Chapter+1


 * __Chapter 11__**

__**Section 1**__
toc ~Car accident scene In the picture I see a pretty severe car accident. One car is flipped over with another on top of it. Before the accident I see a car slamming its breaks to stop the car and the car is skidding down the road. The person in the car's hair is all blown back implying he was speeding.
 * __9/7 What do you see?/What do you think?__**

~What factors affect the time you need to react to an emergency situation while driving? The reaction time can vary on the speed, how far you are from the accident, if you are under the influence, the weather, the size of your car, your tires

12 motions in 10 seconds Time for Ten Motions = 8.57 s econds Time for One Motion = 0.84 seconds 1st time Christine: 5.67 seconds Kristoff: 6.04 seconds 2nd time Kristoff: 5.06 seconds Christine: 5.41 seconds 3rd time Kristoff: 4.21 seconds Christine: 4.67 seconds Average: 5.17 seconds First catch = 43 cm Reaction Time = 0.3 seconds Second catch = 23 cm Reaction Time = 0.2 seconds Third catch = 15 cm Reaction Time = 0.15 seconds Average Reaction Distance: 27 cm Average Reaction Time: 0.23 seconds //1) Fastest Reaction Time of class: .12 seconds, Slowest Reaction Time of class: .84 seconds, Average Reaction Time of class: .32 seconds// //2) Some classmates may have different reaction times depends on whether or not if they were prepared to catch the meter stick, whether they were taking by surprise when it was dropped and if they do any activities that may help them with hand-eye coordination.// First catch = 49cm = 0.3 seconds Second catch = 39cm = 0.27 seconds Third catch = 35cm = 0.25 seconds Average Distance: 41 cm Average Reaction Time: 0.273 seconds //2) Normally when catching the meter stick there were not distractions or complications when you saw the stick drop you had to catch it, but involving decision making made us think before we acted causing a delay in the time.// //3) While trying to avoid road hazards you have to be quick thinking you can't take the time to really think about it you just have to go with your gut and do it quickly.// First catch = 25cm = 0.22 Second catch = 31cm = 0.25 seconds Third catch = 23cm = 0.21 seconds Average Distance: 26.3 cm seconds Average reaction time: 0.23 seconds
 * __9/7 Investigation Reaction Time: Responding to Road Hazards__**
 * Moving foot from pedal to pedal**
 * Stop Watch Test**
 * Catching meter stick**
 * Catching meter stick while making decisions**
 * Catching meter stick while texting and deciding**

1. It takes a fast ball moving 90 miles per hour about half a second to get to home plate. 2. Sometimes you don't hit the ball because even if our reaction time may be under .5 seconds it takes longer for you brain to send the signals to your hands and arms. 4. Yes, it is possible to hit a fastball if you react in time. Where the ball is going is also a factor. 5. To reduce your reaction time when behind the wheel you can cut out all distractions that could distract you and you can act as soon as you notice the hazard. 6. When some yells "hey batter batter swing batter batter" it can distract you and make you not pay attention to the ball coming.
 * __9/9 Reaction Time: Science of the fastball__**


 * __Active Physics Plus 9/9__**

d=1/2at^2 d: distance an object is dropped (cm) t: time it takes the object to drop "d" (s) a: acceleration of the subject (980cm/s^2) d=1/2at^2 ^ ^ ^ cm=cm s^2


 * Distance || Time ||
 * 4.9cm || .1s ||
 * 19.6cm || .2s ||


 * __Reaction Time Ruler 9/9__**
 * Distance || Time ||
 * .3cm || .025s ||
 * 1.2cm || .05s ||
 * 2.8cm || .075s ||
 * 4.9cm || .1s ||
 * 7.7cm || .125s ||
 * 11.0cm || .15s ||
 * 15.0cm || .175s ||
 * 19.6cm || .2s ||
 * 24.8cm || .225s ||

__**9/12 Do Now-**__

~Who would need a faster reaction time? A race car driver or someone in a school zone? I think that a race car driver would need a reaction time because the race car is going 200 miles per hour and crashing into the wall or other cars while someone in a school zone has to watch out for pedestrians and stop signs. The race car driver needs a better reaction time because they are moving so much faster than someone else.


 * __9/12 Reaction Time Ruler Notes-__**



Dollar bill: d=15.4cm Reaction time needed to catch the dollar bill:
 * __9/12 Can you catch a dollar bill?__**

d=1/2at^2 15.4cm=1/2(980cm/s^2)t^2 15.4cm=(490cm/s^2)t^2 .031s^2=t^2 .177seconds=t

6. You might crash into someone or something. 7. It's more expensive for teenagers because they haven't had as much practice driving and they're more likely to get into an accident rather than an experienced driver.
 * __9/14 Do Now: Physics To Go__**

What are the top two causes of accidents on the road? The top two causes of accidents are rubbernecking and driver fatigue. What is rubbernecking? Does rubbernecking constitute a decision or distractions? Rubbernecking is when you stare at an accident or some other distraction which causes a lot of traffic. It causes more of a distraction than a decision.
 * __9/14 Reflecting On the Chapter Challenge__**

__Section 2__
__**Learning Objectives:**__
 * 1) Calibrate the length of a stride
 * 2) Measure a distance by pacing it off and by using a meter stick
 * 3) Identify sources of error in measurement
 * 4) Evaluate estimates of measurements as reasonable or unreasonable

What Do You See? What Do You Think? I see 2 kids walking along a tape measure, one kid is older and one girl is younger. Also, there is one kid taking notes on the 2 kids walking. They're is also a girl and a cat creeping from one of the other rooms. 1. I think that the second student measured in feet rather than meters because 1 foot is a little longer than 1/3 of a meter. 2. I think that the student who got 3.01 meters measured in yards not meters.
 * __9/14 What Do You See/What Do You Think__**


 * __9/15/11 Measurement: Errors, Accuracy, Precision, Investigation__**

Number of strides to pace off distance: about 20 Length of stride: about 54 cm we measured it by having me take a step next to the meter stick a few times length of strides x number of strides = 1,080 cm = 10.8m number of metersticks for distance 13 (+.20cm)

Questions: 1. The measurements listed on the class table are similar. 2. The results vary from one meter to about 0.4 meters. 3. Tere are many differences in the group's chart on the board because they could have moved the meter stick a little back, i.e. slipping the ruler and not noticing, which added or subtracted the distance between the two tape marks. Everyone also has a different stride length. There could also be different stride lengths for every step. 4. 5. If given a tape measure, it would eliminate some randomness, but there is not knowing if people start at different points on the measuring tape. 6. There will always be differences in measurements because there will always be an error barrier. There will always be differences. 7. As a class, if we all made the same mistake, we would be able to fix it (like starting a cm in on the meter stick).

Demo- How long is the tube? (4sided ruler) .82 .73 .72 .76 .71 .79 .72 .74 || .84 .85 .81 .86 .82 .81 .83 .84 .82 || .81 .81 .81 .81 .8 .82 .81 .815 .81 || .815 .814 .815 .8145 .813 .815 .814 .812 .812 ||
 * Group || Strides || Metersticks ||
 * 1 || (13 strides x.93m) =12.09m || 13.16m ||
 * 2 || (20 strides x .54m) = 10.8m || 13.20m ||
 * 3 || (18 strides x .74m) = 13.34m || 13.31m ||
 * 4 || (22 strides x .5m) = 11m || 14m ||
 * 5 || (21 strides x .74m) = 15.54m || 13m ||
 * 6 || (18.3 strides x .56m) = 10.25 || 13.5m ||
 * Interval || 1meter || .1m || .01m || .001m ||
 * || .75

__ **9/18 Homework** __

random error: an error that can't be corrected by a calculation. systematic error: an error produced by using the wrong tool or using the tool incorrectly. this can be fixed 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.

__**9/19 Accuracy&Precision**__

__**9/19 SI System**__
 * Quantity || BaseUnit || Symbol ||
 * Distance || Meters || m ||
 * Mass || Gram || g ||
 * Time || Seconds || s ||

x1000 || 1km=1000m 1m=.001km || x.01 || 1cm=.01, 1m=100cm || x.001 || 1mm=.001m 1m=1000mm ||
 * Prefix || Symbol || Mult.of10 || Exp ||
 * kilo || k || x10^3
 * cent || c || x10^2
 * mili || m || x10^-3

__**9/19** **HOMEWORK:**__ __ Physics To Go __ p. 32-33 3. My friend and I may agree that it takes 25 minutes for us to shower, but we may disagree on how long it takes to drive to school 4. Depending on the size of the oil tanker, five million barrels seems like an inaccurate measurement. If some oil spilled out, then it could lower how much each barrel is worth. 6. Are the following estimates reasonable? Explain your answers. a. A 2-L bottle of soft drink is not enough to serve 12 people at a meeting because it only makes 8cups. b. Yes, a mid-sized sedan would get you from NYC to Boston without refueling. A Honda Accord has a 20 gallon tank, and 34 mpg highway, which would get you 680 miles. NYC is only 225 miles away from Boston, so you would have enough fuel to go both ways. 7. No, being 1 m off in measuring the width of a room is a bigger error than being 1 m off in measuring the distance between my home and my school. In the room, you could measure it incorrectly, but since the distance between the school and my house is larger, the measurement would be easier. 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. 60 mph will guarantee that you will not exceed the speed limit b. A passenger in the vehicle could use the mile markers and wristwatch to time how long it takes to get from one marker to the next.

__**9/20 Measure The Copper Tube**__ Measuring a copper tube
 * Groups || Measurement of Copper Tube ||
 * Group 1 || 66cm ||
 * Group 2 || 64.1 cm ||
 * Group 3 || 66.1 cm ||
 * Group 4 || 64 cm ||
 * Group 5 || 66 cm ||
 * Group 6 || 64.1 cm ||

__**9/20** **Active Physics Plus**__ 1. 10(10 -2 )m = (+,-).1 m --> 50.1 m - 49.9 m (+,-) 1 cm --> (+,-) .01 m --> 50.01 m - 49.99 m (+,-) .001 m --> 5.001 m - 49.999 m

2. (+,-) 1 cm 50.01 m - 49.99 m 50m/25s = 2m/1s 2m/1s = .02m/ts 2t = .02 t = .01s

3. 1500m/15(60 sec) = 1500m/900s = .6 m/ts 150 0t = .6m(900s) t = .36sec

4. 15.36s --- 50.01 m (x30 laps) = 1500.3 m 1500.3m/15.36s = 97.68m/1s

15.35s--- 49.99 m (x30 laps) = 1499.7 m 1499.7m/15.35s = 97.7m/1s

__**9/21 Do Now: Section 2 p.31**__ 1. What does it mean? Systematic because you can go back and change it by calculation. Random error affects precision because they are random and not really fixable. 2. Why do you believe? There will always be an uncertainty but you can do it more than once and average or make it a range 3. Radar gun 75mph in a 30mph No, it is not reasonable because there's no way it can be off by 35mph

__**9/21 Estimations 24 a-g**__ a. It is reasonable because college football players are taller and have more muscle. b. It is not reasonable because no man could be 13 feet tall. c. It is unreasonable because 1440minutes is 24 hours. d. It is not reasonable because a standard poodle does not weight 132 pounds. e. It is not reasonable because the room is closer to 45,000 feet. f. It is not reasonable because its probably longer. g) It's not reasonable because 1/4 mile doesn't give a person driving 50 mph enough time to pass the tractor. You're driving about 3.3 (1/4)mile per minute. h) You must estimate the speed that the dump truck is driving in order to decide whether to stop and wait for the dump truck to cross the bridge or go ahead and squeeze by the truck. i) Some motor homes are bigger than others, so depending on the size, it might not make it safely under the tunnel. But it is reasonable for smaller motor homes.

__Section 3__
__**~ There is a 3 car rear-end collision caused by a distraction.9/21 What Do You Think?**__ ~A safe following distance is one car length for every 10 miles ~You decide a safe following distance by making sure you have enough time to stop if the person in front stops short.

__**9/21 Investigation- Strobe Photo:**__ Strobe Photo: a combined photos of pictures taken at equal intervals of time

Motion Detector: 1. hook up motion detector to USB port 2. open data studio 3. create experiment

1a) Strobe photo of an automobile (rectangle) traveling at 30 mi/h. 2a) Strobe photo of an automobile traveling at 45 mi/h.

2b) The automobile is not the same distance apart between successive photos. The images were farther apart than they were at 30 mi/h. The car traveling 30 mi/h is going 0.75 mi/min and the car traveling 45 mi/h is going 0.5 mi/min.

2c)

Since the car is traveling at 60 mi/h, the distance between them is greater compared to a car traveling 30 mi/h.

3a) In diagram C, the automobile is traveling the slowest. In diagram A, the automobile is traveling the fastest. Diagram C is the slowest because the distance is shorter and diagram A is the fastest because the distance is greater.

3b) Yes, each automobile is traveling at a constant speed because each space to equal to the previous one.

4a) 4b) 4c) 4d)

__//**Checking up Question's :**//__ 1.Explain how the average speed of a vehicle is different from instantaneous speed. The average speed is the total speed while the instantaneous speed is at any given moment. 2.How are the speed and velocity of an object different? Speed is the distance traveled per unit and it has no direction whereas velocity is the speed in a given direction (with direction) 3.If the-time graph shows a straight, inclined line, what does the line represent A fast speed 4.How does reaction time affect reaction distance? If the reaction time is quick than the reaction distance will be shorter. 5. What is the difference between constant speed and average speed?

__//**DoNow 9/27**//__ -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? Vav= d/t Vav+ 90 ft/sec t= .6s d=? (.6s) 90ft/s=d/.6s(.6s) 54ft= d - How much further will the car travel if the driver distracted by texting, so that reaction time is increased to 1.5s?

20m/hr || 40mi || ∆t1=? 2hours || 40m/hr || 40mi || ∆t2=? 1hour || Vaverage=? || 80mi || 3hours || V AVERAGE= 20m/hr + 40 m/hr/2=30m/hr V average= ∆d/∆t V average= 80mi/∆t total
 * //Active Physics Plus p.47//**
 * Part || Distance || Time ||
 * Part1
 * Part2
 * Whole Trip

∆t= 2 hours ∆t=1 hour __//**Paragraph**//__ 1mi/h || 50 miles || 50hrs || 50mi/h || 50miles || 1hr || Vav=? || 100mi || 51hr || Vav= 100mi/51 Vav=1.96mi/h
 * Part 1:** V average+ ∆d/∆t 20m/hr= 40m/∆t1
 * Part 2:** V average= ∆d/∆t 40 mi/hr= 40mi/∆t2
 * Part || distance || time ||
 * part1
 * part2
 * whole trip

__**Physics to Go**__ 1a. The car is moving at a slow and steady speed 1b. The car sped up and then slowed down, changing its speed. 2a. --- --- --- --- --- --- (car starting from rest and then reaching its final constant speed) 2b. --- --- --- --- --- --- --- --- --- (car coming to a stop from a constant speed) 3. The man drove 7,000 feet in 20sec, (350 X 20) 4a. About 48 mph 4b. Its impossible to know how fast the woman was going when she went through Baltimore because we can only find out her average speed 5. One would have to bike 20 mph 6a. The vehicle went fast (it drove away) and then stopped 6b. The vehicle went very fast, driving away, stopped and then drove back very fast 6c. The vehicle drove away slowly, and then increased its speed and drove very fast 6d. The vehicle is driving slowly away 7a. My reaction time is .19s so the vehicle would go 4.75 meters in my reaction time 7b. The vehicle travels 3.04 meters in my reaction time, 1.71 meters less 7c. .38 would be my reaction time, so I would go 9.5 meters. 8a. Experts can use seconds as a way to tell people how close they can be to the car in front of them because they know that normal reaction times are below three seconds so its safe to use this unit. 8b. No, because on a highway you are going much faster than on a small road, so people would have to have more distance between them and the car in front of them. 9a. The vehicle will go 33.33 feet in the time of a sneeze 9b. yes, it is longer than my classroom. 10a. 44 feet 10b. about 3 automobile spaces 10c.22 feet, and about 1 and a half car spaces 10d. 66 feet, about 5 automobile spaces, and it would be about 1/8 of a football field 10e. in one second, the car goes 44 feet at 30mph, the car goes 88 feet at 60mph and 134 feet at 90mph 11.As the reaction time goes up, so does the distance traveled ( i dont know how to make a graph on here)

__**10/3/11 Following Jack**__

1.

2.

__**10/3/11 Distance vs Time Graphs Cyclists**__



a. No they do not because the lines are different. Line B is ahead.

b. Cyclist A because the position is greater.

c. Cyclist A because it has a greater slope than cyclist B

d. No because they travel different speed

e. They're going at the same speed.



a. Same distance, same time.

b. It is going down in distance.(backwards)

c. Cyclist A because it is traveling faster then cyclist B and has more of a slope.

__Section 4__
missing active physics plus || __**10/5/11 What do you see?**__ There is the red car that is taking off at a green light and a guy crossing the street had to run across so he wasn't hit
 * ** Section4 ** || **Points** ||
 * WDYSee/Think: || /10 ||
 * Investigate: || /20 ||
 * PhysicsTalk: || /20 ||
 * PhysicsPlus: || 15/20
 * 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;">**95** ||

__**What do you think?**__ A car and a bus go 0 to 30 mi/h. A car would accelerate faster than the bus; eventually they will both hit 30mi/h

__**!0/5/11 Tangent Lines**__ Tangent Line: a line that only touches a curve at one point side- undefined slope top- 0 slope side- negative slope Slope of the Tangent: the instantaneous velocity at a specific point in time

10/5 __Investigation: Graphing Motion__ Procedure: Questions 1. 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? 2. Below are four different distance vs time graphs.In one of them the cart does not move---Fourth Graph 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 //Motion Detector at top, Cart released from top, stopped at bottom.// PREDICT: a predicted distance versus time graph of the motion of the car
 * No, because it's gaining speed from going down the ramp.
 * __Run 1:__**

TANGENT LINE:

VELOCITY TIME GRAPH: velocity versus time graph made on data studio How is the velocity changing with respect to time? 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?
 * The velocity is increasing as the car speeds up.
 * .5 m/s
 * 1 m/s
 * The times are about the same.

CALCULATE ACCELERATION: Use the velocity versus time graph to calculate the car’s acceleration. A=∆v/∆t A=1/1.4 A= .714 m/s 2

//Motion Detector at the bottom, car pushed from bottom stopped at top.// PREDICT: a predicted distance versus time graph of the motion of the car.
 * __Run 3:__**

TANGENT LINE:

VELOCITY TIME GRAPH: velocity versus time graph made on data studio How is the velocity changing with respect to time? 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?
 * The velocity shoots up at the beginning from the quick acceleration of the push but then it slows as the car looses speed.
 * //.//76 m/s
 * .2 m/s
 * The numbers are close.

CALCULATE ACCELERATION: Use the velocity versus time graph to calculate the car’s acceleration. A=∆v/∆t A= .96/1.85 A= .519 m/s 2

//Motion Detector at the bottom, car pushed from bottom, stopped at bottom.// PREDICT: a predicted distance versus time graph of the motion of the car
 * __Run 4__**:

TANGENT LINE:

VELOCITY TIME GRAPH: screen shot the velocity versus time graph made on data studio How is the velocity changing with respect to time? At what time does the cart make its turn in your velocity vs time graph? How do you know this is the time it makes its turn? Is the acceleration constant or non-constant? How does the graph provide you the answer? Is it possible that the cart has a constant acceleration even though it makes a turn?
 * The velocity is decreasing over time
 * The cart turns at 17.7 seconds
 * The velocity goes negative so that means the car is coming back to the motion detector.
 * Non-constant
 * There is a curve in the graph so it slows down.
 * No because it slows down to come back the opposite way.

CALCULATE ACCELERATION: Use the velocity versus time graph to calculate the car’s acceleration. A=∆v/∆t A= 2/3.5 A= .571 m/s 2

__**DoNow 10/7-**__ 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? a= ∆v / ∆t a = 30 - 0 / 5 a= 6 m/s 2

2) Name 3 vectors and 3 scalars. What is the difference between a vector and a scalar. __Vectors__ __Scalar__
 * A vector is a quantity that has both magnitude and direction. However, a scalar quantity //only// has magnitude, no direction.
 * velocity
 * force (push or pull)
 * speed
 * calories
 * anything you can count

__**Car vs Pickup Truck vs Bus**__



10/10/11 **HOMEWORK:** __ Physics To Go __ p. 68-71 #1-10 1. Can a situation exist in which an object has zero acceleration and nonzero velocity? Explain your answer.
 * No

2. Can a situation exist in which an object has zero velocity and nonzero acceleration, even for an instant? Explain your answer.
 * Yes, but only for an instant.

3. If two automobiles have the same acceleration, do they have the same velocity? Why or why not?
 * Sometimes because two automobiles can have the same velocity but a different change in acceleration.

4. If two automobiles have the same velocity, do they have the same acceleration? Why or why not?
 * Sometimes because two automobiles can have the same change in acceleration but different velocities.

5. Can an accelerating automobile be overtaken by an automobile moving with constant velocity?
 * No, an accelerating automobile cannot be overtaken by an automobile 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?
 * Yes, it is correct to refer to speed-limit signs instead of velocity-limit signs because velocity varies with the direction of a car and speed-limit signs are the signs that are currently used on the roads today. The assumed units for speed-limit signs in the U.S. are miles per hour (mph).


 * 10/12 Predicting Graphs**



__**Physics To Go 10/12-**__ 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? 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 s to rejoin the race. 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 on 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 s. 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. 10/12 **__Classwork: Predicting v vs t graphs from d vs t graphs__**
 * Yes it can, but only for an instant.
 * No, because they can have the same velocity, but different acceleration.
 * No, because they can have the same acceleration, but different velocity.
 * No it can't overtake an accelerating automobile.
 * It's more correct to refer to speed-limit signs because they are used on the roads today and the speedometer in your car can easily be compared to the sign to make sure you're going the right speed. The units for the speed-limit signs are mi/hr.
 * a) What is the race car's acceleration during this time?
 * a= ∆v / ∆t
 * a= (75 - 0) / (9 - 0) = 8.3 m/s^2
 * b) What was the race car's average speed during the acceleration?
 * ∆v= a(∆t)
 * ∆v= (8.3)(9) = 74.7 m/s
 * c) How far does the race go during the time it is accelerating?
 * d= vt
 * d= (74.7)(9) = 672.3 m
 * d) A second race car leaves after its pit stop and accelerates to 75 m/s in 8 s. Compared to the first race car, what is this race car's acceleration, average speed during the acceleration, and distance traveled?
 * a= ∆v / ∆t
 * a= (75 - 0) / (8 - 0)= 9.375 m/s^2
 * ∆v= a(∆t)
 * ∆v= (9.375)(8) = 75 m/s
 * d= vt
 * d= (75)(8) = 600 m
 * a) What is the player's acceleration during the slide?
 * a= ∆v / ∆t
 * a= (.6 - 4.5) / (1.3 - 0) = -3 m/s^2
 * b) What was the distance of her slide?
 * d= vt
 * d= (3.9)(1.3) = 5.07 m
 * c) If she had slid for only 1.1 s, how fast would she have been moving when she reached third base? (assume she had the same acceleration as before)
 * a= ∆v / ∆t
 * a= (4.5 - .6) / (1.1 - 0) = 3.55 m/s^2
 * d) Which of these two trials would get her from second base to third base faster?
 * The second trial would get her from second base to third base faster.
 * a) From the graph, approximately what was the top speed recorded by the astronaut for the falling object?
 * 9 m/s
 * b) What is the acceleration of gravity on this planet?
 * a= ∆v / ∆t
 * a= (9 - 0) / (7.5 - 0) = 1.2 m/s^2
 * 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?
 * a) Which of the graphs would correctly show the boy's velocity versus time as he coasts up the hill?
 * [[image:acpkyoo/GraphA.png caption="GraphA.png"]]
 * b) Which graph shows the distance traveled versus time as he coasts up the hill?
 * [[image:acpkyoo/GraphC.png caption="GraphC.png"]]
 * c) Which graph would show the bike's acceleration as it coasts uphill?
 * [[image:acpkyoo/GraphE.png caption="GraphE.png"]]
 * 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?
 * [[image:acpkyoo/GraphD.png caption="GraphD.png"]]
 * e) Which graph could show the boy's speed versus time graph as the boy coasts up the hill and then down the hill?
 * [[image:acpkyoo/GraphF.png caption="GraphF.png"]]
 * 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?
 * [[image:acpkyoo/GraphB.png caption="GraphB.png"]]
 * a) traveling with constant speed?
 * a to c, e to f
 * b) increasing speed?
 * a to c
 * c) at rest?
 * c to e
 * d) decreasing speed?
 * e to g
 * 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?
 * Back to zero (where it started)
 * a) What is the jet's acceleration?
 * a= ∆v / ∆t
 * a= (250 - 0) / (30 - 0) = 8.33 mi/hr^2
 * b) If after take-off, the jet continues to accelerate at the same rate for another 15 s, how fast will it be going at that time?
 * a=∆v / ∆t
 * a= 375m/hr^2
 * c) How much time does it take for the jet to reach 500 mi/h?
 * 500= (250)(∆t)
 * ∆t= 60 seconds
 * d) How much distance would it take for that same jet to reach 500 mi/h?
 * d= vt
 * d= 15000 m
 * a) If the object falls for 100 m, how fall is it traveling?
 * 44.1m/s
 * b) How much time is required for it to fall this 100 m?
 * 4.5 seconds
 * c) If the object falls for 10 s, how fast is it traveling?
 * a= ∆v / ∆t
 * ∆v= a / ∆t
 * ∆v= (9.8) / (10) = 98 m/s
 * d) How far has it fallen in this 10 s?
 * d= vt
 * d= 490 m
 * e) How would your answer to these questions change for an object falling above the Moon, where the acceleration is about 1/6 g (1.6 m/s^2)?

d = 1/2at 2 +<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;"> V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">t <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">a = ∆v/∆t <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">a = V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">f <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">-V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;"> / ∆t

<span style="color: #ff0000; font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">** 1. ** <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;"> **Given:** <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">f = 20 m/s <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i = 7 m/s a = 3 m/s 2 <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">∆t = ?
 * __sec 4 ACTIVE PHYSICS PLUS worksheet:__**

<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">a = V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">f <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">-V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;"> / ∆t 3 m/s 2 = 20 m/s - 7 m/s / <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">∆t
 * Work:**

<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">∆t (3 m/s 2 ) = (20 m/s - 7 m/s) <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">∆t = 4.33 sec

<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">f = ? <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i = 7 m/s a = 1.5 m/s 2 <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">∆t = 10 sec
 * 2. **<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;"> **Given:**

<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">a = V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">f <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">-V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;"> / ∆t 7 m/s + (10 sec) 1.5 m/s 2 = <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">f - 7 m/s **<span style="color: #ff0000; font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">/ ** 10 sec (10 sec) + 7 m/s <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">f = 22 m/s
 * Work:**

<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">f = 20 m/s <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i = 0 <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">∆t = 5 sec a = ?
 * 3a. Given: **

<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">**Work:** <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">a = V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">f <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">-V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;"> / ∆t a = 20 m/s - 0 m/s ** / ** 5 sec a = 4 m/s 2

<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">d = ? <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">a = 4 m/s 2 t = 5 sec <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i = 0
 * 3b. Given:** d = 1/2at 2 +<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;"> V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">t

d = 1/2 (4 m/s 2 )(5 sec) 2 d = 50 m
 * Work:**

d = 1/2at 2 +<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;"> V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">t <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">d = ? <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">a = 4 m/s <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: super;">2 t = 10 sec <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i = 0
 * 4. ** **Given:**

d = 1/2 (4 m/s 2 )(10 sec) 2 d = 200 m
 * Work:**


 * 10/14** __**Section 4 Quiz**__**:**

solve for instantaneous velocity (slope of tangent line) calculate acceleration from two different slopes of tangent lines acceleration when turning around = constant; never becomes 0 use d = 1/2at 2 use <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">a = ∆v/∆t make a v-t graph from a d-t graph from v-t graph - calculate acceleration
 * <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">a = ∆v/∆t (acceleration = slope of v-t graph)

Section 5
<span style="font-family: 'Arial Black',Gadget,sans-serif; font-size: 16px;">**Negative Accelerati****on: Braking Your Automobile** <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing derivations of eqs of motions did not finish eqs of motion problems || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing preparing for the chapter challenge section of total stopping distance activity || __** 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;">10/10 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">20/20 ||
 * <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;">10/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">15/20
 * <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;">**85** ||
 * **Plan** and carry out an experiment to relate braking distance to initial speed.
 * **Determine** braking distance.
 * **Examine** accelerated motion.

10/18/11 -Do Now What Do You See/What Do You Think?
 * The back wheels of the car are up in the air, so the car is stopping.
 * A moose stopped in the middle of the road, so the driver had to stop short.

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?
 * Mass of the car
 * Your speed
 * Friction between the tires and road
 * The conditions of the tires and brakes
 * If the car has anti-lock brakes or not
 * How old the car is
 * Weather conditions
 * Reaction time

10/18 __Investigation: Relationship between initial speed and braking distance of an automobile__ Objective: to determine the effect of initial speed on braking (stopping) distanc <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i = length of flag<span style="color: #ff0000; font-family: 'Arial Black',Gadget,sans-serif; font-size: 16px;"> **/** time in gate
 * # of textbooks || <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px;">V <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 14.4px; vertical-align: sub;">i (m/s) || Braking Distance (m) ||
 * 1 || 0.835 m/s || 3.048 m ||
 * 1 || 0.844 m/s || 2.896 m ||
 * 2 || 1.212 m/s || 6.462 m ||
 * 2 || 1.960 m/s || 5.944 m ||
 * 3 || 1.110 m/s || 7.087 m ||
 * 3 || 0.949 m/s || 5.639 m ||
 * 3 || 1.077 m/s || 6.401 m ||

__Sec5 Investigation 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 8. Use the data on the sports car provided at the end of this chapter one 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? c) How does this data correspond to what you found in your experiment? 5. If the sports car changed its speed to 30 mph what do you expect its braking distance to be? (//Hint:// if you half the speed by what factor will the braking distance change?
 * x4
 * x9
 * x16
 * Yes, I do expect the ration of the braking distances to also be in the ratio of 80/60 = 1.33.
 * 209/118 = 1.77 (177 percent)
 * 1.33 2 = 1.77
 * This data is similar to our initial velocity.
 * 135/4 = 33.75

read sec5 physics talk p.78-82, answer checking up Qs p.82, record physics words
 * 10/19** **HOMEWORK:**

__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

__Checking Up__ p. 82 1, If a vehicle is traveling at constant velocity and then comes to a sudden stop, has it undergone negative acceleration or positive acceleration? Explain 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?
 * It has undergone negative acceleration because the final velocity is zero when it stops, which is less than the initial velocity.
 * As the velocity of an automobile increases, it increases the braking distance because the car is going faster. The greater speed of an automobile requires more braking distance.
 * Negative acceleration is used instead of deceleration because deceleration means to slow down. However, negative acceleration is decreasing speed in the positive direction or increasing speed in the negative direction.

Vi(2)** / ** Vi(1) ---> ratio of initial velocities Braking Dist(2) **/** Braking Dist(1) ---> ratio of braking distances
 * 10/20/11 Breaking Difference**

[Vi(2) ** / ** Vi(1)] 2 = Braking Dist(2) **/** Braking Dist(1)


 * Honda Civic Stopping Distances **

1) Calculate the stopping distances for Kibala’s Honda Civic 4) If Initial Velocity is doubled how does stopping distance change? When the velocity doubles it multiplies by 4. 5) If the Initial Velocity is multiplied 4 times how does the stopping distance change? The stopping distance is multiplied by 16. 6) If the Initial Velocity is halved how does the Stopping Distance Change? The stopping distance divides by 4. 7) If the initial Velocity is quartered how does the stopping distance change? The stopping distance is divided by 16. 8) What speed would you need to have a stopping distance of a mile? 393.111416 mph
 * Initial Velocity || Stopping Distance ||
 * 10 mph || 3.416 feet ||
 * 20 mph || 13.66 feet ||
 * 30 mph || 30.75 feet ||
 * 40 mph || 54.66 feet ||
 * 50 mph || 85.416 feet ||
 * 60 mph || 123 feet ||
 * 70 mph || 167.4166 feet ||
 * 80 mph || 218.66 feet ||
 * 90 mph || 276.75 feet ||
 * 100 mph || 341.66 feet ||

10/21 __Direction of Accelerations__ **+** < -> **-**

__**10/24 Active Physics Plus-**__ d=1/2(Vi+Vf)t Vf= Vi+at d= Vit+1/2at^2 Vf^2=Vi^2+2ad

1) a= 4.1m/s^2 t= ? d Vi=9m/s Vf= 0m/s
 * Average Acceleration**

Vf= Vi+at 0m/s = 9m/s +(-4.1m/s^2)t -9m/s = (-4.1m/s) t 2.2seconds= t

2) a= 2.5 m/s^ t= ? d= Vi=7.0 m/s Vf= 12 m/s

Vf= Vi + at 12m/s= 7.0m/s + (2.5m/s^2) t 5m/s= (2.5m/s^2)t 2seconds= t

3) a= -0.50m/s^2 t=? d= Vi= 13.5m/s Vf= 0m/s

Vf= Vi + at 0m/s= 13.5m/s + (-0.50m/s^2) t -13.5m/s= (-0.50m/s^2) t 27 seconds= t

4) a=? t= 25m=1500s d= Vi= -1.2m/s Vf= -6.5m/s

Vf= Vi + at -6.5m/s= 1.2m/s + a(1500s) -5.3m/s= (1500s)a a= -.0035 m/s^2

5a) a= 4.7 x 10^-3 m/s^2 t= 5min = 300s d= Vi= ? Vf= 0 m/s

Vf= Vi + at 0= Vi + (4.7 x 10^-3) (300) 0= Vi + 1.41 Vi= 1.41 m/s

5b) a=4.7 x 10^-3 m/s^2 t= 300 s d Vi= 1.7 m/s Vf=?

Vf= Vi + at Vf= 1.7 + (4.7 x 10^-3) (300) Vf= 3.11m/s


 * Displacement With Constant Uniform Acceleration**

1) a= t= 6.5s d= ? Vi= 0m/s Vf= 23.7km/h= 6.58m/s <-- (23.7 km)/(hr) x (1000m)/(1km) x (1hr)/(60min) x(1min)/(60sec)= 6.58m/s

d=1/2 (Vi+Vf) t d= 1/2 (0/ms+6.58m/s) 6.5 d= 21.4 m

2) a= t= 2.5s d=? Vi= 15m/s Vf= 0m/s d= 1/2 (Vi+Vf) t d= 1/2 (15m/s+0m/s) 2.5 d= 18.75m

3) a= t= -5.0m/s^2 d= ? Vi= 100m/s Vf= 0m/s 1000m=no

4) a= t= ? d= 99m Vi= 78km/h= 21.67m/s Vf= 0 m/s

9.14m/s


 * Velocity and Displacement with Uniform Acceleration**

1) a=.92 m/s^2 t= 3.6s d=? Vi= 23.7km/h= 6.583 m/s Vf=?

Vf= Vi + at Vf= 6.583 m/s + (.92m/s^2)(3.6) Vf= 9.89m/s

d=1/2 at^2+Vi t d= 1/2 (.92m/s^2) (3.6s)^2 + (6.583m/s)(3.6s) d= 29.7m

2) a= 3.0m/s^2 t= 5.0s d= ? Vi= 4.30m/s Vf=?

Vf= Vi +at Vf= 4.30m/s + (3.0m/s^2)(5.0s) Vf= 19.3 m/s

d=1/2 at^2 + Vi t d= (1/2) (3.0m/s^2) (5.0s)^2 + (4.30m/s) (5.0s) d= 59 m

3) a=-1.5 m/s^2 t= 5.0 d=? Vi= 0 m/s Vf=?

Vf= Vi + at Vf= 0m/s + (-1.5m/s^2) (5.0s) Vf= -7.5 m/s

d= 1/2 at^2 +Vi t d= (1/2) (-1.5m/s^2) (5.0 s)^2 + (0 m/s) (5.0s) d= -18.75m

4) 2.5 s 31.25m


 * Final Velocity After Any Displacement**

2a) a= .8m/s^2 t= d=245 m Vi=7 m/s Vf= ?

Vf^2= Vi^2 + 2ad Vf^2= (7m/s)^2+ 2 (.8m/s^2) (245m) Vf^2= 441m/s Vf= 21m/s

2b) 15.78m/s 2c) 12.497m/s

3a) 15.9 m/s

4) d= 86.5m

5) a= 2.27m/s^2

6) d= 7.4m

Quiz Stopping Distance effected by Vi direction of acceleration equations of motion

__**10/25 HOMEWORK:**__ Physics To Go p. 88-89 #1-8 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. >
 * Larger initial speed creates an increase braking distance.

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?

Automobile A because at a certain point B and A cover the same distance but B covers it in a less amount of time. The less time a car travels, the greater speed it travels which makes it unsafe for braking distances.

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) 5 m

B) 60 mi/h? (twice 30 mi/h) 80 m

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?

speed=10m/s reaction=.9s v=d/t vt=d (10m/s) (.9s)=d 9m= dreaction

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 now know about stopping that will make you a safer driver? -How good your breaks are -The speed before breaking jfjeih __**11/26 Practice Conversion km/h to m/s**__ 100km/h x 1000m/1km x 1h/60min x 1min/60sec= 27.7m/s

25km/h x 1000m/1km x 1h/60min x 1min/60sec= 6.94m/s

220km/h x 1000m/1km x 1h/60min x 1min/60sec= 61.1m/s


 * __Total Stopping Distance (TSD)__**

Vi^2/2a

If Vi increases then TSD increases

If treaction increases then TSD increases

If a(really good brakes) increases then TSD decreases

__**11/28/11- Total Stopping Distance Activity-**__

2. There is a negative acceleration in the picture above because the car is slowing down to avoid the reindeer

__Calculating Reaction Distance-__ 1. no acceleration during the dreaction so: t (V) = d/t (t) Vt=d Vi treaction=dreaction

2. Vi treaction=dreaction (10)(1)=dreaction d= 10 m

3. Vi treaction= dreaction (20)(1)= dreaction d=20 m

4. The faster you are going the bigger the reaction time will be. The slower you go the less distance you'll go

__Calculating Braking Distance-__ 'a' is always negative making 'd' a positive 1. Vf^2= Vi^2 + 2ad 0= Vi^2+2a dbraking -(Vi)^2= 2a dbraking dbraking= -(Vi)^2/2a

5. a=-11 m/s^2 Vi= 51m/s Vf= 0 Db=?

Vf^2 = Vi^2 + 2aDb (0)^2= 51m/s^2 + 2(-11m/s) Db -(51m/s)^2/ -22= Db Db=118.23

6. a= -24m/s^2 Vi= 51m/s Vf=0 Db=?

Vf^2= Vi^2 + 2aDb 0^2= 51m/s^2 + 2(-24m/s^2) Db (-51m/s)^2/48m/s= Db Db= 54.19m

__Section 6__
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing WDYThink || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">+ 20 EC <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing #4 in 1st part of investigation || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing stop zone and go zone notes ||
 * <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;">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;">15/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">15/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/20

missing #1-4 for active physics plus || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing stop zone and go zone of a real intersection ||
 * <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;">**76*.75 = 57/75** ||

Red car stopped at a red light while the green car tried to run the light
 * What do you see?**

__**11/2/11 Create a problem-**__

Assume an old lady's reaction time is 1.6seconds. She is traveling down a one way street at 60m/s and her car's deceleration is -4.2m/s^2. What is her reaction distance, braking distance, and total stopping distance?

tr= 1.6sec Vi= 60 m/s a= -4.2m/s^2 dr= ?

V=dr/tr 60= dr/1.6 dr= 96

Vf^2= Vi^2 + 2ad 0= (60)^2 + 2(-4.2) db -(60)^2/ -8.4= db -3600/-8.4= db db= 428.57

T.S.D.= dr+db 96+428.57 T.S.D.= 524.57

__**Investigation 11/4 pg. 91 3&4-**__

3 a) Will automobile B be able to make it through during the yellow light? Yes car B is in the GO Zone. 3 b) Is automobile B in the GO Zone? Explain your answer? Yes automobile B is in the GO Zone because its right before the intersection. 3 c) Would any automobile closer to the intersection than automobile A be in the GO Zone Yes it would be in the GO Zone. 3 d) Is automobile C in the GO Zone? What might happen if automobile C decides to continue? No, automobile C is not in the GO Zone because if automobile C decided to continue then it wouldn't make the light, unless it sped up.


 * 11/7/11 Go Zone Prediction Sheet**

//Directions: List Expand, Shrink, or no effect in each box//
 * Variable |||| Change || Predicted shrink or expansion of GO ZONE ||
 * ty || yellow-light time || Increase ty || expand ||
 * ^  ||^   || Decrease ty || shrink ||
 * tr || reaction time || Increase tr || none ||
 * ^  ||^   || Decrease tr || none ||
 * v || speed limit || Increase v || expand ||
 * ^  ||^   || Decrease v || shrink ||
 * a || negative acceleration || Increase a || none ||
 * ^  ||^   || Decrease a || none ||
 * w || width of intersection || Increase w || shrink ||
 * ^  ||^   || Decrease w || expand ||

moves it back || moves it forward ||
 * Variable |||| Change || **Predicted** shrink or expansion of STOP ZONE ||
 * ty || yellow-light time || Increase ty || none ||
 * ^  ||^   || Decrease ty || none ||
 * tr || reaction time || Increase tr || expand/
 * ^  ||^   || Decrease tr || shrink/
 * 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 || none ||
 * ^  ||^   || Decrease w || none ||

Page 95 part B 1-4
 * __11/8/11 Yellow Light Dilemma-__**

1. a-no stop b- yes because its halfway threre c- yes d- no stop

2. e- no not safe f- no too late g- yes h- stop or go depending on the speed of the person in front

3. j- no not safe to go k- yes go zone l- no too late to go m- can't do either without unsafety

4. a. different speeds, different light times, different

__go zone-__ v=d/t v=d/ty v=(w + gz)/ty gz=vty-w <- the furthest distance at which you can go safely (smaller than original)
 * __11/8/11 Go Zone and Stop Zone Equations-__**

__stop zone-__ sz=TSD sz=dr+db sz= vtr + -(Vi)^2/ 2a sz= vtr + Vi^2/ 2a <- the closest position to the intersection in which you can brake safely (bigger number than original)

2) __5 variable changes__ i. tr decrease and ty increase ii. V decrease and W decrease ty increase iii. decrease A increase ty decrease width
 * __11/9/11__**

1. The go zone is greater than what was 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 = increase -tr = no effect -w = increase -Vi = increase -a = no effect 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 worn out brakes will have to brake first. -The driver with worn out brakes has a larger stopping distance. -The driver with worn out brakes has a stop zone that is pushed further back from the intersection. 4. The faster driver has a larger total stopping distance. The driver who's slower has a stop zone that is pushed further back from the intersection. 5. The sober driver has a larger stopping distance. The drunk driver has a stop zone which is pushed further back from the intersection. 6. Speed affects the stop zone more drastically than the go zone because it must be considered with the person's reaction time
 * __11/13/11 Active Physics Plus-__**

__Section 7__
__**11/14/11-**__ The car is tipping over the edge turning too fast.
 * <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** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**22.5/75** ||
 * What do you see?**

Theres a sign there because you can spin out if you are going too fast. How much you slow down depends on your current speed, the sharpness of the turn, and what size car you have.
 * What do you think?**

__Centripetal Acceleration__: a change in velocity caused not buy a change in speed but just by a change in direction Acceleration and force are unidirectional. they point in the same direction.
 * __11/16/11-__**