Chapter+1

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Chapter 1

Section 1
Grade: 100

__WDY SEE/THINK__ toc I see a flipped orange car with a yellow car driving over it. A blue car is driving toward the accident and trying to slow down. The road curves and slopes slightly. There is oil on the road coming out of the car. There is a bunny and grass next to the road. Mountainous background. Reaction time is affected by the noises, possibly bright lights or explosions of some sorts. Tire friction on the road, When the driver sees what is happening, How long it takes for him to comprehend the situation How hard he can hit the brake. Distractions such as scenery or animals. He might have the radio on. Decisions and distractions which way to swerve, whether or not to stop or avoid, surroundings

__INVESTIGATE__ gas to break- 2.82 seconds for 10 .282 seconds for 1

stopwatch reaction time- .37 seconds .26 seconds .33 seconds

Classes Fastest/Slowest
 * Ruler Reaction time || Red Green || Red Green With Texting ||
 * 30 cm / 0.25s || 16 cm / 0.18s || 14 cm / 0.16s ||
 * 20 cm / 0.2s || 18 cm / 0.19s || 10 cm / 0.14s ||
 * N/A || 13 cm / 0.16s || 19 cm / 0.19s ||
 * 15 cm / 0.17s || 3 cm / 0.8s || 12 cm / 0.15s ||
 * N/A || 14 cm / 0.16 || 10 cm / 0.14s ||
 * 21.6 cm / 0.21s || 12.6 cm 0.155 || 13.2 cm 0.16s ||

What factors may affect reaction time for people of the same age, like your classmates? Level of focus, natural reflexes, distractions, experience, personality.
 * Investigation Questions**

How does your reaction time with needing to make a decision compare with your reaction time without needing to make a decision? You react slower when making a decision.

What does this difference in reaction time when making a decision apply to your ability to avoid road hazards? You react slower when making a decision and will make it harder for you to avoid hazard.

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? You react slower when distracted.

What does this difference in reaction time when distracted apply to your ability to avoid road hazards while texting? Your reaction is slowed making it harder to avoid hazard.

__PHYSICS TALK Date? __ Checking Up Questions 1) How do distractions affect reaction time? Distractions delay your reactions.

2) Why is driving under the influence of alcohol or drugs illegal? Alcohol and drugs significantly slow your reaction time.

3) Name three factors in addition to distractions and drugs or alcohol that can affect reaction time. Age, Gender, Personality, and more also affect your reaction time. Title? 1) How long does it take a fastball to reach home plate? (The application tells you if you click “continue”) Half a second

2) Why if you reaction time is under .5 seconds do you still sometimes not hit the ball? Swing time

3) Explain what else might take some time before you a 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? You need to decide where to swing. swing time (distance/gas to break)

4) Can you really hit a 90 mph fastball? What factors are __not__ included in the program that decide whether you could hit the ball in real life? (Think back to the reaction time investigation) Distractions, aiming your swing, how fast you swing, where the ball is pitched

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? You want to stay focused.

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? It is a distraction to break your focus

__ ACTIVE PHYSICS PLUS 9/9__ d=1/2at 2 d: distance any object falls in a time "t" (s) t: time any object falls a distance "d" (cm) a: acceleration of any object falling (980cm/s^2)


 * Time || Distance ||
 * .025s || .31cm ||
 * .05s || 1.23cm ||
 * .075s || 2.76cm ||
 * .1s || 4.9cm ||
 * .125s || 7.66cm ||
 * .15s || 11.03cm ||
 * .175s || 15.01 ||
 * .2s || 19.6cm ||
 * .225s || 24.81cm ||

(my parents) 25cm, 23cm, 14cm, 24cm, 27cm.
 * Reaction Time Ruler 9/12**

__9/12 Do now/physics to go__

Although a race car driver is going much faster than someone driving in a school zone, the driver in a school zone needs a faster reaction time. The reason for this is in a school zone there are kids present and a driver may need to stop as soon as possible if necessary to safe a childs life. a race car driver needs to react to other drivers, a school zone driver needs to react to other drivers, parked cars, speed limit, children, parked cars.

a faster object will cover more distance in the same amount of time compared to a slower object.

Can you catch a dollar Bill? 9/12 Max distance dollar bill can drop: 15.55cm=d reaction time to catch it: t=0.178s

__Physics to go 9/14__

6. If a driver's reaction time is slow rather than quick the consequences can be a collision. 7. Even though teenagers often have good reaction times auto insurance is more expensive for teenagers because they don't have as much driving experience and tend to be unreliable behind the wheel.

__Reflecting on the section 9/14__ Rubbernecking and driver fatigue are the top two causes of accidents on the road. Rubbernecking is when you look off the road at something interesting such as a crash. It is a distraction.

__Active Physics Plus 9/14__ __distance vs time__



I agree with Steven and Zack because they too believe that it is more dangerous to drive while under the influence.

Section 2
Grade: 100
 * Measurement:**
 * Errors, Accuracy, and Precision**

__WDY See/Think__ Room numbers, tape measurer, person recording, girl walking, guy walking, stairs, dog or cat, person's head, people are walking along the tape measurer (possibly counting steps) notebook... They are doing an experiment for class to see the length of the boys steps compared to the girls steps. (walking speed)

If two students measure the length of an object and one gets 3m and the other gets 10m. one of them made an error because it was such a difference in distance. If one gets 3m and the other gets 3.01 meters it can be the same distance just how accurate they were.

pace it off 19 strides.

Measure your stride(cm) 64, 69, 70

Distance from line to line 13.165m

Strides = .69m

Class table
 * Group || Strides || Meter sticks ||
 * 1 || (19)(.74) = 14.06 m || 13.17 m ||
 * 2 || (22)(0.42) = 9.24 m || 12.28 m ||
 * 3 || (18)(0.5) = 9 m || 13.1 m ||
 * 4 || (20)(0.71) = 14.2 m || 13.12 m ||
 * 5 || (21)(0.64) = 13.44 m || 12.73 m ||
 * **6 (our group)** || (19)(0.69) = 13.11 m || 13.165 m ||

Questions 1) Do the measurements listed on your class table agree? For the strides they do not agree, For the meter sticks they agree with each other.

2) By how many meters do the results vary? All within 1 meter for meter sticks. 5 for strides

3) Why are there differences in the measurements made by different groups? List as many reasons as you can think of Height of person, Different stride length, different method, steady hands, rounding, focus, possible error. When you measure your stride, its not true that every stride will be that exact length.

4) Suggest a method of making the class’ measurements more precise. If all groups use your suggested method how will this reduce the range of measurements collected. Two meter sticks, two people measure, two people stride.

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? It will not be exact because people still make measuring error and calculation error and rounding error.

6) Can you develop a system that will produce measurements that would agree exactly or will there always be differences in measurements? Justify your answers There will always be a difference in answers because people see differently and think differently and work differently.

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) a) We did did have a systematic error. We started at the back of the tape and had to subtract half the distance of the tape at the end. b) The stride was random error.

8) If you did not have any systematic errors then name 3 ways a systematic error may have been introduced. It could have been introduced by the starting or ending point. The distance of a persons stride while thinking about the stride. difference in multiple strides vs one.

Calibrate- Measure- Identify- Evaluate-

Physics Talk Questions 1. A systematic error is a mistake in writing down data or reading the measurement can be corrected with calculation. Random errors are mistakes that cannot be fixed with calculation..

2. Not everyone is going to have the exact same measurement. Each individual person will have different rounding or methods on the measurement and therefore a different number.

3. The arrows would have to be all of the target not in a group to show a non-accurate and non precise shot.

Systematic error- an error produced by using the wrong tool or using the wrong tool incorrectly for measurement and can be corrected by calculations. 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 result.

__DO NOW 9/19__



__Demo- How long is the tube? / 4 sided ruler__ .8m .85m .832m .85m .84m .83m .8m || .85m .83m .82m .84m .83m .83m .82m .817m .81m .825m || .81m .812m .813m .809m .81m .82m .815m .82m .812m ||  || __Physic Talk Notes- SI system 9/19__ __Measuring a copper tube__ __Active Physics Plus 9/20__
 * Interval || 1 meter || 0.1 meter || 0.01 meter || 0.001 meter ||
 * || .8m
 * Quantity || Unit || Symbol ||
 * distance || meter || m ||
 * mass || grams || g ||
 * time || seconds || s ||
 * Groups || Measurements ||
 * 1 || 66m ||
 * 2 || 64m ||
 * 3 || 66m ||
 * 4 || 64.1m ||
 * 5 || 66m ||
 * 6 || 64m ||

1) Uncertainty __RANGE__ +/- 10 cm --> +/- 10(.01)m --> +/- .1m (49.9 - 50.1)m +/- 1cm --> +/-1(.01)m --> +/- .01m (49.99 - 50.01)m +/- 1mm --> +/- 1(.001)m --> +/- .001m (49.999 - 50.001)m

2) Faster s=d/t s=2m/s d=50m t=25s (2=.02/t) = (t=.02/2) ( t= .01seconds )

3) 30 laps .02m *30laps = .6m (1500.3 -- 1499.7) s=d/t = 1500m/15(60)s = 1.66m/s t=d/s t=.6/1.66 = .36seconds

pg.24 #9 Investigation: Estimation Activity a. True --> Most football players are around 220lb b. False -->No the tallest people ever are around 7ft, 13 is double that c. False --> It is impossible to work 24hrs a day d. False --> Poodles are not over 100lbs e. Trye --> No the room would have to be like 53x10x10 which is reasonable f. True --> When you bring in sports fields and other it is at least 1/2mile and that is reasonable for most schools g. False --> You would move 1.3 k/m in one minute h. Estimate how long it will take for the car to get over the bridge by estimating its speed and the distance.

9/21/11 - ABSENT

Section 3
Grade: 9/23 Average Speed: Following Distance and Models of Motion

__WDY See/Think__ Two cars going fast on one side but not too close together. Three car collision on the other side. back car going fast with dog in the car woman driver, middle care with blond woman driver with child in back, front car with woman driver looks sad. There is a rabbit on the side of the road. 2 cars length following distance (3 second rule) Give yourself time to be able to react to the car in front of you. 1 car for every 10mph you are both going.

__Investigation__ pg 34-36 1a. 2a-b. 2c 3a. The slowest is C while the fastest is A. The fastest car covers the most distance while the slowest coovers less. 3b. Yes because the distance between the cars is constant. 4a. walking towards the detector(above)

4b. walking away from the detector (above)

4.c walking away from detector slowly.(above)

4.d

Walking towards the detector at a fast speed.(above)

4.d con. walking away from detector at a fast speed (above)

4.e The person who walked faster would have a stepper incline rather than someone walk at average speed which would take long to get to the same point. Walking away and towards the detector would have different slopes because of the distance change. Walking slower than average would make the graph longer.

5a.

5.b test:

6.a

6b. You can tell the difference because you can compare the two lines. The one with the steeper incline is the fast walking speed than the red line.

7a. From our graph we have walked 3.3m in total. 7b. 2.5/3.0= .589m and .9/3.0= .3m 7c. 3.3m / 5.6sec= .58928 7d. You would multiply the answer by two, therefore you would cover twice the distance, in twice the time.

8a. 30 ft 8b. 90 ft 8c. 25 ft, then 75 ft 8d. 35ft, then 105ft 8e. 20ft 8f. 5 cars

Physics talk 9/26/11: 1. The average speed of the vehicle is different than the instantaneous speed because instantaneous speed is the speed you are going at that particular moment, while the average is all the speeds that the car has gone in average. 2. Velocity is the rate and direction of an object while the speed is the rate of change in a particular distance. 3. The time it takes to move the distance is represented on the graph. It is inclining to the top right corner of the graph which means that it is going farther away while he takes time to do it. **Constant Speed** 4. The longer it takes the person to react the more distance that you will take up and this makes your reaction distance longer. Longer time means longer distance needed. Created Checking Up Question: What does it mean when a graph shows a straight line moving the same distance of a long period of time?

__Do Now 9/27__ 90*.6= 54ft 90*1.5= 135ft.

Go cart test run 9/28

__Active Physics Plus 9/28 page 47__ V=80/3mi/hr (26.67 mi/hr) || 80 miles || 3hr || You go 20 more than you go 40.
 * Part || Distance || Time ||
 * part 1 20mi/hr || 40 miles || 2hr ||
 * part 2 40mi/hr || 40 miles || 1hr ||
 * Whole trip

100 mile strobe /\
 * Part || Distance || Time ||
 * 1 mi/hr || 50 miles || 50 hr ||
 * 50 mi/hr || 50 miles || 1 hr ||
 * 1.96 mi/hr || 100 miles || 51 hr ||

100 mile trip /\ 80 mile trip /\ 80 mile strobe /\

3a. Average speed is about 28.7 miles/hour 3b. Actual average speed is 18.75 miles/hour

4) If you are going 20 mi/hr for the first 40 miles. You must go 2400 mi/hr for the next 40 miles to get an average speed of 40mi/hr. (if you go 80 miles total). If you were to go a total of 120 miles and the first part you go 20 mi/hr for 40 miles. You would need to go the last 80 miles at a speed of 80 mi/hr to get an average speed of 40 mi/hr. 20 mi/hr for 3 hr. 100mi/hr for 1 hr = 160/4 = 40 mi/hr.

Physics to Go: 10/2/11 1a. In this strobe photo the vehicle was moving at the same constant speed for the total time. 1b. The vehicle started of at a slow speed and then moved up to a faster speed represented by the large space in between the diagram. Then slows back down.

2a-2b.

3. V=d/t 350=d/20 20(350) d= 7,000 ft.

4a. =47.77 4b. You will not be able to determine the speed while traveling through Baltimore because we are only calculating the average speed so it ranges around 47.7 but not the exact speed.

5. 5 miles=s/15min(or .25) s=5/.25 s=20 miles/hour

6a. This is showing at first the vehicle was moving fast and then came to a stop as time continued. 6b. This one is showing that at first the vehicle was moving really fast and then came to a stop and then went backwards to the original starting distance. 6c. The vehicle is moving ,at first, fast and then picks up the speed to be moving faster than originally. 6d. This graph shows a constant increase of speed at a gradual rate.

7a. (0.3)(25)= 7.5m 7b. (0.3)(16)= 4.8m 7c. (0.6)(25)= 15m

8a. They can calculate the time need to properly stop the vehicle a safe distance away from the vehicle in front of you. It is also making sure you have enough reaction time to even stop the car. 8b. Three seconds on the highway will not be helpful because you are going a lot faster on the highway or interstate. You would need more reaction time and space to find the problem and stop the vehicle.

9a. 100/(1/3)= 33.33 feet 9b. No 33.33 feet is not as long as my classroom.

10a. 88/.5= 176 feet 10b. 176/15= 11.73 or 12 car lengths 10c. 44/.5= 88 and then 88/15= 5.86 10d. 132/.5= 264 and then 264/15=17.6. This is about two and a half football fields in length. 10e. 30 miles/hours will change from 4 car lengths to 8, 60 miles/hours will change from 8 to 16, and 90 miles/hour will change from 12 to 24 car lengths. 11.

__Following Jack Part 1 10/3__

The drips show that jack started out walking slowly, then walked faster and continued to walk at that same speed because the dots were close together to start then spread out to a constant distance. Graph D is the correct graph because it starts slow and then hits a constant slope.

__Distance Vs Time Graphs Cyclists__ 3. a) Cyclist B starts ahead of cyclist A b) Cyclist A is ahead of B at 7 seconds because his line is higher on the distance axis. c) Cyclist A is going faster at 3s because he has a greater slope d) No they do not go the same velocity at any time because their slopes are never the same. e) Thee intersection means they are at the same distance at the same time.

4. a) Cyclist A goes the same motion in this graph. b) Cyclist B is going backwards slowly in this graph. c) Cyclist A has a greater speed because he has a greater slope d) At 5 seconds the intersection means they are at the same distance at the same time. e) cyclist A has traveled further in the first 5 seconds because his line shows that he covered more distance than cyclist B.

Section 4
What do you See/Think The red car is accelerating too fast and burning out. The traffic light is green. The guy and his dog are running away in fear. There is a garage but it doesnt look very functional. There is water which is a distraction and possibly a destination of interest for these people. There is a fire hydrant. The yellow car with her pet seems very calm.The red car driver lost his hat. The red car has a fast acceleration rate but friction is not giving enough energy for the car to move effectively. A car has a greater acceleration than a bus.
 * ** Section4 ** || **Points** ||
 * WDYSee/Think: || 10/10 ||
 * Investigate: || 20/20 ||
 * PhysicsTalk: || 20/20 ||
 * PhysicsPlus: || 20/20 ||
 * PhysicsToGo: || 20/20 ||
 * Wiki || 10/10 ||
 * **TOTAL POINTS** || **100** ||

__Learning Outcome__

Measure- a change in velocity of a cart on a ramp using a motion detector. Construct-graphs of 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.

__Tangent Line__ 10/5 Tangent line is a line that touches a curve at only one point. Slope of tangent line on a distance vs time graph: the instantaneous velocity at any point in time

Investigation:10/5/11



Pre-lab 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?

No they would move at different speeds because it is gradually accelerating. 2. Top left graph: In this graph, the car travels faster at the beginning and slows toward the end. Top right graph: In another, the car travels slower at the beginning and speeds up. Bottom right graph: In one of them the cart does not move Bottom left graph: In another the car moves at constant velocity

Run 1: The velocity increases as time goes on.

.5 seconds tangent

1.5 seconds tangent Velocity Graph above

Change in velocity/change in time so 1.1/1.9 =.5789 m/s^2

Run 3: Prediction (above)

0.5 tangent line from start



.7 tangent line from start



velocity of run #2

change of velocity/change in time 2.7/69 .0391m/s^2

Run 4: prediction (below)

0.5 seconds: (below)

1.5 seconds: (below)

Velocity Graph:



change of velocity/change of time .17/3.0 .5666m/s^2

__Physics Talk__ (This was not titled) 1. The equation of acceleration. ( A = ∆v/∆t ) 2. The SI unit for measuring acceleration = m/s^2. 3. A vector is a quantity that has both magnitude and direction. Scalar is a quantity that has only magnitude, but //not// direction. 4. Graphs: 5. Slope of the velocity-time graphs represent velocity. /\

__Words (This was not titled)__ acceleration: the change in velocity with respect to time. vector: a quantity that has both magnitude and direction. negative acceleration: a decrease in velocity with respect to time. positive acceleration: an increase in velocity with respect to time.

__Physics talk review do now 10/7__ 1) 30/5 = 6m/s^2

2) Pressure, force, and velocity are vectors. Speed, and calories are scalars. A vector has size and direction, scalar doesn't have direction.

fan Car drawings/\

[[image:slowing_down.jpg width="512" height="384"]]
slowing down/\

speeding up/\

Front /\

Back/\

__Active Physics Plus 10/12__

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? A=3 m/s^2. Vi= 7m/s Vf= 20m/s T=? a=(20-7)/(t) 3=(13)/t 3t=13 T=13/3seconds = 4.33 sec.

2) If a car accelerates from 7 m/s to with an acceleration of 1.5 m/s2 for 10 seconds what speed is it now going? Vi=7 m/s a=1.5 m/s^2 t=10 sec. Vf=? 1.5=(Vf-7)/10 15=Vf-7 Vf=22m/s

3) A car accelerates from rest at a stop light to a speed of 20 m/s in 5 sec. a) What is the car’s acceleration? a=20/5 a=4m/s^2 b) What distance has the car gone? d=(.5) (4)(25) d=50m

4) A car accelerates from rest at a stop light to a speed of 40 m/s in 10 sec. That is an acceleration of 4 m/s2. What distance has the car gone? d=(.5)(4)(100) d=200m

5) A car rolls up to a stop light with initial speed 3 m/s and accelerates to 23 m/s in 5 sec. a) What is the car’s acceleration? a= (23-3)/5 a=4m/s^2 b) What distance has the car gone? d=(.5)(4)(25)+(3)(5) d=65m

6) A car slows down coming to a red light from a speed of 25 m/s to 0 m/s in 4 sec. a) What is the car’s acceleration? a=(0-25)/4 a= -6.25m/s^2 b) Why is it negative? Slowing down c) How far did it take the car to stop? d=(.5)(-6.25)(16) + (25)(4) d= -50+100 = 50m

7) A car slows down coming to a red light from a speed of 50 m/s to 0 m/s in 8 sec. a) What is the car’s acceleration? a= (0-50)/8 a= -6.25m/s^2 b) Why is it negative? Slowing down c) How far did it take the car to stop? d=(.5)(-6.25)(64) + (50)(8) d= -200+400 = 200m d) When the speed was doubled how did the braking distance change. It was doubled.

8) A car accelerates at a rate of -7 m/s2 up to a deer in the road which is 40 m away in 3 seconds from a speed of 21 m/s to a stop. Will the car hit the deer? a=-7m/s^2 t=3 sec. Vi=21m/s Vf=0 d=? d=(.5)(-7)(9) + (21)(3) d= -31.5+63 d=31.5m The car will stop 8.5m from the deer.

9) Now the car is moving twice as fast, 42 m/s and slows down with the same acceleration of -7 m/s2. a) How much time will it take the car to come to a stop? -7=42/t t= 6s b) If the car is attempting to avoid a deer which is 80 m away will the car hit the deer? d=(.5)(-7)(36) + (42)(6) d= -126 +252 d=126m HE WILL HIT THE DEER.

10) A ferrari whizzes by a cop at a red light at constant velocity of 40 m/s (80 mph), clearly speeding. The cop tries to catch our red light running speeding friend who never notices the cop behind him so continues at simply 80 mph. a) If the cop car accelerates toward the speeder at a rate of 7 m/s2 how long will it take the cop to catch the speeder. D=VT .5at^2=40T b) How far will the cop have gone to catch the speeder? d=(.5)(7)(32.65) d=114.274m

11) A mysterious biker whizzes by a cop at a red light at constant velocity of 50 m/s(100 mph), clearly speeding. The cop tries to catch our red light running speeding friend who never notices the cop behind him so continues at simply 100 mph. a) If the cop car accelerates toward the speeder at a rate of 6 m/s2 how long will it take the cop to catch the speeder. 6=(50--0)/t t=8.333seconds b) How far will the cop have gone to catch the speeder? d=(.5)(69.44)(6) d=208.33m

Physics to go 1) This can not happened because in order for acceleration to happen you need speed or velocity to go along with it.

2) This is not possible because in order for there to be acceleration there needs to be a force of speed/ velocity behind it.

3) They may not have the same velocity because one car might be heavier than the other, and it depends in what direction becasue one car might be going up a hill be enacted upon by gravity more.

4) They might not have the same acceleration becasue one car can take longer time getting up to the speed rather than quickly.

5) Yes it can be overtaken because it takes a few seconds for the vehicle to get up to that certain speed and the car moving at the constant velocity already will take no time since they are already at the speed.

6) They are different than each other, so it would be wrong to refer to them as the same thing. The speed limit is miles per hour while velocity is purely on the speed. the speed limit signs in the United States are in Miles/ Hour.

8)A. a=∆v/∆t a= (75m/s-0m/s)/ (9s-0s) a= 8.33 m/seconds^2 B. 37.5 m/s C. d= (1/2)x a x t^2 d= (1/2) x 8.33 x9^2 d= 337.365 meters D. a= 9.375 seconds speed= 37.5 m/s d= 300 meters

9)A. a= ∆v/∆t a= -3 m/seconds^2 B. d of slide= 3.43 meters C. 4.03 seconds D. if she slide for 1.1 seconds she would arrive slightly quicker.

10) A. 11 m/s B. a=∆v/∆t a=(11 m/s- 0m/s)/ (9.0-0) a= 1.22 seconds C. The objects velcoity would be a lot higher and greater since it has more space and time to speed up before hitting the ground.

11) A. Graph F would represent it the best. B. Graph D. C. Graph B. D. Graph A. E. Graph F F. Graph C.

12) A. c B. b C. d D. e E. A little bit over 1,000 meters. F. Back at the start of the track since it has no distance and no time.

13)A. a= ∆v/∆t a= (250-0)/(30-0) a=8.33 m/s B. 374.95 m/s ? C. It would take 60 seconds or 1 minute for it to reach 500 m/s D. d= (1/2) x a x t^2 d= 91/20 x 8.33 x 60^2 d= 14994 meters

14) A. 9.8=(Vf-0)/4.5 Vf = 48.02m/s B. a=9.8m/s^2 Vi=0m/s 100=(.5)(9.8)(t^2) 100=4.9(t^2) t^2=20.4 t=4.5seconds C. t=10 seconds Vf=? m/s Vi=0m/s a=9.8m/s^2 9.8=Vf/10 Vf=98m/s D. d=(.5)(98)(10) d= 490m

15) A. a=∆v/∆t a= (282.52928-0)/(0-1.4) a= -201.806 m/s^2 B. -20.59 g's C. d=(1/2) x a x t^2+ Vi x t d= (1/2) x -201.806 x 1.4^2 + 282.52928 x 1.4 d= 197.77 meters

16) A. d=(1/2) x a x t^2+ Vi x t d= (1/2) x 4.0 x 1^2+ 4.0 x 1 d= 6 meters B. (1/2) x 4.0 x 2^2+ 4.0 x 2 d= 16 meters

C. d= 30 meters D. 48 meters E. F. G. They are relatively similar to each other. It takes a while for the vehicles to get up to speed but and the distance increases as the speed increases. Velocity is constant since the distance is gradually increasing.

Quiz Friday. Instantaneous velocity (slope of tangent) Acceleration distance

Section 5
__NEGATIVE ACCELERATION: BREAKING YOUR AUTOMOBILE__ missing derivations of equations of motion || __Learning Outcomes 10/18__
 * <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;">/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** ||
 * Plan** and carry out an experiment to relate breaking distance to initial speed.
 * Determine** breaking distance.
 * Examine** accelerated motion.

__WDY SEE/Think 10/18__
 * SEE -** A car is going down a road and hits a moose. The driver failed to slow down in time. The moose looks like he wants to get hit. Back tires are up so he is clearly trying to stop.
 * Think** - Speed ( velocity ) distance to the moose, honking at the moose, reaction time, under the influence, distracted, swerve? Decisions, weather,

__**Investigation**__ __Objective:__ determine the effect of initial speed on braking (stopping) distance.


 * Vi (m/s) || Breaking Distance (ft) ||
 * 0.55 || 25 ||
 * 0.68 || 27 ||
 * 0.69 || 30 ||
 * 0.66 || 29 ||
 * 0.77 || 26 ||
 * .53 || 26 ||

__1st period Data__
 * Vi (m/s) || Breaking Distance (m) ||
 * .59 || 2.8 ||
 * 1.27 || 11.32 ||
 * .94 || 5.15 ||
 * 1.06 || 6.5 ||
 * .8 || 4 ||
 * .97 || 5 ||
 * 1.23 || 11 ||



(1.23/.59)= 2.08

(11/2.8)= 3.92

(Vi(2)/Vi(1))= Breaking distance(2) / Breaking distance(1) (x2)^2 = x4 (x3)^2 = x9 (x4)^2 = x16 (x10)^2 = x100

__Physics Talk pg 78-82 10/19/11__ __Checking up questions__ 1. If a vehicle traveling at constant velocity comes to a sudden stop, it has positive acceleration. It is not decreasing or increasing speed, its coming to a complete stop, which means it has positive acceleration when it stops. 2. I know that increasing the velocity of an automobile will increase its braking distance because the more speed an automobile has, the longer (time) it will take for the braking mechanism to kick in, and eventually stop the automobile. 3. The term negative acceleration is used rather than deceleration because negative acceleration can apply to an object increasing speed in a negative direction OR an object decreasing speed in a positive direction.

__Physics Words:__
 * __negative acceleration:__** a change in the velocity with respect to the time of an object by decreasing speed in the positive direction or increasing speed in the negative direction.

__Honda Civic Stopping Distance 10/20/11__ (60mph/30mph)^2 = (123ft/BD) 4=123/BD 123/4 = 30.75ft
 * Initial Velocity || Stopping Distance ||
 * 10 mph || 3.42 ft ||
 * 20 mph || 13.67 ft ||
 * 30 mph || 30.75 ft ||
 * 40 mph || 54.67 ft ||
 * 50 mph || 85.42 ft ||
 * 60 mph || 123 feet ||
 * 70 mph || 167.42 ft ||
 * 80 mph || 218.66 ft ||
 * 90 mph || 276.75 ft ||
 * 100 mph || 341.67 ft ||

4) If Initial Velocity is doubled how does stopping distance change? x4 5) If the Initial Velocity is multiplied 4 times how does the stopping distance change? x16 6) If the Initial Velocity is halved how does the Stopping Distance Change? 1/4 7) If the initial Velocity is quartered how does the stopping distance change? 1/16 8) What speed would you need to have a stopping distance of a mile? (60/x)^2=(123/5280) 123x^2=5280(60^2) x^2=15436.5854 x= 393.11 mph

__Active Physics Plus 10/24__ __equations:__ d=.5(Vi+Vf)t Vf=Vi+at d=Vit + .5at^2 Vf^2=Vi^2 + 2ad

Average Acceleration 1. t=?s Vi=9 m/s Vf=0 m/s a= -4.1m/s^s 0= 9+(-4.1)t -9= -4.1t -9/-4.1=t t= 2.20s

2. a= 2.5m/s^2 Vi= 7.0m/s Vf= 12.0m/s t=? 12= (7)+(2.5)t 5= 2.5t 5/2.5=t t=2s

3. a= -0.5m/s^2 Vi=13.5m/s Vf=0m/s t=? 0=13.5 +(-.5)t -13.5= -0.5t 13.5/.5= t t=27s

4. Vi= -1.2m/s t= 1500s Vf= -6.5m/s a=? -6.5=-1.2+a(1500) -5.3= 1500a -5.3/1500=a a=.004m/s^2

5. a=4.7x10^-3 m/s^2 a) t=300s (Vf-Vi)=at 4.7x10^-3(300) ∆V= 1.41m/s b) Vf-1.7=1.41 Vf= 3.11m/s

Displacement with constant uniform acceleration 1. Vi=0 Vf=23.7 km/h = 23,700m/h = 6.6m/s t=6.5s d=? d=.5(6.6)(6.5) d=21.45m

2. Vi=15m/s Vf= 0 t=2.5s d=? d=.5(15)(2.5) d=18.75m

3. Vi=100m/s Vf=0 a= -5 d=? 0^2=100^2 +2(-5)d 10d=10000 d=1000m NO(800m)

4. Vi=21.67m/s Vf=0 t=? d=99m 99=.5(21.67)(t) t=9s

5. Vi=6.4m/s d=3200m t=210s Vf=? 3200=.5(6.4+Vf)(210) 6400=210(6.4+Vf) 30.48=6.4+Vf Vf=24.08m/s

Velocity and displacement with uniform acceleration 1. Vi= 6.6m/s a=.92m/s^2 t=3.6s Vf=? Vf=6.6+.92(3.6) Vf=9.912m/s

2. Vi=4.3m/s a=3m/s^2 t=5s Vf=? Vf=4.3+(3)(5) Vf=19.3m/s

3. Vi=0 t=5s a=-1.5m/s^3 Vf=? d=? Vf=(-1.5)(5) Vf=-7.5m/s d=.5(-7.5)(5) d=-18.75m

4. Vi=15m/s a=-2m/s^2 t=? Vf=10m/s d=? 10=15+(-2)t -5/-2 t=2.5s d=15(2.5) + .5(-2)(6.25) d=31.25

Final Velocity after any displacement 1. No info

2. Vi=7m/s a=.8m/s^2 d=245m a) Vf=49 + 2(.8)(245) Vf=441m/s b) Vf=49 + 2(.8)(125) Vf=249m/s c) Vf=49 + 2(.8)(67) Vf=156.2m/s

3. Vi=0 a=2.3m/s^2 a) d=55m Vf=2(2.3)(55) Vf=253m/s b) 55=.5(253)t t=.43s

4. a=.85m/s^2 Vi=23.06m/s Vf=26.11m/s 26.11=23.06+.85t t=3.59s 26.11=531.76 + (2).85d d=-297.44

5. d=240m Vi=0 Vf=120km/h = 33.33m/s a=? 33.33=0+480a d= 2.31m/s^2

6. Vi=6.5m/s Vf=1.5m/s a= -2.7m/s^2 d=? 2.25=42.25 + 2(-2.7)d -40= -5.4d d= 7.41m

__Physics To Go 10/26__ 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. 2.
 * Initial Speed || Breaking Dist. ||
 * 5 m/s || 4 m ||
 * 10 m/s || 15 m ||
 * 15 m/s || 35 m ||
 * 20 m/s || 62 m ||
 * 25 m/s || 98 m ||
 * 30 m/s || 140 m ||

Automobile A is safer because it clearly has a smaller braking distance. Safer means the car that can stop the quickest. Regardless of how fast the car is going initially, braking distance determines how "safe" the automobile actually is.

3. a. 20m/30mi/h = x/15mi/h x=10m. If a car is able to stop in 20 meters at a speed of 30 miles per hour, then the car would be able to stop 10 meters while going at 15mi/hr.

b. 20m/30mi/h = x/60mi/h x=40meters If a car is able to stop in 20 meters at a speed of 30 miles per hour, then the car would be able to stop in 40 meters while going 60mi/hr.

c.20m/30mi/h = x/45mi/h x=30m. If a car is able to stop in 20 meters at a speed of 30 miles per hour, then the car would be able to stop in 30 meters while going 45 mi/hr.

d. 20m/30mi/hr = x/75mi/hr x=50m. If a car is able to stop in 20 meters at a speed of 30 miles per hour, then the car would be able to stop in 50 meters while going 75 mi/hr.

4. 3s/30m = 0.9s/ x x=9 meters. If an automobile going 10m/s stops in 30 meters, and the driver has a reaction time of 0.9 seconds, then the total distance traveled during this reaction time is 9m. The total stopping distance is 39 meters.

Total Stopping Distance (TSD) TSD=Dreaction+ Dbraking TSD=Vtr+Vi^2/2a Dreaction V=Dreaction/Treaction V(treaction) = d reaction
 * if Vi -> || then TSD -> || Vi - TSD ||
 * if tr -> || then TSD -> || tr - TSD ||
 * if a -> || then TSD <- || TSD - 1/a ||

__Mini Challenge Plan__ 1) paragraph explaining theme/storyline. 2) 3 ways you will use equations. 3) 3 ways you will use graphs.

Story line - A Zamboni is on the ice when it shouldn't be because the guy was trying to leave work early. There was a skater still on the ice and the Zamboni driver had to slow down before it hits him.

3 ways you will use equations- Will he hit the person. (distance equation) The zamboni is racing a skater. (acceleration equation) How long will the Zamboni take to get across the entire rink (velocity?)

3 ways you will use graphs- Graph to compare skater and Zamboni speeds/accelerations. Graph to show how longit takes to go a distance. Graph to show how ice effects slowing down

__Total Stopping Distance 10/28__ 1)

2) (-). braking.

__Calculating Reaction Distance:__ 1) V=d/t tv=d

2)d=(t)(v) d=10m

3) d=(t)(v) d= 20m

4) An increase in speed makes you go farther while reacting. A decrease in speed makes you take shorter to react. It is a long time. If you go 20m/s and your reaction time is 1 whole second, you go a whole 20 meters.

__Calculating Braking Distance:__ 1) Vf^2=Vi^2 + 2ad d=-(Vi^2 ) / (2a)

2) d=-100/-10 d=10m

3) d= - 400/-10 d=40m

4) When speed increases so does the distance. It increases exponentially. doubled = quadrupled.

5) d= -2601/-22 d= 118.23m

6)d= -2601/-48 d=54.19m

7) Brakes change acceleration by making it a more negative number. It will change stopping distance by lowering the distance.

__Mini Challenge Plan__ Story line - A Zamboni is on the ice when it shouldn't be because the guy was trying to leave work early. There was a skater still on the ice and the Zamboni driver had to slow down before it hits him.

3 ways you will use equations- Will he hit the person. (stopping distance equation) The zamboni is racing a skater, who will get to a higher speed first. (acceleration equation) How long will the Zamboni take to get across the entire rink (distance)

3 ways you will use graphs- Graph to compare skater and Zamboni speeds/accelerations. Graph to show how long it takes to go a distance. Graph to show how ice effects slowing down

Section 6
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">+16 EC || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing diagrams || ABSENT 11/2 and 11/3, **get the makeup work/notes** __WDY SEE/ THINK__ <span style="font-family: 'Lucida Sans Unicode','Lucida Grande',sans-serif;">- One car is speeding up trying to beat the yellow light <span style="font-family: 'Lucida Sans Unicode','Lucida Grande',sans-serif;">- Another car is try to stop on the line so that it doesn't blow the red light. <span style="font-family: 'Lucida Sans Unicode','Lucida Grande',sans-serif;">- Red car is braking and up in the air <span style="font-family: 'Lucida Sans Unicode','Lucida Grande',sans-serif;">- Red car might have reacted late <span style="font-family: 'Lucida Sans Unicode','Lucida Grande',sans-serif;">- Red car might be speeding <span style="font-family: 'Lucida Sans Unicode','Lucida Grande',sans-serif;">- Affected braking distance
 * <span style="font-family: 'Courier New',Courier,monospace;">** Section6 ** || <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
 * <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;">/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;">**111*.75 = 83.25/75** ||

<span style="font-family: 'Lucida Sans Unicode','Lucida Grande',sans-serif;">What is wrong with the intersection?

<span style="font-family: 'Lucida Sans Unicode','Lucida Grande',sans-serif;">- Red car was speeding and might not have programed the yellow light to be very short. While the green car is speeding up to get through the red light.





pg. 91 3, 4 3a.) Yes, automobile B will be able to make it through the intersection 3b.) Automobile B is in the Go Zone because it is closer to the interesection than automobile A 3c.) Yes, any automobile closer, would be able to into the intersection 3d.) Automobile may not make it through the intersection, if it tried to do so, it may run a red light, get a ticket, or cause an accident 4a.) Yes, it is in a stop zone, because it is farther behind than automobile A 4b.) No, automobile F is in the go zone, as it is close to the interesection
 * __Investigation 11/4__**

Go Zone predictions Actual: expand, shrink no effect, no effect expand, shrink no effect, no effect shrink, expand Stop Zone predictions actual: no effect, no effect back, forward back, forward forward, back no effect, no effect
 * Input || Output ||
 * Vi (speed limit) || Go Zone ||
 * t(y) (yellow light time) || Stop Zone ||
 * w (width of intersection) ||  ||
 * a (negative acceleration) ||  ||
 * t(r) (reaction time) ||  ||

__Part B: Yellow-Light Dilemma page 95 #1-4 11/8/11__ 1) a - stop b - go c - go d - stop

2) e - stop f - stop g - go h - go

3) j - stop k - go l -stop m - go

4) a - The space between go and stop zones is different. b - You have the choice to go or to stop it would be safest to stop but you can go if you think you can. c - You have the choice to go or to stop it would be safest to stop but you can go if you think you can. d - II has an overlap zone and III has a dilemma zone.

__Go zone and Stop zone Equations__

Go Zone: v=d/t v= d/t(y) v= (w + GZ) / t(y) GZ = (V)(Ty) - (W)

Stop Zone = SZ= TSD (dr + db) SZ= (Vi)(Tr) - (Vi)^2/(2a)

__Checking up questions__ 1. It is model because it uses mathematics to symbolize the relationship between each of the variables. 2. Go zone is the distance where it is safest to go through the intersection instead of stopping. 3. Stop Zone is the area where it is safest to stop before the intersection instead of trying to beat the light. 4. Overlap zone is the zone where the go and stop zone overlap giving you a choice where both are safest. 5. Dilemma zone is the area that the driver must make a decision of whether to go or not where neither choice is clearly safest.

__ACTIVE PHYSICS PLUS HW__ 1. Since the intersection is wider, the go zone became smaller 2. ty, vi could have been increased to make go zone bigger 3. The driver with worse brakes will have to brake first and will have a larger TSD. The driver with good brakes has his go zone pushed farther back 4. The one moving faster has larger TSD. THe other driver's stop zone is pushed farther back 5. The drunk one has larger TSD because he will have a slower tr. His stop zone is pushed farther back

__Active Physics Plus 11/9/11__ 1) w= 25m Ty= 3 sec Vi = 20 m/s a = -5 m/s^2 Tr = .7 sec

GZ = (20)(3) - (25) GZ= 35m

SZ= (20)(.7) - (20)^2)/2(-5) SZ= 54m

2) A) w= 10m Ty= 4 sec Vi = 15 m/s a = -5 m/s^2 Tr = .6 sec

GZ= (15)(.6) - (10) GZ= 50m

SZ = (15)(.6) - (15)^2)/2(-5) SZ= 31.5m Overlap = 18.5

B) GZ = (15)(5)-(14) GZ = 61m
 * ty || 5s ||
 * tr || 0.6s ||
 * v || 15m/s ||
 * a || -5m/s^2 ||
 * w || 14m ||

SZ = (15)(.6) - 15^2/2(-5) SZ= 31.5 Overlap = 29.5

C) GZ = (20)(5)-(20) GZ = 80m
 * ty || 5s ||
 * tr || 0.6s ||
 * v || 20m/s ||
 * a || -5m/s^2 ||
 * w || 20m ||

SZ = (20)(.6) - 20^2/2(-5) SZ= 52 Overlap = 28

3) GZ = (17)(5)-(20) GZ = 65m
 * ty || 5s ||
 * tr || 0.6s ||
 * v || 17m/s ||
 * a || -6 m/s^2 ||
 * w || 20m ||

SZ = (17)(.6) - 17^2/2(-6) SZ= 34.3 Overlap = 30.7

4) GZ = (4)(2)-(2) GZ = 6m
 * ty || 2s ||
 * tr || 0.1s ||
 * v || 4 m/s ||
 * a || -4m/s^2 ||
 * w || 2m ||

SZ = (4)(.1) - 4^2/2(-4) SZ= 2.4m Overlap = 3.6m

__Piermont Ave. and Kinderkamack__ width of intersection: about 20.2 m yellow light time: 3 seconds speed limit: 8.94 m/s reaction time: .5 seconds negative acceleration: -4 m/s^2
 * **Go Zone and Stop Zone of a Real Intersection (11/10)**

GZ: vty - w SZ: Vi(tr) - (Vi)^2/2a

GZ = (8.94)(3) - (20.2) GZ = 6.62 meters long SZ = (8.94)(.5) - (8.94^2)/2(-4) SZ = 14.46 meters long

__Calculations for double speed limit:__ GZ = (17.88)(3) - (20.2) GZ = 33.44 meters long SZ = (17.88)(.5) - (17.88)^2/2(-4) SZ = 48.90 meters long If you double the speed limit, the intersection's dilemma zone is much greater therefore, more dangerous. A ticket for traveling at double the speed limit is such a large fine because creates a lot more danger for everyone than just if you were barely exceeding the speed limit. With a greater dilemma zone, there is a higher chance of accidents.The stop or go zone changes more when the speed is doubled because the stop and go zones are quadrupled. || z ||

Section 7
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing centripetal force notes || __WDY See/Think__
 * <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;">0/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/20
 * <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;">**56.25/75** ||

__Learning outcomes__ In this section, you will • **Recognize** the need for a centripetal force when rounding a curve. • **Predict** the effect of an inadequate centripetal force. • **Relate** speed to centripetal force.

__Physics Words__ Centripetal Force --> A force directed to the center to keep the object moving in a circle Centripetal Acceleration --> A change in the direction of the velocity in respect to time

__Checking Up__ 1. The direction is towards the center of the circle 2.Centripetal force keeps objects moving in a circle 3. Friction keeps an car turning on a road, but it is n relation to centripetal force 4. Velocity can change if it changes in direction rather speed 5. Three situations are speeding up, slowing down, and turning 6. The force is centripetal force being pulled in by the sun.

Centripetal acceleration: Change in the direction of an object's velocity but not a change in its speed.

Centripetal Acceleration and Centripetal Force are unidirectional. "towards the center"

__Physics To Go__ 1. t=24h r=6400km s=c/t c=2pir c=2pi6400 c=40212.39 km s=40212.39/24
 * s = 1675.52 km/hr** (1hr/3600s)(1000m/1km) **= 465.42 m/s**

2. r=1.5x10^8 km, s=? c=2pir = 9424777961 km
 * s = 4753 m/s = 1329.3 km/h**

3. s = 7200pi, r = 15cm s=v^2/t
 * s = 56.5 m/s**

4. a) if the curve is tighter, then the car will spin off b) if the road surface becomes slippery, the car will spin off and go straight because friction is no longer keeping it in control c) if the road is slippery AND the curve is tighter, the the car will spin off because there is little friction 5. A baseball bat travels in a circular path when being swung, and the muscles of the batter produce the force that keeps the object towards the center of the curve. 7. "The driver may turn the wheels but it is the road that turns the automobile." This means that although the driver of a car chooses either to turn left or right through the steering wheel, without the friction of the road that is a centripetal force, the car would not be able to turn either way at all. 8. s=270m/s, r=1000m, a=? a = (270^2)/1000 9. Both explanations are correct because they are describing the same thing how since there was not enough friction between them and the seat, they kept going in a straight line because nothing was holding them back. The car including the seat did turn though because of friction from the road. 10. The force is friction. 11. These types of turns that get tighter and tighter are especially dangerous because you do not always see it coming, and in the beginning when it is gentle you may be going faster, but when it gets tighter you don't have enough time to significantly decrease your speed, and you could end up driving off the road. 12. If you are curving to the right, you will end up on the other side of the road because the car would want to keep moving straight. If you are curving to the left, you would end up on the right side.
 * a = 72.9 m/s^2**

__Active Physics Plus__ __1-4__ F = m* v^2 / r (N) = (kg)*(m/s)^2/ (m) N: newton; unit of force 1) r=10 m= 2000 N= 13720 v=8.3? 13720= 2000x^2/10 x^2=68.51 V=8.3m/s

2) 6860=2000x^2/10 x^2=34.3 x=5.9m/s

3) V=10 M=3000 N=20580 r=? 20580=3000(100)/r r=14.6m

4) V=5 M=2200 N=6000 r=? 6000=2200(25)/r r=9.2m

__5-7__ a=v^2/r

5) a=? v=10 r=12 a=(100)/12 a=8.33m/s^2

6) a=? v=20 r=12 a=(400)/12 a=33.33m/s^2

7) a=? v=10 r=24 a=(100)/24 a=4.2m/s^2