hour is 15 min?) 6. For each graph below, describe the motion of the automobile. The vertical axes are labeled with the distance the automobile traveled, denoted d.
 * a) The car moves at a constant speed.
 * b) The car's speed becomes slower in the beginning, but there is a big gap in the middle, which indicates that the car had sped up. The car is shown being slow again at the end.
 * [[image:Automobilespeeds.jpg]]
 * (350ft / 1s) * (x ft/ 20 s)
 * x = 350 * 20
 * x = 7000 ft/s
 * a) What was her average speed?
 * 215 mi / 4.5 h
 * Average = 47.77 mph
 * b) Do you know how fast she was going when she passed through Baltimore? Explain your answer.
 * Baltimore is in the middle of the distance between NYC and Washington DC so when she passed through Baltimore, she was going 23.85 mph.
 * 5 miles / 15 minutes = 0.3 mph
 * 15 min / 60 hr = .25 hours
 * ∆t = .25 hours
 * ∆d = 5 miles
 * V av = ∆d / ∆t
 * 5 mi / .25 hr = 20 mi / hr
 * a) The automobile is speeding up and then moving at a constant speed.
 * b) The automobile speeds up quickly and continues to move at a constant speed. Then it slows down.
 * c) The automobile is moving slowly, gaining speed, but then it speeds up.
 * d) The automobile is moving steadily, gaining speed slowly.

7. Use your average reaction time from Section 1 to answer the following:
 * a) How far does your automobile travel in meters during your reaction time if you are moving at 55 mi/h (25 m/s)?
 * 25 m / s = x m/ 0.5 s
 * x = 12.5 m
 * b) How far does your automobile travel during your reaction time if you are moving at 35 mi/h (16 m/s)? How does the distance compare with the distance at 55 mi/h?
 * 16 m / s = x m / 0.71 s
 * x = 11.36 m
 * The car goes farther because the average speed is longer.
 * c) Suppose you are very tired and your reaction time is doubled. How far would you travel at 55 mi/h during your reaction time?
 * 0.71s * 2 = 1.42 s
 * 25 m / s = x m / 1.42 s
 * x = 35.5 m

8. According to traffic experts, the following distance between your automobile and the vehicle in front of you should be three seconds. As the vehicle in front of you passes a fixed point, say to yourself "one thousand one, one thousand two, one thousand three." Your automobile should not reach that point before you complete the phrase. 9. A sneeze requires you to close your eyes for one third of a second. 10) Imagine you are driving your automobile at 60 mi/h (88 ft/s) moving in a straight line and your reaction time is 0/5 s. 11) Consider an automobile traveling at 60 mi/hr. Sketch a graph showing distance traveled versus reaction time, with reaction times of 0.25 s, 0/50 s, 0.75 s, and 1.00 s.
 * a) A second is a unit of time. How can traffic experts be sure that this is a safe following distance?
 * They could make sure it is a safe following distance because they can use equations that have distance and time in them.
 * b) Will three seconds following "distance" be equally as safe on an interstate highway as on a rural road? Explain your answer.
 * No, it is not because on a highway, cars are moving much faster and the reaction time will not be enough to prevent an accident.
 * a) if you are driving at 70 mi/h (100 ft/s), how far will you travel with your eyes closed during a sneeze?
 * 70 mi / hr = x mi / (1/3 s)
 * x = 23.33 ft
 * b) Is this longer than the length of your classroom?
 * No it is not.
 * a) How far does your automobile travel in this time?
 * 60 mi / hr = x mi / 0.5 s
 * x = 30 ft
 * b) How many automobile spaces is this for an automobile that is 15 ft long?
 * Two automobile spaces.
 * c) Answer questions A and B when you travel 30 mi/h.
 * 30 mi / hr = x mi / 0.5 hr
 * x = 15 ft
 * d) Answer questions a and b when you travel 90 mi/h. What fraction of a football field is this distance?
 * 90 mi / hr = x mi / 0.5 hr
 * x = 45 ft
 * This distance is 1/8 of a football field.
 * e) If talking on the cell phone while driving at this speed doubles your reaction time, how do these distance numbers change at 30 mi/hr, 60 mi/h, and 90 mi/h?
 * The distance becomes smaller.
 * (60 mi / hr) * (1 hr / 3600 s) * (5280 ft / 1 mi)
 * 316,800 ft / 3600 s
 * 88 ft / s
 * .25 seconds
 * Vav = ∆d / ∆t
 * 88 ft / s = ∆d / 0.25 s
 * ∆d = 22 ft
 * .50 seconds
 * Vav = ∆d / ∆t
 * 88 ft / s = ∆d / 0.5 s
 * ∆d = 44 ft
 * .75 seconds
 * Vav = ∆d / ∆t
 * 88 ft / s = ∆d / 0.75 s
 * ∆d = 66 ft
 * 1.00 seconds
 * Vav = ∆d / ∆t
 * 88 ft / s = ∆d / 1 s
 * ∆d = 88 ft




 * __9/28 Active Physics Plus Notes__**

40 mi/hr || 40 mi || ∆ t 2 = 1 hour || V av = ? || 80 mi || ∆t total = 3 hours || __Total__ V av = (20 mi/hr +40 mi/hr) / 2 = 30 mi/hr V av = ∆d / ∆t Vav = 80 mi / ∆t total
 * Part || Distance || Time ||
 * 120 mi/hr || 40 mi || ∆ t 1 = 2 hours ||
 * 2
 * Whole Trip

__Part 1__: Vav = ∆d / ∆t 20 mi / hr = 40 mi / ∆t 1 ∆t 1 = 2 hours

__Part 2:__ Vav = ∆d / ∆t 40 mi / hr = 40 mi / ∆t 2 ∆t 2 = 1 hour

__Total:__ Vav = 80 mi / 3hrs = 26.67 mi/hr

Does this answer make sense? Why should the average speed be 27 mi/h instead of 30 mi/h?
 * __Active Physics Plus Questions--page 47__**
 * The answer makes sense because the average speed should be 27 mi/h instead of 30 mi/h because the question is finding the average. If you divide 80 mi / 3 hrs, you will not get 30 mph. 30 mph isn't the average speed.

Paragraph: Vav = ∆d / ∆t Vav = 100 mi / 51 hours Vav = 1.96 mi/hr
 * Make a table of the trip
 * Find Vav
 * Explain why V av is closer to 1 mph than 50 mi/hr
 * Part || Distance || Time ||
 * 1 = 1 mi/hr || 50 mi || 50 hours ||
 * 2 = 50 mi/hr || 50 mi || 1 hour ||
 * Total = 1.96 mi/hr || 100 mi || 51 hours ||

1. Draw a distance versus time graph for both situations described above (the 80 mi trip and the 100 mi trip)

__ 80 mi trip __

__100 mi trip__



2. Draw a strobe sketch for both situations described above (the 80 mi trip and the 100 mi trip)
 * __ 80 mi trip __**

__ Route 1 __ __Route 2__ __Whole Trip__
 * [-]---[-]---[-]---[-]---[-]---[-]
 * ^40 miles 20 mi/hr
 * [-][-][-]
 * ^40 miles 40 mi/hr

[-]---[-]---[-]---[-]---[-]---[-]---[-][-][-]

80 mi 27 mi/hr 3 hours

**__100 mile trip__**

__Route 1__ __Route 2__ __Whole Trip__
 * __[__-]-[-]-[-]- [-]-[-]-[-]
 * 50 mi 1 mi/hr
 * [-]--[-]--[-]--[-]
 * 50 mi 50 mi/hr

[-]-[-]-[-]-[-]-[-] -[-]- [-]--[-]--[-]--[-] 100 mi 1.96 mi/hr 51 hours

3. Suppose someone travels 50 mi at 50 mi/h, then travels 50 mi at 25 mi/h, then travels 50 mi at 10 mi/hr. a.) Estimate their average speed. b.) Calculate the average speed. How close was it to your estimate? __**10/3/11**__ __Following Jack- Part 1__ __Following Jack- part 2__ __**Distance vs Time Graph- Cyclists worksheet**__
 * V av = ∆d / ∆t
 * About 18 mi/hr
 * V av = ∆d / ∆t
 * V av = 150 mi / 8 hr
 * V av = 18.75 mi / hr

__**Section 3 Quiz--Wednesday 10/5/11 notes**__ __3 forms of motion:__
 * Vav = ∆d / ∆t
 * Strobe Photo
 * Distance vs Time Graph

__Section 4__
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing learning outcomes || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing acceleration calculations || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing #8,9 ||
 * <span style="font-family: 'Courier New',Courier,monospace;">** Section4 ** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 20px;">**Points** ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">WDYSee/Think: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">7/10
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">16/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">20/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">18/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">20/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**91** ||

__**Graphing Motion, Distance, Velocity, and Acceleration**__

__What Do You See?__ The picture is showing a scene at a traffic light with one side having the red light and the other having the green light. On the Green light side, the red car looks like it was speeding up once the green light had turned green. The person and his dog near the fire extinguisher look like they ran away from the cars because the red car was gaining speed. The yellow car on the side is stopped completely, backing up my opinion that the red car had started accelerating once the green light had appeared.

__What Do You Think?__ An automobile and a bus are stopped at a traffic light. What are some differences and similarities of the motion of these two vehicles as each goes from a stop to the speed limit of 30 mi/hr?
 * Differences: The bus would take longer to reach 30 mi/hr but a car would get to it faster because it's smaller than a bus. The car has greater acceleration.
 * Similarities: Same change in speed and velocity.

__**Investigate**__ In this Investigate, you will use a motion detector to explore motion. You will produce distance-time and velocity-time graphs for a cart as it moves down and up an inclined ramp. You will also use the defining equation to calculate acceleration.
 * Set a motion detector at the top of a ramp along with a cart. Before collecting the data, you will make several predictions.

In this section, you will:
 * __Learning Outcomes__**
 * **Measure** a change in velocity (acceleration) of a cart on a ramp using a motion detector
 * **Construct** graphs on the motion of a cart on a ramp.
 * **Define** acceleration using words and an equation
 * **Calculate** speed, distance, and time using the equation for acceleration.
 * **Interpret** distance-time and velocity-time graphs for different types of motion.

Questions

> In another the car moves at constant velocity--Third Graph > In another, the car travels faster at the beginning and slows toward the end.-Second Graph > In another, the car travels slower at the beginning and speeds up-First graph Predicted DIstance vs Time graph: __Tangent Line:__ How is the velocity changing with respect to time? Provide evidence. Screen shot 2 pictures of your graph with tangent lines at t = .5 seconds into your run and 1.5 seconds respectively into your wiki.
 * If you were to place the cart at the top of the ramp and release it to freely move down the ramp would it move through the first half of the distance in the same amount of time as the second half of the distance? Why or why not?
 * No, because it's gaining speed from going down the ramp
 * Below are four different distance vs time graphs.In one of them the cart does not move---Fourth Graph
 * **__Run 1:__**[[image:graphs.png width="591" height="540"]]
 * As the time passes the velocity increases. From 0.5 seconds the line start to go up and then it stops at 2 seconds.



__Velocity Time Graph:__ Screen shot the velocity versus time graph made on data studio. How is the velocity changing with respect to time? Provide evidence. What is your velocity at t = .5 sec, t = 1.5 sec? How does the velocity relate to the slope of the tangent line found for those same times?



__Calculate Acceleration__: Use the velocity versus time graph to calculate the car's acceleration. Remember to show your whole calculation in your wiki.

Run 2: Motion detector is at the bottom of the incline __PREDICT:__ On a piece of scrap paper draw a predicted distance versus time graph of the motion of the car. Photo Into Wiki.

TANGENT LINE: How is the velocity changing with respect to time? Provide evidence. Screen shot 2 pictures of your graph with tangent lines at t=.5 seconds into your rune and 1.5 seconds respectively into your wiki.



VELOCITY TIME GRAPH: screen shot the velocity versus time graph made on data studio. How is the velocity changing with respect to time? Provide evidence. What is your velocity at t=.5 sec, t = 1.5 sec? How does the velocity relate to the slope of the tangent line found for those same times?

CALCULATE ACCELERATION: Use the velocity versus time graph to calculate the car’s acceleration. Remember to show your whole calculation in your wiki. Velocity for 0.5s = -0.1m/s Velocity for 1.5s = -0.7m/s The velocity is almost the same as the slope

__PREDICT:__ On a piece of scrap paper draw a predicted distance versus time graph of the motion of the car. Photo Into Wiki. ^Krista's amazing rainbow picture complete with birds and clouds __TANGENT LINE:__ How is the velocity changing with respect to time? Provide evidence. Screen shot 2 pictures of your graph with tangent lines at t=.5 seconds into your rune and 1.5 seconds respectively into your wiki.
 * __Run #4__**

__VELOCITY TIME GRAPH:__ screen shot the velocity versus time graph made on data studio. How is the velocity changing with respect to time? Provide evidence. What is your velocity at t=.5 sec, t = 1.5 sec? How does the velocity relate to the slope of the tangent line found for those same times?

__CALCULATE ACCELERATION:__ Use the velocity versus time graph to calculate the car’s acceleration. Remember to show your whole calculation in your wiki.

__**10/5/11**__ __Tangent Line:__ A line that only touches a curve at one point. __Slope of the tangent:__ the instantaneous velocity at a specific point in time.

__**10/6/11 Homework**__

__acceleration__: the change in velocity with a respect to the change in time. __vector__: a quantity that has both magnitude and direction __negative acceleration:__ a decrease in velocity with respects to time. The object can slow down ( 20 m / s to 10 m / s) or speed up (-20 m / s to -30 m / s) __positive acceleration:__ an increase in velocity with respects to time. The object can speed up (20 m / s to 30 m / s) or slow down (-20 m / s to -10 m / s) __tangent line__: a straight line that touches a curve in only one point
 * Physics Words:**

1. Give the defining equation for acceleration in words, and by using symbols.
 * Checking Up Questions:**
 * Acceleration is the change in velocity with a respect to the change in time.
 * Acceleration = Change in velocity / Change in Time
 * A = ∆V / ∆T

2. What is an SI unit for measuring acceleration? Use words and unit symbols to describe the unit. 3. What is the difference between a vector and a scalar quantity? 4. Sketch a distance-time graph for 5. What does the slope of a velocity-time graph represent?
 * Meters per second (m/s) or kilometers per hour (km/h).
 * (m/s) / s = m / s^2
 * A vector quantity has both magnitude and direction but scalar quantity only has magnitude and no direction.
 * Constant Velocity
 * [[image:Photo_on_2011-10-07_at_00.33.jpg]]
 * Constant Acceleration
 * It represents the change in velocity with respect to time or acceleration.
 * Slope = rise / run
 * V = ∆d / ∆t
 * A = ∆v / ∆t

1. From a stop light a car accelerates when the light turns green from 0 m/s to 30 m/s(60 mph) in 5 seconds. What is the acceleration of the car? 2. Name 3 vectors and 3 scalars. What is the difference between a vector and a scalar?
 * __Do Now--10/7/11 Physics Talk Review__**
 * A = ∆v / ∆t
 * A = (30 - 0 m/s) / 5 s
 * A = (30 m/s) / 5 s
 * A = 6 m/s^2
 * Vectors (magnitude and direction) **(2 for now)**: Velocity, force(push or pull)
 * Scalars (Magnitude but no direction): Speed, calories, anything you can count (amount of people on an airplane)

Car = fan cart Bus = fan cart + 1 km Truck = fan cart + 1 km + a little less than 1 km
 * __Car vs Truck vs Bus__**

__**Physics To Go Homework: 1-16**__ 1. Can a situation exist in which an object has zero acceleration and nonzero velocity? Explain your answer. 2. Can a situation exist in which an object has zero velocity and nonzero acceleration, even for an instant? Explain your answer. 3. If two automobiles have the same acceleration, do they have the same velocity? Why or Why Not? 4. If two automobiles have the same velocity, do they have the same acceleration? Why or why not? 5. Can an accelerating automobile be overtaken by an automobile moving with constant velocity? 6. Is it correct to refer to speed-limit signs instead of velocity-limit signs? Why or why not? What units are assumed for speed-limit signs in the United States? 7. Suppose an automobile were accelerating at 2 mi/h every 5 seconds and could keep accelerating for 2 min at that rate. 8. At an international auto race, a race car leaves the pit after a refueling stop and accelerates uniformly to a speed of 75 m/s in 9 seconds to rejoin the face. 9. During a softball game, a player running from second base to third base reaches a speed of 4.5 m/s before she starts to slide into third base. When she reaches third base 1.3 s after beginning her slide, her speed is reduced to 0.6 m/s. 10.) Suppose an astronaut on an airless planet is trying to determine the acceleration of an object that is falling toward the ground. She has a motion detector in place that records the graph to the right for the falling object until just before it strikes the ground. 11.) A boy riding a bike with a speed of 5 m/s across level ground comes to a small hill with a constant slope and lets the bike coast up the hill. All graphs have time on the x-axis. 12.) An automobile magazine runs a performance test no a new model car and records the graph of distance versus time as the car goes around a track. During which segment or segments of the graph is the car 13.) A jet taking off from an aircraft carrier goes from 0 to 250 mi/h in 30 seconds. 14.) Whenever air resistance can be neglected or eliminated, an object in free-fall near Earth's surface accelerates vertically downward at 9.8 m/s^2 due to Earth's gravity. This acceleration is also called 1 g. 15.) In 1954, in a study of human endurance prior to the manned space program, Colonel John Paul Stapp rode a rocket-powered sled that was boosted to a speed of 632 mi/hr (1017 km/h). The sled and he were then decelerated to a stop in 1.4 s. 16.) An automobile accelerates from rest at 4.0 m/s every second (4.0 m/s^2).
 * Yes, because any object that moves at a constant velocity has zero acceleration.
 * No, because the velocity would change.
 * No, because they might increase speed even though they have the same acceleration and also they might have started from a
 * No, because they might not end up at the same point at the same time.
 * No it cannot.
 * It is correct to refer to speed-limit signs because many cars use speed. In the United States, we use miles per hour (mph).
 * a) How fast would it be going at t = 2 min?
 * 2 mi/h * 1 h / 60 min = 2 mi / 60 min
 * 5 sec/60 sec = 0.083 min
 * (2 mi / 60 min) / 0.083 min = x mi/ 2 min
 * (2 mi / 60 min)* (2 min) = 0.083x mi/min
 * 4/60 = 0.083x
 * x = .803 mi/min
 * .803 mi/min * 60 min / 1 h
 * 48.2 mi/h
 * b) How far would it be from the starting line?
 * 48 mi
 * a) What is the race car's acceleration during this time?
 * A = ∆V / ∆T
 * A = (75 - 0) m/s / (9 - 0) seconds
 * A = 75/9
 * A = 8.33 m/s^2
 * b) What was the race car's average speed during the acceleration?
 * 37.5 m/s
 * c) How far does the race car go during the time it is accelerating?
 * D = (((Vi +at) +Vi) / 2)*t
 * D = (((0 +(8.33*9)) +0)/2) *9
 * D = (75/ 2) *9
 * D = 337.5 m
 * d) A second race car leaves after its pit stop and accelerates to 75 m/s in 8 seconds. Compared to the first race car, what is this race car's acceleration, average speed during the acceleration, and distance traveled?
 * Acceleration: 75 m/s / 8 s
 * 9.375 m/s^2
 * Average Speed
 * 37.5 m/s
 * Distance: (((0 +(9.375)(8)) +0)/2) (8)
 * (75/2) (8)
 * 300 m
 * a) What is the player's acceleration during the slide?
 * A = V / T
 * A = .6 - 4.5 / 1.3
 * A = -3.9 / 1.3
 * A = -3
 * b) What was the distance of her slide?
 * D = v*t
 * D = (4.5)(1.3)
 * D = 5.85 m
 * c) If she had slid for only 1.1 seconds, how fast would she have been moving when she reached third base? (Assume she had the same acceleration as before.)
 * A = 4.5 - .6 / 1.1
 * A = 3.55 m/s^2
 * A = 1.2 m/s
 * d) Which of these two trials would get her from second base to third base faster?
 * The second trial because it has a bigger acceleration rate than the first trial.
 * a) From the graph, approximately what was the top speed recorded by the astronaut for the falling object?
 * 12 m/s
 * b) What is the acceleration of gravity on this planet?
 * A = v / t
 * A = (9-0)* (7.5 -0)
 * A = 1.2 m/s
 * c) If the astronaut had dropped the object from a greater height, what would happen to the object's acceleration as it falls and the object's final velocity before striking the ground?
 * Its acceleration would be faster and its final velocity would be larger.
 * a) Which of the graphs would correctly show the boy's velocity versus time as he coasts up the hill?
 * Graph B
 * b) Which graph shows the distance traveled versus time as he coasts up the hill?
 * Graph D
 * c) Which graph would show the bike's acceleration as it coasts uphill?
 * Graph E
 * d) Which graph shows after reaching the top of the hill, the speed of the boy as he coasts down the hill on the bike?
 * Graph A
 * e)Which graph could show the boy's speed versus time graph as the boy coasts up the hill and then down the hill?
 * Graph F
 * f) Starting from the top of the hill, which graph could correctly show the boy's distance vs. time as he goes down the hill?
 * Graph C
 * a) Traveling with constant speed?
 * D
 * b) Increasing speed?
 * B and E
 * c) At rest?
 * D
 * d) Decreasing speed?
 * C and F
 * e) How far did the car travel during the total test?
 * 1000 m
 * f) According to the graph, where was the car when the test was completed?
 * At the starting point
 * a) What is the jet's acceleration?
 * A = v / t
 * Vf = 250
 * Vi = 0
 * A = (250 - 0) / 30
 * A = 8.33 (m/hr)/s
 * Increasing 8.33 miles an hour per second.
 * b) If after take-off, the jet continues to accelerate at the same rate for another 15 seconds, how fast will it be going at that time?
 * ∆T = 30+15 = 45
 * Vf = ?
 * Vi = 0
 * A = 8.33 mph/s
 * A = ∆V / ∆T
 * 8.33 = (Vf - 0) / 45
 * Vf = 45(8.33)
 * Vf = 374.85 m/s
 * c) How much time does it take for the jet to reach 500 mi/h?
 * Vf = 500 mi/h
 * Vi = 250 mi/hr
 * ∆T = ?
 * A = 8.33 mph/s
 * A = ∆V / ∆T
 * A = (Vf - Vi) / ∆T
 * 8.33 = (500 - 250) / ∆T
 * ∆T = 250 / 8.33
 * ∆T = 30 s
 * 30 s*2 = 60 seconds
 * d) How much distance would it take for that same jet to reach 500 mi/h?
 * A = 8.33 mph/s
 * ∆T = 60 seconds
 * d = 1/2*a*t^2
 * d = 1/2(8.33) (60^2)
 * d = (8.33 * 3600)/2
 * <span style="color: #ff0000; font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">d = 15000 m
 * 1 g = 9.8 m/s^2
 * a) If the object falls for 100 m, how fast is it traveling?
 * Vf = ?
 * A = 9.8 m/s^2
 * Vi = 0
 * ∆T = 4.5 s
 * A = ∆V / ∆T
 * 9.8 = (Vf - 0) / 4.5
 * Vf = 44.26 m/s
 * b) How much time is required for it to fall this 100 m?
 * A = 9.8 m/s^2
 * ∆T = ?
 * d = 100 m
 * d = 1/2at^2
 * 100 = 1/2(9.8m/s^2) *t^2
 * 100 = 4.9t^2
 * 20.4 = t^2
 * 4.5 s = t
 * c) If the object falls for 10 seconds, how fast is it traveling?
 * a = v/t
 * v = at
 * v = 9.8*10
 * v = 98
 * d) How far has it fallen in this 10 seconds?
 * <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">d = vt
 * d = 10 * .98
 * d = 9.8
 * 490 m
 * e) How would your answers to these questions change for an object falling above the Moon, where the acceleration is about 1/6 g (1.6 m/s^2)?
 * The acceleration would be slower.
 * a) What was the acceleration of this stop?
 * 632/ 1.4 = A
 * A = 451.43
 * b) What is this acceleration in terms of g's?
 * 1 g / 9.8 m = x / 632
 * 9.8x = 632
 * <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">x = 64.48
 * c) In what distance did the speed of the sled travel in its speed changed from 1017 km/h to 0?
 * d = v*t
 * d = (1017 km) * (1.4 s)
 * d = 1423.8 km
 * a) How far does it travel after 1.0 s?
 * 4 m
 * 2 m
 * b) How far does it travel after 2.0 s?
 * 8 m
 * 8 m
 * c) How far does it travel after 3.0 s?
 * 12 m
 * 18 m
 * d) How far does it travel after 4.0 s?
 * 16 m
 * 32 m
 * e) Complete a d-t graph for this automobile.
 * [[image:Photo_on_2011-10-12_at_23.43.jpg]]
 * f) Complete a v-t graph for this automobile.
 * [[image:Photo_on_2011-10-12_at_23.43_#2.jpg]]
 * g) How does the motion of this automobile compare with the motion of a real automobile (as you investigated previously)?
 * This one has constant speed.

__**Speeding Up and Slowing Down worksheet :Work and Notes**__ __Speeding Up: Acceleration__
 * A = ∆V / ∆T = (V(3.5 sec) - V(2 sec) ) / 3.5 s - 2 s
 * A = (.4 m/s - .25 m/s) / (3.5 s - 2 s)
 * A = .15 m/s / 1.5 s = + .1 m/s^2

__Slowing Down__
 * A = ∆V / ∆T = ( V(3.5s) - V (2 s) ) / (3.5 s - 2 s)
 * Greater Number
 * A = (.3 m/s - .7 m/s) / (3.5s - 2s)
 * A = (.4 m/s) / 1.5 s
 * A = -0.2667 or - 0.27 m/s^2

__**Predicting V vs T graphs from D vs T graphs**__

__**10/12/11 Notes**__ Acceleration:
 * Slope
 * Constant gravity does not shut off.
 * Active Physics Plus**
 * d = 1/2at^2 +V i (t)
 * 1) A car accelerates from a stop light with acceleration 3 m/s2 from an initial rolling speed of 7 m/s to a speed of 20 m/s. How long does it take the car to do this?
 * V f = 20 m/s
 * V i = 7 m/s
 * A = 3 m/s^2
 * ∆T = ?
 * A = (Vf - Vi) / ∆T
 * 3 = (20 - 7) / ∆T
 * (∆T) (3) = (20 - 7)
 * ∆T = 4.33 seconds
 * 1) If a car accelerates from 7 m/s with an acceleration of 1.5 m/s^2 for 10 seconds, what speed is it now going?
 * Vf = ?
 * Vi = 7 m/s
 * A = 1.5 m/s^2
 * ∆T = 10 seconds
 * A = (Vf - Vi) / ∆T
 * 1.5 = (Vf - 7) / 10
 * 15 = Vf - 7
 * 22 m/s = Vf
 * 1) A car accelerates from rest at a stop light to a speed of 20 m/s in 5 seconds.
 * 2) What is the car's acceleration?
 * Vf = 20 m/s
 * Vi = 0 m/s
 * A = ?
 * ∆T = 5 seconds
 * A = (Vf - Vi) / ∆T
 * A = 20/5
 * A = 4 m/s^2
 * 1) What distance has the car gone?
 * d = 1/2at^2 +Vi(∆T)
 * d = 1/2(4 m/s^2) (5^2 s) + 0(5s)
 * d = (25*4m/s)/2
 * d = (100m/s)/ 2
 * d = 50 m/s


 * 1) A car accelerates from rest at a stop light to a speed of 40 m/s in 10 seconds. That is an acceleration of 4 m/s^2. What distance has the car gone?
 * Vf = 40 m/s
 * Vi = 0 m/s
 * A = 4 m/s^2
 * ∆T = 10 seconds
 * d = 1/2at^2 +Vi(∆T)
 * d = 1/2 (4 m/s^2) * (10^2) + 0(10)
 * d = 400/2 m
 * d = 200 m
 * 1) A car rolls up to a stop light with initial speed 3 m/s and accelerates to 23 m/s in 5 seconds.
 * 2) What is the car's acceleration?
 * Vf = 23 m/s
 * Vi = 3 m/s
 * A = ?
 * ∆T = 5 seconds
 * A = (Vf - Vi) / ∆T
 * A = (23 - 3) / 5
 * A = 4 m/s^2
 * 1) What distance has the car gone?
 * d = 1/2at^2 +Vi(∆T)
 * d = 1/2 (4 m/s^2) * (5^2 s) + (3*5)
 * d = 200 + 15
 * d = 215 m/s

__**Section 4 quiz notes**__
 * 1) A car slows down coming to a red light from a speed of 25 m/s to 0 m/s in 4 seconds.
 * 2) What is the car's acceleration?
 * Vf = 0 m/s
 * Vi = 25 m/s
 * A = ?
 * ∆T = 4 seconds
 * A = (Vf - Vi) / ∆T
 * A = (0 - 25) / 4
 * A = -25/4
 * A = -6.25 m/s^2
 * 1) Why is it negative?
 * It is decreasing speed.
 * 1) How far did it take the car to stop?
 * d = 1/2at^2 +Vi(∆T)
 * d = 1/2 (-6.25 m/s^2) (4^2 s) + (25* 4)
 * d = 1/2 (-6.25 m/s^2) (16 s) + 100
 * d = (-50) +100
 * d = 50 m
 * Solve instantaneous velocity/ slope for the tangent line
 * Acceleration from 2 different slopes of the Tangent Line
 * Describe the object's velocity when it turns around is 0
 * Use the equation d = 1/2at^2
 * Be able to make a Velocity vs Time Graph from a Distance vs Time Graph
 * Change in velocity is the Rise
 * change in time is the run

__Section 5__
__What Do You See?__ __What Do You Think?__ __**Learning Outcomes**__
 * <span style="font-family: 'Courier New',Courier,monospace;">** Section5 ** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 20px;">**Points** ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">WDYSee/Think: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**100** ||
 * The driver in the red car is stepping on the brakes hard so he wouldn't hit the giant moose. He is stepping on the brakes hard because in the picture the back of his car is raised up and there is smoke coming out from the front wheels. The driver looks really surprised and then scared that they will hit the moose, but the moose looks calm.
 * What factors must you consider to determine if you will be able to stop in the distance between you and the animal to avoid hitting it?
 * Weather, how big the car is, if there is a slope or something that will cause the car to swerve or slide forward on the road, if the animal is running.
 * **Plan** and carry out an experiment to relate braking distance to initial speed.
 * **Determine** braking distance.
 * **Examine** accelerated motion

__Objectives:__ to determine the effect of initial speed on braking (stopping) distanc
 * __Investigate:__**
 * # of textbooks || <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">V i (m/s) || Braking Distance (m) ||
 * 1 || 0.83462 m/s || 3.048 m ||
 * 1 || 0.844075 m/s || 2.8956 m ||
 * 2 || 1.21156 m/s || 6.46176 m ||
 * 2 || 1.9604 m/s || 5.9436 m ||
 * 3 || 1.10984 m/s || 7.0866 m ||
 * 3 || 0.949474 m/s || 5.6388 m ||
 * 3 || 1.07650 m/s || 6.4008 m ||

d = 1/2at^2

__Questions__: 1. If initial velocity is doubled by what factor does braking distance increase? 2. If initial velocity is tripled by what factor does braking distance increase? 3. If initial velocity is quadrupled (x4) by what factor does braking distance increase? 4. Do #8 in book on page 77 5. If the sports car changed its speed to 30 mph what do you expect its braking distance to be? (Hint: If you cut the speed by half what factor will the braking distance change?) 6. If the touring sedan changed its speed to 30 mph what do you expect its braking distance to be? 7. If each car (Sports car and touring sedan) decreased its speed to 15 mph what would their braking distances be? (Hint: If you multiply the speed by a quarter by what factor will the braking distance change?)
 * 4
 * 9
 * 16
 * Use the data on the sports car provided at the end of this chapter on pages 116- 117 to answer the following:
 * b) The braking distance is shown for two speeds. The ratio of the two speeds is 80 mi/h : 60 mi/h. This ratio is 80 / 60 = 1.33. This is an increase of 133 percent. Do you expect the ratio of the braking distances to also be in the ratio of 80 / 60 = 1.33? What is the ratio of the braking distances? How does it compare with the ratio of the two speeds?
 * Yes I do expect it.
 * 60 mph --> 80 mph
 * Vi = 1.33 = (80/60) mph
 * Ratio of Increase is 1.33
 * Stopping Distance = 118 ft--> 209 ft
 * Distance = 209 /118 = 1.77
 * 1.33^2 = 1.77
 * c) How does this data correspond to what you found in your experiment?
 * This data is similar to our initial velocity.
 * 30 mph / 4 = 7.5

V i = Length of the Flag / Time in Gate

__**Checking Up Questions**__ 1. If a vehicle is traveling at constant velocity and then comes to a sudden stop, has it undergone negative acceleration or positive acceleration? Explan your answer. 2. Explain how you know that increasing the velocity of an automobile increases the braking distance. 3. Why is the term negative acceleration used instead of deceleration?
 * Yes it has, because the final velocity is zero when it stops, which is less than the initial velocity.
 * Increasing the velocity increases the braking distance because it would need more time to stop from the faster speed.
 * Deceleration means to slow down, but negative acceleration is decreasing speed in the positive direction or increasing speed in the negative direction.

__**Physics Words**__ __Negative Acceleration:__ a change in the velocity with respect to time of an object by decreasing speed in the positive direction or increasing speed in the negative direction. ( Vi 2 / Vi 1 ) = Ratio of initial velocities (Braking Distance 2 / Braking Distance 1 ) = Ratio of braking distances
 * A = (0 - Vi) / ∆T = -Vi / ∆T
 * __10/19/11 Notes__**

(Vi 2 / Vi 1 )^2 = Braking Distance 2 / Braking Distance 1

__Example:__ ( 60 mph / 30 mph)^2 = (123 / Braking Distance 1)
 * __Honda Civic Stopping Distances__**

1) Calculate the stopping distances for Kibala’s Honda Civic


 * Initial Velocity || Stopping Distance ||
 * 10 mph || 3.42 feet ||
 * 20 mph || 13.67 feet ||
 * 30 mph || 30.75 feet ||
 * 40 mph || 54.67 feet ||
 * 50 mph || 85.42 feet ||
 * 60 mph || 123 feet ||
 * 70 mph || 167.42 feet ||
 * 80 mph || 218.67 feet ||
 * 90 mph || 276.75 feet ||
 * 100 mph || 341.67 feet ||

2) Graph Stopping Distance (y-axis) vs Initial Velocity

3) Does Stopping Distance Increase Linearly as Initial Velocity Increases or does it increase exponentially?

4) If Initial Velocity is doubled how does stopping distance change? It is multiplied by 4. 5) If the Initial Velocity is multiplied 4 times how does the stopping distance change? It is multiplied by 16 6) If the Initial Velocity is halved how does the Stopping Distance Change? The stopping distance is divided by 4 or quartered. 7) If the initial Velocity is quartered how does the stopping distance change? It is divided by 16. 8) What speed would you need to have a stopping distance of a mile? 393.1 mph
 * (60 mph / ? mph )^2 = (123 ft / 5280 ft)
 * (3600 mph / x^2 mph) = (123 ft / 5280 ft)
 * 123x^2 = 19008000
 * x^2 = 154536.59
 * x = 393.1 mph

Positive: Forward Negative: Backwards
 * __10/21 Direction of Accelerations__**

Same Direction : Speeding Up Different Direction : Slowing Down
 * A>
 * V>
 * A>
 * <V
 * __10/24__**
 * __Active Physics Plus__**

Vf = Vi + at 0 = 9 + (-4.1)(t) 4.1t = 9 t = 9 / 4.1 t = 2.2 seconds
 * __Worksheet: Average Acceleration__**
 * 1.) As the shuttle bus comes to a sudden stop to avoid hitting a dog, it accelerates uniformly at -4.1 m/s^2 as it slows from 9.0 m/s to 0 m/s. Find the time interval of acceleration for the bus.**
 * a = -4.1 m/s^2
 * t = ?
 * d is not asked for
 * Vi = 9 m/s
 * Vf = 0 m/s

Vf = Vi + at 12.0 = 7.0 + (2.5)(t) 5.0 = 2.5t t = 2 seconds
 * 2. A car traveling at 7.0 m/s accelerates uniformly at 2.5 m/s^2 to reach a speed of 12.0 m/s. How long does it take for this acceleration to occur?**
 * a = 2.5
 * Vf = 12.0 m/s
 * Vi = 7.0 m/s
 * t = ?

Vf = Vi + at 0 = 13.5 + (-0.50)(t) 0.50t = 13.5 t = 27 seconds
 * 3. With an average acceleration of -0.50 m/s^2, how long will it take a cyclist to bring a bicycle with an initial speed of 13.5 m/s to a complete stop?**
 * a = -0.50 m/s^2
 * Vi = 13.5 m/s
 * Vf = 0
 * t = ?

Vf = Vi + at -1.2 = -6.5 + 1500a 5.3 = 1500a a = .004 m/s^2
 * 4. Turner's treadmill runs with a velocity of -1.2 m/s and speeds up at regular intervals during a half-hour workout. After 25 min, the treadmill has a velocity of -6.5 m/s. What is the average acceleration of the treadmill during this period?**
 * Vf = -1.2 m/s
 * Vi = -6.5 m/s
 * a = ?
 * t = 25 min ---> 1500


 * 5. Suppose a treadmill has an average acceleration of 4.7 * 10^-3 m/s^2.**
 * **a. How much does its speed change after 5 min?**
 * a = 4.7 * 10^-3
 * t = 5 min --> 5*60 = 300 seconds
 * Vi = ?
 * Vf = 0
 * Vf = Vi + at
 * 0 = x + (4.7*10^-3)(300)
 * 0 = x + (1.41)
 * x = -1.41 (?)
 * **b. If the treadmill's initial speed is 1.7 m/s, what will its final speed be?**
 * Vi = 1.7 m/s
 * Vf = ?
 * a = 4.7 * 10^-3 m/s^2
 * t = 5 min --> 300 seconds
 * Vf = Vi + at
 * x = 1.7 + (4.7 * 10^-3)(300)
 * x = 1.7 + 1.41
 * x = 3.11 m/s (?)

d = 1/2(Vi + Vf)t d = 1/2 (0 + 6.6) (6.5) d = 1/2(42.8) d = 21.4 meters
 * __Displacement with Constant Uniform acceleration__**
 * 1. A car accelerates uniformly from rest to a speed of 23.7 km/h in 6.5 seconds. Find the distance the car travels during this time.**
 * Vf = 23.7 km/h
 * 23.7 km/h * 1000 m/km * 1/3600 hr/s
 * 6.6 m/s
 * Vi = 0
 * t = 6.5 seconds
 * d = ?

d = 1/2(Vi + Vf)t d = 1/2(15 + 0) (2.5) d = 1/2(37.5) d = 18.75 meters
 * 2. When Maggie applies the brakes of her car, the car slows uniformly from 15.0 m/s to 0 m/s in 2.5 seconds. How many meters before a stop sign must she apply her brakes in order to stop at the sign?**
 * Vi = 15 m/s
 * Vf = 0 m/s
 * t = 2.5 seconds
 * d = ?

Vf^2 = Vi^2 +2ad 0^2 = 100^2 + 2(-5.0)d 0 = 10000 + -10d 10d = 10000 d = 1000 m No.
 * 3. A jet plane lands with a speed of 100 m/s and can accelerate uniformly at a maximum rate of -5.0 m/s^2 as it comes to rest. Can this plane land at an airport where the runway is 0.80 km long?**
 * Vi = 100 m/s
 * Vf = 0
 * a = -5.0 m/s^2
 * d = ?

d = 1/2(Vi + Vf)t 99 = 1/2( 21.7 + 0) t 99 = 10.85t t = 9.12 seconds
 * 4. A driver in a car traveling at a speed of 78 km/h sees a cat 101 m away on the road. How long will it take for the car to accelerates uniformly to a stop in exactly 99 m?**
 * d = 99 m
 * Vi = 78 km/h
 * 78 km/h * 1000 m/km * 1/3600 h/s
 * 21.7 m/s
 * Vf = 0 m/s
 * t = ?

d = 1/2(Vi + Vf) t 3200 m = 1/2( 6.4 + x) (3.5) 3200 = 1/2( 22.4 + 3.5x) 3200 = 11.2 + 1.75x 3188.8 = 1.75x x = 1822.17 (????)
 * 5. A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3.2 km in 3.5 min. How fast is the car moving after this time?**
 * Vi = 6.4 m/s
 * Vf = ?
 * d = 3.2 km
 * 3.2 km * 1000 m = 3200 m
 * t = 3.5 min

Vf = Vi + at Vf = 6.583 + (.92)(3.6) Vf = 6.583 + 3.312
 * __Velocity and Displacement with Uniform Acceleration__**
 * 1. A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s^2 for 3.6 seconds. Find the final speed and the displacement of the car during this time.**
 * a = .92 m/s^2
 * t = 3.6 seconds
 * d = ?
 * Vi = 23.7 km/h --> 6.583 m/s
 * Vf = ?
 * Vf = 9.895 m/s**

d = Vit + 1/2at^2 d = (6.583)(3.6) + 1/2(.92)(3.6)^2 d = 23.6988 + 5.9616
 * d = 29.7 m**

Vf = Vi + at Vf = 4.3 + (3)(5)
 * 2. An automobile with an initial speed of 4.3 m/s accelerates uniformly at the rate of 3 m/s^2. Find the final speed and the displacement after 5 seconds.**
 * a = 3 m/s^2
 * Vi = 4.3 m/s
 * Vf = ?
 * d = ?
 * t = 5 seconds
 * Vf = 19.3 m/s**

d = Vit + 1/2at^2 d = (4.3)(5) + 1/2(3)(5^2) d = 21.5 + 37.5
 * d = 59 m**

t = 5 seconds a = -1.5 m/s^2 Vf = ? d = ? Vi = 0 Vf = Vi + at Vf = 0 + (-1.5)(5)
 * 3. A car starts from rest and travels for 5 seconds with a uniform acceleration of -1.5 m/s^2. What is the final velocity of the car? How far does the car travel in this time interval?**
 * Vf = -7.5 m/s**

d = Vit + 1/2at^2 d = (0)(5) + 1/2 (-1.5)(5^2)
 * d = -18.75 m**

a = (Vf - Vi) / t -2 = (10 - 15) / t -2t = -5
 * 4. A driver of a car traveling at 15 m/s applies the brakes, causing a uniform acceleration of -2 m/s^2. How long does it take the car to accelerate to a final speed of 10 m/s? How far has the car moved during the braking period?**
 * Vi = 15 m/s
 * Vf = 10 m/s
 * a = -2 m/s^2
 * d = ?
 * t = ?
 * t = 2.5 seconds**

d = 1/2(Vi + Vf)t d = 1/2 (15 + 10)(2.5) d = 1/2(62.5)
 * d = 31.25 m**

a = .8 m/s^2 d = 245 m Vi = 7 m/s Vf = ? a = 2.3 m/s^2 Vi = 0 Vf = ? t = ? Vf^2 = Vi^2 +2ad 26.1^2 = 20.8^2 + 2(.85)d 681.21 = 432.64 + 1.7d 248.57 = 1.7d
 * __Final velocity after any displacement__**
 * 2. A car traveling initially at +7 m/s accelerates uniformly at the rate of + 0.8 m/s^2 for a distance of 245 m.**
 * **a. What is its velocity at the end of the acceleration?**
 * Vf^2 = Vi^2 + 2ad
 * Vf^2 = (7^2) + 2(.8)(245)
 * Vf^2 = 441
 * **Vf = 21 m/s**
 * **b. What is its velocity after it accelerates for 125 m?**
 * Vf^2 = Vi^2 + 2ad
 * Vf^2 = (7^2) + 2(.8)(125)
 * Vf^2 = 249
 * **Vf = 15.8 m/s**
 * **c. What is its velocity after it accelerates for 67 m?**
 * Vf^2 = Vi^2 + 2ad
 * Vf^2 = (7^2) + 2(.8)(67)
 * Vf^2 = 156.2
 * **Vf = 12.5 m/s**
 * 3. A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s^2.**
 * **a. What is the speed of the car after it has traveled 55 m?**
 * Vf^2 = Vi^2 + 2ad
 * Vf^2 = 0 + 2(2.3)(55)
 * Vf^2 = 253
 * **Vf = 15.9 m/s**
 * **b. How long does it take the car to travel 55 m?**
 * d = 1/2(Vi + Vf)t
 * 55 = 1/2(0 +15.9)t
 * 55 = 7.95t
 * **t = 6.92 seconds**
 * 4. A certain car is capable of accelerating at a uniform rate of 0.85 m/s^2. What is the magnitude of the car's displacement as it accelerates uniformly from a speed of 73 km/h to one of 94 km/h?**
 * a = 0.85 m/s^2
 * Vi = 73 km/h ---> 20.3 m/s
 * Vf = 94 km/h ---> 26.1 m/s
 * d = ?
 * d = 146.22 m**
 * 86.5 meters **

Vf^2 = Vi^2 + 2ad 33.3^2 = 0 + 2(240)a 1108.9 = 480a
 * 5. An aircraft has a liftoff speed of 120 km/h. What minimum uniform acceleration does this require if the aircraft is to be airborne after a take-off run of 240 m?**
 * Vf = 120 km/h --> 120 km/h * 1000 m/km * 1/3600 h/s = 33.3 m/s
 * Vi = 0
 * a = ?
 * d = 240 m
 * a = 2.31 m/s^2**

Vf^2 = Vi^2 + 2ad 1.5^2 = 6.5^2 + 2(-2.7)d 2.25 = 42.25 + -5.4d -40 = -5.4d
 * 6. A motorboat accelerates uniformly from a velocity of 6.5 m/s to the west to a velocity of 1.5 m/s to the west. If its acceleration was 2.7 m/s^2 to the east, how far did it travel during the acceleration?**
 * Vi = 6.5 m/s
 * Vf = 1.5 m/s
 * a = -2.7 m/s^2
 * d = ?
 * d = 7.41 m**

d = 1/2(Vi +Vf)t Vf = Vi + at d = Vit + 1/2at^2 Vf^2 = Vi^2 +2ad
 * __Physics To Go #1-8__**

V = d/t d = V*t d = 0.9 (10) d = 9 meters
 * 1. A student measured the braking distance of her automobile and recorded the data in the table. Plot the data on a graph and describe the relationship that exists between initial speed and braking distance.**
 * As the initial speed increases, the braking distance increases also.
 * 2. Below is a graph of the braking distances in relation to initial speed for two automobiles. Compare qualitatively (without using numbers) the braking distances when each automobile is going at a slow speed and then again at a higher speed. Which automobile is safer? Why? How did you determine what “safer” means in this question?**
 * [[image:Screen_shot_2011-10-25_at_12.22.32_AM.png]]
 * Automobile B is faster because it took less time to go farther than Automobile A. B's braking distance would be faster than A's.
 * A is safer than B because its going slower and its braking distance would react quickly.
 * 3. An automobile is able to stop in 20 m when traveling at 30 mi/h. How much distance will it require to stop when traveling at the following:**
 * **a) 15 mi/h? (half of 30 mi/h)**
 * Vi = 30 mi/h / 2 = 15
 * BD = 20 m / 4 = 5
 * 5 m
 * **b) 60 mi/h? (twice 30 mi/h)**
 * 80
 * **c) 45 mi/h? (three times 15 mi/h)**
 * 45 m
 * **d) 75 mi/h? (five times 15 mi/h)**
 * 125 m
 * 4. An automobile traveling at 10 m/s requires a braking distance of 30 m. If the driver requires 0.9 s reaction time, what additional distance will the automobile travel before stopping? What is the total stopping distance, including both the reaction distance and the braking distance?**
 * Vi = 10 m/s
 * BD = 30 m
 * t = 0.9 s
 * d = ?


 * 5. Consult the information for the sports car at the end of this chapter. This shows the stopping distance. How far would you expect this automobile to travel until coming to rest when brakes are applied at a speed of 30 mi/h?**
 * 6. Use the information for the sedan at the end of this chapter. Find the braking distances for 50 mi/h and 25 mi/h. Draw a graph using the different braking distances. Plot the speeds on the horizontal axis and the braking distances on the vertical axis.**
 * 7. Does the braking information for the sedan include the driver's reaction time? If it does not, then how much distance is added to the total braking distance, supposing that the driver has a 1/2 reaction time? Who should let the consumer know about the 1/2 second reaction time ---The information sheet or a driver training manual?**
 * 8. Apply what you learned in this section to write a statement explaining the factors that affect stopping distance. The total stopping distance includes the distance you travel during your reaction time, plus the braking distance. What do you know know about stopping that will make you a safer driver?**
 * You should drive slower to reduce your reaction time.

Total stopping distance: D reaction + D braking Total stopping distance: Vtreaction and Dbraking Total stopping distance: Vt reaction and Vi^2 / 2a If Vi increases then the Total stopping distance increases also. Vi ~ TSD If Reaction Time increases, then Total stopping distance increases also. Vi ~ TSD If acceleration increases(good brakes), then total stopping distance decreases. TSD ~ 1/a __**Quiz notes for Wednesday 10/26**__
 * Vf^2 = Vi^2 + 2adbraking
 * 0 m/s = Vi^2 + 2anegative dbraking
 * (-Vi^2 = 2(-a negative) dbraking )
 * Vi^2 / 2a neg = dbraking
 * Total stopping distance = reaction distance + braking distance
 * Vi = initial velocity before braking
 * t reaction: reaction time
 * a: acceleration
 * Stopping distance effected by Vi
 * Direction of acceleration
 * Slowing down
 * Speeding up

Practice Conversion km/h to m/s

__100 km/h__ 100 km/h * 1000 m/km * 1/60 h/min * 1/60 min/sec
 * 27.8 m/s**

__25 km/h__ 25 km/h * 1000 m/km * 1/60 h/min * 1/60 min/sec
 * 6.94 m/s**

__220 km/h__ 220 km/h * 1000 m/km * 1/60 h/min * 1/60 min/sec
 * 61.1 m/s**


 * __Total Stopping Distance Activity 10/28/11__**
 * 1.)**
 * 2.)** **Assuming we are traveling forward and have a +v, what is the sign (+ or -) of the acceleration when braking? Explain why the acceleration has that sign (+ or -).**
 * The acceleration is braking(stopping) so it should be - acceleration.

d = vt d = 10(1)
 * __Calculating Reaction Distance:__**
 * 1. Derive the equation for reaction distance using your diagram above and one of the equations above.**
 * No acceleration during d reaction
 * Use v = d/t or d = vt or V i t reaction = d reaction
 * 2. An older person is driving a buick up to a yellow light. How far does the person go while deciding if they should brake (reaction distance). They are traveling at 10 m/s (22 mph) and have a reaction time of 1 second.**
 * V = 10 m/s
 * t = 1 second
 * d = 10 m**

d = vt d = (20)(1)
 * 3. An older person is driving a buick up to a yellow light. How far does the person go while deciding if they should brake( reaction distance). They are traveling at 20 m/s (44 mph) and have a reaction time of 1 second.**
 * V = 20 m/s
 * t = 1 second
 * d = 20 m**


 * 4. How does an increase in speed change reaction distance? How does a decrease in speed change reaction distance? Is one second a long reaction time or not? List reasons why the reaction time may be so large.**
 * It will be a shorter distance.
 * Yes it is a long reaction time.
 * The reaction time might be large because of the old people, yellow lights, and how they slow down.

d = vt d = (2)(.5)
 * 5. A child rides a tricycle down her driveway. All of a sudden a squirrel jumps in front of the tricycle. The child is traveling at 2 m/s and has a reaction time of .5 seconds. What is the child's reaction distance?**
 * V = 2 m/s
 * t = .5 seconds
 * d = 1 m**

__**Calculating Braking Distance**__
 * 1. Derive an equation for braking distance using one of the equations above and your diagram.**
 * V f ^2 = V i ^2 +2ad
 * (-Vi^2) / 2a = d braking
 * Note: A is negative "slowing"

Vf^2 = Vi^2 +2ad 0 = 51^2 + 2(-11)d 0 = 2601 - 22d -2601 = -22d
 * 5. A drag racer usually brakes using a parachute to help him or her slow down. Without a parachute the drag racer obtains a deceleration of -11 m/s^2. If the drag racer reaches a top speed of 51 m/s (121 mph) before it applies the brakes. How long will its braking distance be without the parachute?**
 * a = -11 m/s^2
 * Vi = 51 m/s
 * Vf = 0
 * d = ?
 * d = 118.23 meters**

Vf^2 = Vi^2 +2ad 0 = 51^2 + 2(-24)d 0 = 2601 - 48d -2601 = -48d
 * 6. The parachute acts like better brakes on a car. Now with the parachute the drag racer has a deceleration rate of -24 m/s&2. If the drag racer reaches a top speed of 51 m/s (121 mph) before it applies the brakes, how long will its braking distance be with the parachute?**
 * a = -24 m/s^2
 * Vi = 51 m/s
 * Vf = 0
 * d = ?
 * d = 54.19 meters**

V = d(reaction) / t(reaction) 11 = d(reaction) / .5
 * __Preparing for the Chapter Challenge:__**
 * 1. Assume you have a reaction time of .5 seconds including moving your foot from gas to brake. You are traveling in a school zone at 11 m/s (25 mph) and your car has a maximum deceleration of -4 m/s^2. What is your reaction distance, braking distance, and total stopping distance?**
 * T = .5 seconds
 * Vi = 11 m/s
 * a = -4 m/s^2
 * Reaction Distance**
 * d(reaction) = ?
 * d(reaction) = 5.5 meters**

Vf^2 = Vi^2 + 2ad 0 = 11 m/s^2 + 2(-4 m/s^2) d(braking) -(11^2) / -(8) = d(braking) -121 / -8 = d(braking)
 * Braking Distance**
 * 15.13 meters = d**

TSD = d(reaction) + d(braking) TSD = 5.5 + 15.13
 * Total Stopping Distance**
 * TSD = 20.63 meters**

V = d(reaction) / t(reaction) 27 = d(reaction) / .5
 * 2. Now drive on the highway with a speed of 27 m/s(60 mph). What is your reaction distance, braking distance, and total stopping distance?**
 * Vi = 27 m/s
 * T(reaction) = .5
 * A = -4 m/s^2
 * Reaction Distance**
 * d(reaction)
 * d(reaction) = 13.5 meters**

Vf^2 = Vi^2 + 2ad 0 = 27^2 + (2)(-4) d(braking) -729 = -8d(braking)
 * Braking Distance**
 * 91.13 = d(braking)**

TSD = d(reaction) + d(braking) TSD = 13.5 + 91.13
 * Total Stopping Distance**
 * TSD = 104.63 meters**

V = d(reaction) / t(reaction) 27 = d(reaction) / 1
 * 3. Assume you are now still driving on the highway but are suffering from driver fatigue and your reaction time is 1 second. What is your reaction distance, braking distance, and total stopping distance?**
 * Vi = 27 m/s
 * T(reaction) = 1 second
 * A = -4 m/s^2
 * Reaction Distance**
 * d(reaction) = ?
 * d (reaction) = 27 meters**

Vf^2 = Vi^2 + 2ad 0 = 27^2 + 2(-4)d -729 = =8d
 * Braking Distance**
 * d(braking) = 91.13**

TSD = d(reaction) + d(braking) TSD = 27 + 91.13
 * Total Stopping Distance**
 * TSD = 118.13 m**

V = d(reaction) / t(reaction) 27 = d(reaction) / .5 seconds
 * 4. Assume now you have worn out brake pads instead of suffering from driver fatigue. Put your reaction time back to .5 seconds but now set your acceleration to .2 m/s^2 What is the reaction distance, braking distance, and total stopping distance?**
 * **T(reaction) = .5 seconds**
 * A = .2 m/s^2
 * Vi = 27 m/s
 * Reaction Distance**
 * d(reaction) = 13.5 meters**

Vf^2 = Vi^2 + 2ad
 * Braking Distance**
 * 0 = 27^2 + 2(.2)d**
 * -729 = .4d**
 * d = -1822.5**

TSD = 13.5 + -1822.5 TSD = 1809 meters
 * Total Stopping Distance**

__**Create a Problem**__ -Solve for Reaction Distance, Braking distance, and TSD -Embed givens in a problem. -Don't embed givens that are not needed. -Finish your own question.

Assume that you have a reaction time of .7 seconds. You are driving a car with a maximum deceleration of -3 m/s^2 and is moving at a speed of 20 m/s. What is your reaction distance, Braking distance, and Total stopping distance? V = d/t 20 = d / .7 d (reaction) = 14 meters

Vf^2 = Vi^2 + 2ad 0 = 20^2 + 2(-3)d -400 = -6d 66.7 meters = d(braking)

TSD = 14 + 66.7 TSD = 80.7 meters

You're reaction time is 2.5 seconds. You are going to Candy Land, but on your way you see a few reindeer crossing your path. You are traveling 110 mi/hr on a magical speeding, talking donkey. The donkey a deceleration of -6 m/s^2. What is your reaction distance, braking distance, and total stopping distance?
 * __Kim's question:__**
 * Vi = 110 mi/hr ---> 110 mi/hr * 1609.3 m / mi * 1/ 3600 h/s * = 49.2 m/s
 * T = 2.5 seconds
 * A = -6 m/s^2

Reaction Distance: V = d / t 49.2 = d / 2.5 d = 123 meters

Braking Distance: Vf^2 = Vi^2 + 2ad 0 = 49.2^2 + 2(-6)d -2420.64 = -12d d = 201.72 meters

Total stopping distance TSD = 123 + 201.72 TSD = 324.72 meters

__**Section 6**__
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">+12 EC || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">needed to do 2 more intersections for #2 on active physics plus || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing stop zone and go zone of a real intersection || __What Do You See?__ The Red car is stopping and the green car is speeding up. The dog in the red car is flying out of the car and since the signal is red, the red car seems to have been braking to not run the red light. The green car just kept increasing speed after running the red light. There is a police officer on the side of the road so the red car was probably braking hard to not get caught and get a ticket from the officer. The green car driver seems to have not noticed the police officer.
 * <span style="font-family: 'Courier New',Courier,monospace;">** Section5 ** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 20px;">**Points** ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">WDYSee/Think: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">15/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">0/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**87*.75 = 65.25/75** ||

__What Do You Think?__
 * **If all traffic lights stayed yellow the same amount of time, how would this affect drivers' decisions at intersections?**
 * Drivers would have to think fast or just stop before the yellow light turns red.
 * **How could an intersection with a traffic light be dangerous?**

__Learning Outcomes__ In this section you will:
 * Investigate** the factors that affect the __STOP__ and __GO__ Zones at intersections with traffic lights.
 * Investigate** the factors that result in an Overlap Zone or a Dilemma Zone at intersections with traffic lights.
 * Use** a computer simulation to mathematically model the situations that can occur at an intersection with traffic lights.


 * __Mini Challenge Plan__**
 * 1. Paragraph explaining your storyline**
 * 2. 3 ways you will use equations in the storyline**
 * 3. 3 ways you will use graphs in storyline**
 * email text to Kibala for #1-3**

Summary: Driving downhill and there are bunnies (a bunny family complete with a bunny daddy, bunny mommy and bunny baby) on the side of the road. There is a sharp turn on a narrow road. The weather is sunny and the sun is blinding the driver. The passenger stares at the bunnies and doesn't tell the driver because she's too distracted by the bunnies outfit(tuxedos and a little apron). Another car comes going and they both are about to clash into each other. They're both going different speeds. Driver A (the one being blinded by light) is going 40 mph and can't see what's in front of him. Driver A is 50 meters away from the turn but he doesn't realize that until he is 20 meters into the turn. Driver B (a nice old lady) is going 25 mph and is turning __now__. The turn is 15 meters in length. Will the nice old lady __die__? WILL __DRIVER A__ SLOW DOWN IN TIME TO NOT **KILL** THE __NICE__ __OLD__ __LADY__?

Equations: Time, braking distance, and reaction time Graphs: Distance vs Time, Distance vs Velocity, Velocity vs Time

__**Stop and Go Zone notes**__

__**Investigation 11/4 pg 91 #3, 4**__
 * 3.The diagram below shows the position of three automobiles at the moment a light turns yellow. Assume al three automobiles are moving in the same speed. Automobile A is able to make it through the intersection before the light turns red. It is in the GO zone. Automobile C may not be able to make it through during the yellow light. The light may turn red before Automobile C gets to the intersection.**
 * **a. Will automobile B be able to make it through during the yellow light?**
 * Yes
 * **b. Is automobile B in the GO zone? Explain your answer.**
 * Yes it's in the Go zone because it's about to go through the intersection.
 * **c. Would any automobile closer to the intersection than automobile A be in the GO Zone?**
 * Yes.
 * **d. Is automobile C in the GO Zone? What might happen if automobile C decides to continue?**
 * No because if they decided to continue, they would run a red light.
 * 4. In the following diagram of an intersection, automobile D is able to come to a safe stop when the light turns red because it is in the STOP Zone. In this zone, automobiles can stop safely before they reach the intersection. Automobile F is closer to the intersection than automobile D. If the driver of automobile F tries to stop the automobile, he or she may not be able to stop in such a short distance. Again, assume that all the automobiles have the same initial speed.**
 * **a. Is automobile E in the STOP Zone? Explain your answer.**
 * Yes because it's behind automobiles D and F which are in the Go Zone and Total stopping distance area.
 * **b. Is automobile F in the STOP zone? Explain your answer. What might happen if automobile F decides to stop?**
 * Automobile F is not in the STOP zone because it's about to pass the yellow light. If they decide to stop, Automobile F is going to waste the yellow light and red light time.

__**Go Zone Prediction Graph**__


 * Variable |||| Change || **Predicted** shrink or expansion of GO ZONE ||
 * ty || yellow-light time || Increase ty || Expand ||
 * ^  ||^   || Decrease ty || Shrink ||
 * tr || reaction time || Increase tr || No effect ||
 * ^  ||^   || Decrease tr || No effect ||
 * v || speed limit || Increase v || Expand ||
 * ^  ||^   || Decrease v || Shrink ||
 * a || negative acceleration || Increase a || No effect ||
 * ^  ||^   || Decrease a || No effect ||
 * w || width of intersection || Increase w || Shrink ||
 * ^  ||^   || Decrease w || Expand ||

__**Stop Zone Prediction Graph**__ 1. Imagine that you are at Intersection I shown in the diagram below. 2. Imagine that you are at Intersection II. 3. Imagine you are at Intersection III. 4. Compare the Go zone and the stop zone for Intersections 1, 2, and 3. __Go Zone:__ V = d / t V = d / t y t y (V = (width + Go zone) / t y ) Vt y = w + GZ GZ = the furthest distance at which you can go safely. "edge" of GZ.
 * Variable |||| Change || **Predicted** shrink or expansion of STOP ZONE ||
 * ty || yellow-light time || Increase ty || No effect ||
 * ^  ||^   || Decrease ty || No effect ||
 * tr || reaction time || Increase tr || Moves it back ||
 * ^  ||^   || Decrease tr || Moves it forward ||
 * v || speed limit || Increase v || Moves it back ||
 * ^  ||^   || Decrease v || Moves it forward ||
 * a || negative acceleration || Increase a || Moves it forward ||
 * ^  ||^   || Decrease a || Moves it back ||
 * w || width of intersection || Increase w || No effect ||
 * ^  ||^   || Decrease w || No effect ||
 * __Page 95 Part B: Yellow Light Dilemma__**
 * a. Would you go or stop if the light turned yellow when you were driving in automobile A? Automobile B? Automobile C? Automobile D?
 * Automobile A: I would stop because it is in the stop zone.
 * Automobile B: I would go because it's already half past the stop zone.
 * Automobile C: I would go because the car is in the front of the Go zone area and closest to the intersection.
 * Automobile D: I would stop because it's in the stop zone and at the front of it.
 * a. Would you go or stop if the light turned yellow when you were driving in automobile E? Automobile F? Automobile G? Automobile H?
 * Automobile E: I would Stop because it's in the beginning of the stop zone.
 * Automobile F: I would stop because it's in the stop zone.
 * Automobile G: I would go because it's the closest to the intersection
 * Automobile H: It depends on the speed of the car, but I would go. It's in the overlap zone.
 * a. Would you go or stop if the light turned yellow when you were driving in automobile J? Automobile K? Automobile L? Automobile M?
 * Automobile J: I would stop because it's in the stop zone
 * Automobile K: I would go because it's in the go zone.
 * Automobile L: It's in the stop zone so I would stop.
 * Automobile M: It depends on the speed of the car and the person in front of you.
 * a. How are the intersections different?
 * Intersection 1 shows only stop and go zones. Intersection 2 shows stop and go zones and also stop/go zones. Intersection 3 shows stop, go, and a zone that depends on the person in the front of them.
 * b. In intersection II, if the light turned yellow when you were between the Go zone and Stop zone, what would your choices be? Which choices would be safe? Explain your answer.
 * Your choices would be go or stop but if you go, it would depend on your speed.
 * c. In Intersection III, if the light turned yellow when you were in the space between the STOP Zone and the GO Zone, what would your choices be? Which choice(s) would be safe? Explain your answer.
 * Your choices would be either to stop or go but if you go, it would have to depend on the car in front of you. If the person spends a long time turning, there will be no time for your car to go through the intersection without running a red light.
 * d. When both choices are safe, the space between the GO and STOP Zones is called the Overlap Zone. When neither choice is clearly safe, it is called the Dilemma Zone. Intersections with a Dilemma Zone are not safe. Which intersection has an Overlap Zone and which has a Dilemma Zone?
 * Intersection 2 has an overlap Zone and Intersection 3 is a Dilemma Zone.
 * __Go Zone and Stop Zone Equations__**
 * Vt y - w = GZ**
 * If you go faster, you have a greater go zone. (Bigger V, Bigger GZ)

__Stop Zone:__ Stop Zone = Total Stopping Distance SZ = TSD SZ = d r + d b SZ = Vt r + (-Vi^2) / (-2a) SZ = the closest position to the intersection in which you can brake safely. "edge" of SZ.
 * SZ = Vt r + Vi^2 / 2a**

1. In this section, the spreadsheet is referred to as a model. What makes it a model?
 * __Checking Up Questions__**
 * It shows the different variables in going and stopping at an intersection.

2. In your own words, describe what is meant by the Go Zone.
 * The Go zone is where a car can go through an intersection safely.

3. In your own words, describe what is meant by the Stop Zone.
 * It is the region where a car has to stop because it can't make it through an intersection safely.

4. Describe what is meant by the Overlap Zone.
 * It is the region where it would be safe for the car to either go or stop.

5. Describe what is meant by the Dilemma Zone.
 * The Dilemma Zone is the region where it is not safe for a car to go or stop.

1) Given a car approaching an intersection of width 25 m and yellow light time 3 seconds on a road of speed limit 45 mph (20 m/s) with a maximum deceleration of -5 m/s2 and a driver with reaction time .7 seconds. Diagram the intersection, calculate the Stop Zone and Go Zone, and decide if the intersection is safe or dangerous and explain why.
 * __Active Physics Plus Questions:__**

2) Change one of the 5 variables at a time in the excel sheet to make 3 separate realistic and safe intersections. A realistic intersection has an overlap zone that is 10-30 m long. Re-compute the Stop Zones and Go Zones using the stop zone and go zone formulas then diagram the 5 safe intersections and photo them along with the calculations into your wiki. Each intersection should be labeled by what variable was changed and how changing that variable affected the Stop or Go Zone. Check your intersections using the excel sheet to make sure its safe before you calculate the stop zone and go zone using the formulas.

3) Design 1 safe intersections for a cognizant driver of reaction time .5 sec and maximum deceleration of -6 m/s2. Choose the other 3 variables(speed limit, yellow light time and width of intersection) to make a safe and realistic intersection. You can use the excel sheet and do not have to use the manual formulas to calculate Stop Zone and Go Zone. Diagram them and list the 5 input variables under the photo of the diagram in your wiki.

4) Design 1 safe intersections for a fatigued driver with reaction time 2 seconds and maximum deceleration of -4 m/s2. Choose the other 3 variables(speed limit, yellow light time and width of intersection) to make a safe and realistic intersection. You can use the excel sheet and do not have to use the manual formulas to calculate Stop Zone and Go Zone. Diagram them and list the 5 input variables under the photo of the diagram in your wiki. 1) Approaching an intersection in the Midwest you realize it intersects another wide highway. Since the intersection is wider than you anticipated how has the Go Zone changed compared to what you had anticipated?
 * __Active Physics Plus Homework:__**
 * The Go zone has expanded and became bigger than anticipated.

2) Approaching an intersection you realize the engineer of the intersection could have compensated for wide intersection by increasing what variable below to extend the Go Zone? What could the driver have changed to increase the GoZone? ty (yellow light time) Increase tr (reaction time), No effect w(width of intersection), Increase a(negative acceleration), No effect vi(initial speed before braking). Increase

3) A driver with worn out brakes approaches an intersection while a driver with superior brakes approaches an intersection next to him. Which driver will have to brake first? Which drive has a larger total stopping distance? Which driver has a stop zone that is pushed further back from the intersection?
 * The driver with the worn out brakes would have to brake first because they have a bigger stopping distance. The driver with the worn out brakes would also have a stop zone that’s pushed further back from the intersection.

4) Two drivers approach an intersection equal in all respects except one is moving faster than the other. Which one has a larger total stopping distance? Which one has a stop zone that is pushed further back from the intersection?
 * The driver that goes faster has a bigger total stopping distance and the slow driver has the stop zone that pushes further back from the intersection.

5) Two drivers, equal in all respects except one drunk and one sober approach an intersection. Which one has a larger stopping distance? Which on has a Stop Zone which is pushed further back from the intersection?
 * The sober one has a larger stopping distance and the drunk one has a stop zone pushed further back from the intersection.

6) Why does speed affect the Stop Zone more drastically than the Go Zone? You can use the formulas for Stop Zone and Go Zone to explain. Consider also how speed affects braking distance compared to how it affects the GoZone.
 * The Stop zone is affected by speed more because it is harder to brake in time when you go faster. In the go zone, you would have to go through the intersection because you’re in the go zone.

__Section 7__

 * <span style="font-family: 'Courier New',Courier,monospace;">** Section7 ** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 20px;">**Points** ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">WDYSee/Think: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || **75/75** ||

__What do you see?__ __What do you think?__
 * The red car in the picture is turning on a curve on a mountain. It is leaning over into the turn because of its weight and side that it is turning on. It looks like as if it is going to fall over from leaning over too much. There's a triangular shaped sign that looks like it's warning drivers about turning on a mountain.
 * Why is the sign indicating to slow down?
 * The picture is showing everything on a mountain top/side, and there aren't any side railings to keep cars from falling over, so it is warning drivers about the dangerous curves and turns.
 * How is the amount you should slow down determined?
 * Sharp turns, size of the car, conditions of the road(wet roads, wet leaves, etc

__Learning Outcomes__
 * Recognize** the need for a centripetal force when rounding a curve.
 * Predict** the effect of an inadequate centripetal force.
 * Relate** speed to centripetal force.



__**11/14/11 Go and Stop Zone of a real Intersection**__ 2) Measure the width of the intersection using Google maps(there is a scale in the bottom left corner).
 * 50 feet, 20 meters

3) Estimate a realistic reaction time based on what you measured in section 1 and estimate a realistic negative acceleration for an automobile coming to a short stop (you can Google this or look at accelerations we have used in our problem sets).
 * A realistic reaction time would be .4 seconds
 * A realistic negative acceleration for a car to stop would be -4.5 m/s^2.

4) Calculate and Diagram the Go Zone and Stop Zone for your intersection. Include the width of the intersection in your diagram and convert speeds from mph to m/s for your calculations.

5) Double the speed limit and calculate a new Go Zone and Stop Zone and diagram it just as before. Does this make a more or less safe intersection if you travel at double the speed limit? Why do you think a ticket is such a large fine for traveling at double the speed limit? Does the Stop Zone or Go Zone change more when the speed is doubled? Explain why.
 * Doubling the speed limit causes a bigger risk for drivers to get into car accidents. It would take them longer to brake in time.
 * This is a reason as to why a ticket is a big fine for traveling at double the speed limit. Someone (or possibly a lot of people) could get into a terrible car accident.
 * Both zones get expanded

__**Physics Words**__
 * Force:** A push or pull
 * Centripetal Force**: A force directed toward the center to keep an object in a circular path.
 * Centripetal Acceleration**: A change in the direction of the velocity with respect to time. (A change in velocity caused by a change in direction)

__**Checking Up Questions:**__ 1. What is the direction of the force that keeps an object moving in a circle? 2. What is the name of the force that keeps an object moving in a circle? 3. Name the force that keeps an automobile moving in a ciruclar path on a road. 4. Explain how the velocity of an object can change even if the speed is not changing. 5. Describe three situations in which acceleration can take place. 6. What is the force that keeps Earth moving in a circle around the sun?
 * There is no certain direction because it always changes from moving east, south, west, then north, and then starts a new revolution.
 * Centripetal Force
 * The friction between an automobile and the road.
 * It changes when the direction changes.
 * It takes place when an object speeds up, slows down, or changes directions.
 * Gravity

__**Investigation**__

__**Part 1: Can a car go faster around a wide turn or a sharp turn?**__

1) Choose two radii to place your circular mass on, one towards the center of your Lazy Susan and one towards the edge. The radius towards the center will simulate the car on a sharp turn while the radius towards the edge will simulate the car on a wide turn. 2) Spin the Lazy Susan with the mass at the inner radius for 10 revolutions then spin the Lazy Susan with the mass at the outer radius for 10 revolutions, recording the time each takes to make 10 revolutions. 3) Calculate the time it took the Lazy Susan to make one revolution for each radii. 4) Use the formula C=2 πr to solve for the circumference of the path that the weight traveled for each radii. Next, Divide the Circumference for each path by the time for the weight to make one revolution for the inner and outer radius. This will tell you the speed of the weight while it traveled around the Lazy Susan.
 * C = 2π(18.25) = 36.5π = 114.7
 * 114.7 / 1.724 = 66.5
 * C = 2π(3) = 6π = 18.85
 * 18.85 / 1.279 = 14.74

5) Create a table similar to below to organize your data for Part 1 (cm) || Max speed (cm/s) || 6) Can you achieve a larger maximum speed on a wide turn (large radius) or a sharp turn (small radius)? How does your data show this?
 * Radius
 * 18.25 cm || 18.25/1.724=10.59 cm/s ||
 * 3 cm || 3/1.279 = 2.34 cm/s ||
 * Yes because the speed becomes greater when the radius increases. Our data shows this because the smaller radius (3 cm) has a small speed (2.34 cm/s) but the bigger radius (18.25 cm) has a greater speed (10.59 cm/s)

__**Part 2 - Can a car go faster turning on an icy surface or a dry surface?**__

Hypothesis: Answer the Question posed in the Title of Part 2 and give a reason for this hypothesis.

Procedure: 1) Spin the Lazy Susan with the mass on the Wood Surface for 10 revolutions then spin the Lazy Susan with the mass at the same radius on the sandpaper for 10 revolutions, recording the time each takes to make 10 revolutions. 2) Calculate the time it took the Lazy Susan to make one revolution for each surface 3) Use the formula C=2 πr to solve for the circumference of the path that the weight traveled for each radii. Next, Divide the Circumference for each path by the time for the weight to make one revolution for the inner and outer radius. This will tell you the speed of the weight while it traveled around the Lazy Susan. 4) Create a table similar to below to organize your data for Part 2
 * C = 2π(16.5) = 103.8 / 1.365 = 76.04
 * C = 2π(14.5) = 91.11 / 1.651 = 55.2
 * Radius (cm) || Max Speed (cm/s) ||
 * 16.5 cm || 16.5/1.365 = 12.09 cm/s ||
 * 14.5 cm || 14.5/1.651 = 8.78 cm/s ||

5) Can you achieve a larger maximum speed on sandpaper (normal asphalt) or on the wood surface (slippery road conditions)? How does your data show this?
 * Yes, because it has friction. Our data shows this because we have a greater speed on sandpaper and a smaller time.

__**Part 3 - Can a more massive vehicle go faster turning or a less massive vehicle go faster turning?**__

Hypothesis: Answer the Question posed in the Title of Part 3 and give a reason for this hypothesis.

Procedure: 1) Spin the Lazy Susan with the lighter mass on the Wood Surface for 10 revolutions, then spin the Lazy Susan with the heavier mass at the same radius on the sandpaper for 10 revolutions, recording the time each takes to make 10 revolutions. 2) Calculate the time it took the Lazy Susan to make one revolution for each mass. 3) Create a table similar to below to organize your data for Part 2 4) Can you achieve a larger maximum speed before skidding out in a heavier vehicle or a lighter vehicle? How does your data show this?
 * Mass || Max Speed (cm/s) ||
 * Light || 12.5 / 1.542 = 8.11 ||
 * Heavy || 12.5 / 1.461 = 8.56 ||
 * You can achieve it in a heavier vehicle. Our data shows this because the heavier vehicle has a smaller time with the same radius as the light vehicle's.

__Centripetal Acceleration__ : A change in **velocity** caused not by a change in speed but just by a change in **direction**. Acceleration and Force are __uni-directional__ (Go in the same direction all the time). They point in the same direction

__**Physics to Go**__ 1. A person at the equator travels once around the circumference of Earth in 24 h. The radius of Earth is 6400 km. How fast is the person going? Compute the speed in kilometers per hour (km/h) and in meters per second (m/s). Recall that 1 km is equal to 1000 m. 2. Earth travels in a circular motion around the Sun. The radius of Earth’s motion is about 1.5 × 10^8 km. What is the speed of Earth around the Sun? Compute the speed in km/h and m/s.
 * V = 2πr
 * V = 2π(6400) / 24
 * **V = 1675 km/h**
 * 1675 km/h * 1000 / 1 m/km * 1 / 3600 h/s = **465.3 m/s**
 * V = 2πr / t
 * V = 2π(1.5*10^8) / 1 year
 * V = 942477796.1 / (365.25 days/year * 24 hrs/day)
 * V = 942477796.1 / 8766
 * **V = 107515.2 km / h**

3. A fan turns at a rate of 60 revolutions per second. If the tip of the blade is 15 cm from the center, how fast is the tip moving? 4. Friction can hold an automobile on the road when it is traveling at 20 m/s and the radius of the turn is 15 m. What happens if: 5. Think about other examples in which objects travel in curved paths, such as the clothes in a spin dryer, or the Moon traveling around Earth. For each example, explain what produces the force that is constantly being applied to the object toward the center of the curve. 6. Sketch a graph that shows the radial distance and the maximum speed at which the block remains on a turntable for one type of surface. 7. Explain the following statement: “The driver may turn the wheels but it is the road that turns the automobile.” 8. Active Physics A jet pilot in level flight at a constant speed of 270 m/s (600 mi/h) Plus rolls the airplane on its side and executes a tight circular turn that has a radius of 1000 m. What is the pilot’s centripetal acceleration? Draw a sketch of the acceleration’s direction relative to the ground. 9. Below you will find alternate explanations of the same event given by a person who was not wearing a seat belt when an automobile went around a sharp curve. “I was sitting near the middle of the front seat when the automobile turned sharply to the left. A force made my body slide across the seat toward the right, outward from the center of the curve, and then my right shoulder slammed against the door on the passenger side of the automobile.” “I was sitting near the middle of the front seat when the automobile turned sharply to the left. My body kept going in a straight line while, at the same time due to insufficient friction, the seat slid to the left beneath me, until the door on the passenger side of the automobile had moved far enough to the left to exert a centripetal force against my right shoulder.” Are both explanations correct? Explain your answer in terms of both explanations. 10. Race cars can make turns at 150 mi/h. What forces act on a race car as it moves along a circular path at constant speed on a flat, horizontal surface? 11. Why are highway curves that have radii that decrease as you go into them especially dangerous? In other words, curves that start out as gentle turns but become tighter and tighter as you get into them. 12. In the United States, vehicles drive on the right-hand side of a two-lane road. If the curve bends to the right and you lose traction in the turn, would you end up in the ditch on your side of the road, or into the lane of oncoming traffic? What if the curve bends to the left?
 * 107515.2 km/h * 1000 / 1 m/km * 1/3600 h/s
 * = **29865.33 m/s**
 * V = 2πr
 * V = 2π(15) / 60
 * **V = 1.6 cm/s**
 * a) the curve is tighter?
 * It should go slower to not hit any other car.
 * b) the road surface becomes slippery?
 * The car should go slower to avoid accidents
 * c) both the curve is tighter and the road is slippery?
 * It should go slower.
 * A merry-go-round in a playground and in a carnival spins around the middle. The carnival merry-go-round has a machine that creates the force and acceleration for it to go around.
 * [[image:Photo_on_2011-11-17_at_01.11.jpg]]
 * If the roads are slippery, the automobile can go out of control and get into an accident. If they have friction, the automobile won't have trouble turning.
 * [[image:Photo_on_2011-11-17_at_01.11_#3.jpg]]
 * The first explanation is correct because objects would always lean or go the direction opposite of where the automobile is curving. Also, someone without a seatbelt would slide to the right as described in the explanation.
 * The second explanation does not seem correct because objects always go another way when an automobile turns around, and also, seats in automobiles don't usually slide around or move.
 * Friction keeps the race car on the ground and since it is not slippery, the race car would have less of a risk of crashing. Without friction, it wouldn't turn. Also centripetal acceleration allows the race car to move around the curves.
 * You would have to watch the speed you are going at or else it would be very dangerous.
 * If the curve bends to the right, I would end up in the lane of oncoming traffic, but if the curve bends to the left, I would end up in the ditch.

Fc = Centripetal Force m = mass V = speed r = radius
 * __Active Physics Plus__**
 * F c = mv^2 / r**

Ac = Centripetal Acceleration V = speed r = radius
 * A c = v^2 / r**

Force measured in Newtons (N)

1) A car makes a left turn at a small intersection with turning radius 10 m. The mass of their car is 2000 kg. The pavement provides a frictional force of 13,720 N. What maximum speed can the car make the turn with? 8.3 m/s 2) Now the pavement is wet and made on the same turning radius and with the same car. The wet pavement causes the frictional force to cut in half. What maximum speed can the car make the turn with? 5.6 m/s 3) What minimum radius can a car make a turn with if it is traveling at 10 m/s, has a mass of 3000 kg and has a provided frictional force of 20,580 N? What can the car do to make the turn at an even sharper radius? 14.6 m 4) What minimum radius can a car make a turn with if it is traveling at 5 m/s, has a mass of 2200 kg and has a provided frictional force of 6,000 N? What can the car do to make the turn at an faster speed? 9.2 m 5) Calculate the centripetal acceleration of a car making a 12 m radius turn at 10 m/s. 8.33 m/s2 6) Now calculate their centripetal acceleration at 20 m/s. Twice the speed. How many times more centripetal acceleration is produced? 33.3 m/s2 7) Use the Givens in number 5 but now change the radius to twice the radius. What is the centripetal acceleration? What fraction is the centripetal acceleration of the centripetal acceleration in number 5 when the radius was still 12m? 4.17 m/s2
 * Fc = mv^2 / r
 * 13720 N = (2000 kg) v^2 / 10 m
 * 137200 N = 2000v^2
 * 68.6 = v^2
 * **8.28 m / s= v**
 * 13720 / 2 = 6860
 * 6860 N = (2000 kg) v^2 / 10 m
 * 68600 N = 2000v^2
 * 34.3 = v^2
 * **5.86m/s = v**
 * 20580 N = (3000 kg) (10^2) / r
 * r = 300000 / 20580
 * **r = 14.6 m**
 * 6000 N = (2200) (5^2) / r
 * r = 11000 / 6000
 * **r = 9.17 m**
 * Ac = 10^2 / 12
 * Ac = 100 / 12
 * **Ac = 8.33** **m/s^2**
 * Ac = 20^2 / 12
 * **Ac = 33.33 m/s^2**
 * Ac = 10^2 / 24
 * **Ac = 4.17 m/s^2**
 * It's 1 / 2 because 4.17 is half of 8.33.