Chapter+1

//__CHAPTER 1__//

__//**SECTION 1-Reaction Time**//__
__//**Grade:100**//__ toc //**Do Now:**// 9/7: What Do You See/What Do You Think? There are two cars that have obviously crashed. There is a speeding car trying to slow down. The person is trying to slam on the breaks. What factors affect the time you need to react to an emergency situation while driving? The person is trying to slam on the breaks because they were speeding but yet realized theres an accident ahead. The cad smoke coming from the tires probably because the car lost its' transition from slamming on the breaks. He is slamming on his breaks because the back wheels are slightly off the ground. If a person is speeding the person's reaction time is shorter than a person not speeding. The size of the car is also a factor because smaller cars slow down faster. The age of the tires and the direction of where the car is going, uphill or downhill.

//**Investigation:**// 9/7: Investigation Reaction Time: Responding to Road Hazards

Starting and Stopping Stop Watches: Average: .173 seconds
 * .20sec ||
 * .02sec ||
 * .19sec ||
 * .30sec ||
 * .05sec ||
 * .26sec ||
 * .17sec ||
 * .20sec ||
 * .17sec ||
 * .17sec ||

Time From Gas to Break: Average: .527 seconds
 * 1.09sec ||
 * .87sec ||
 * .67sec ||
 * .54sec ||
 * .53sec ||
 * .42sec ||
 * .53sec ||
 * .62sec ||
 * .57sec ||
 * .38sec ||

Dropping and Catching Ruler: Distance Time Average Time- .18 seconds Average Distance- 20.2 cm
 * 17cm - .17sec ||
 * 22 cm- .22sec ||
 * 21cm - .21sec ||
 * 13cm - .16sec ||
 * 18 cm- .18sec ||
 * 23 cm- .23sec ||
 * 24 cm- .23sec ||
 * 27cm - .24sec ||
 * 24cm - .23sec ||
 * 13cm - .16sec ||

Reaction time with Decisions: Average: .268 seconds
 * 28cm - .25sec ||
 * 20cm - .20sec ||
 * 28cm - .25sec ||
 * 22cm - .22sec ||
 * 17cm - .17sec ||

Reaction Time with Distractions and Descisions Average: .19 seconds
 * 24cm - .23sec ||
 * 19 cm- .19sec ||
 * 16cm - .17sec ||
 * 19cm - .19sec ||
 * 15cm - .17sec ||

1. Record the class' Fastest, Slowest, and Average Reaction Time. The class' fastest time is .12 seconds, the class's slowest time is. 527 seconds, and the average time is .279 seconds. 2. What factors may affect reaction time for people of the same age, like your clasmates? Factors could include fatigue and texting while driving.
 * Questions:**

2. How does your reaction time with needing to make a decision compare with your reaction time without needing to make a decision? The reaction time needing to make a decision takes a longer time to react compared to the reaction time without needing to make a decision. The average reaction time was 10.6 and without making a decision, the average was 13.3. 3. What do es this difference in reaction time when making a decision apply to your ability to avoid road hazards? Well if there are road hazards and you are making a decision, the reactions will be slower because if you are making a decision when a road hazard occurs, the decision would have to be quick since you deciding what to do about the road hazard.

1. 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? My reaction time would take a longer time to react while texting because yet you are making a decision while texting. This would happen because you are distracted and on concentrating on something else other than the road. 2) What does this difference in reaction time when distracted apply to your ability to avoid road hazards while texting? The difference in the reaction time when distracted applies to my ability to avoid road hazards while texting. The timing is slower and can cause careless problems because you are doing multiple things at once while the person should be focusing on the road.

//**Checking Up Questions**// 1. How do distractions affect reaction time? Distractions affect the reaction time by slowing it down and making it longer to react to an emergency. 2. Why is driving under the influence of alcohol or drugs illegal? Driving under the influence of alcohol or drugs are illegal because by being under the influence and drugs all slow down the reaction time. You are being distracted and your mind is in another place and you are not yourself. 3. Name three factors in addition to distractions and drugs or alcohol that can affect reaction time Age, practice, and fatigue are other factors in addition to distractions, drugs, and alcohol all can affect the reaction time of a driver.

//** 9/9 Do Now: Reaction TIme: Science of the Fastball **//

1. It takes a 90-mile per hour fastball about half a second to reach home plate. 2. If your reaction time is under a half a seconds you sometimes miss the ball because you could swing and miss the ball if you swing to early. 3. Deciding whether to swing at the ball or not. Besides reacting to an obstacle is the reaction time to your brain down your nervous system down to your foot. Another example is what kind of pitch the pitcher is throwing and have time to react. 4. Yes, a batter can actually hit a 90 mph fastball and I think I could. A batter just needs to have a fast reaction time. Factors of screaming fans, the score, whether its another kind of pitch, and the other players on the field. 5. To reduce reaction time while driving, drivers could slow down, drivers can hover their foot over the break to reduce your reaction time. 6. People would yell comments from the bench when the batter is trying to hit because it distracts the batter and his concentration. This is similar to texting and driving because texting distracts your concentration just like a player from the other team is distracting the batter at the plate.

//** Active Physics Plus 9/9 **// d=1/2at^2 d:distance an object is dropped t: time it takes an object to drop "d" (seconds) a: acceleration of the object (speeds up towards ground) 980cm/s^2 d=1/2at^2 cm=1/2(cm/ seconds )( seconds ^2) (units) .1=doubles .2=times four .3=times 9 .5=times 25
 * Distance || Time ||
 * 4.9cm || .1s ||
 * 19.6cm || .2s ||

//**Reaction Time Ruler**// //**d=1/2at^2**//
 * Distance || Time ||
 * .3 cm || .025sec ||
 * 1.2cm || .05sec ||
 * 2.73cm || .075sec ||
 * 4.9cm || .1sec ||
 * 7.66cm || .125sec ||
 * 11.03cm || .15sec ||
 * 15.01cm || .175sec ||
 * 19.6cm || .2sec ||
 * 24.81cm || .225sec ||


 * //Do Now 9/12//**

A race car driver needs a faster reaction time than someone driving in a school zone. A race car driver is going faster and if there is a danger occurring, it takes longer for the car to stop. They also cover more of a distance in the same amount of time as a person driving in a school zone. A race car driver has many dangers such as car crashes, car malfunctions, and the car is going 200 miles per hour. Someone driving in a school zone also has many dangers as well such as, children running into the street, stop signs, speed limit to follow, and other drivers doing crazy speeds.


 * //Reaction Time Ruler Notes 9/12//**



15.4cm-Dollar Bill Distance Reaction time needed to catch the dollar bill: d=1/2at^2 15.4cm=1/2(980cm/s^2)**t^2** 980/2=490cm/s^2 490cm/s^2 (15.4cm)=t^2 15.4cm/490=.0314s^2 square root(.0314s^2) 0.177s=t^2
 * //Can You Catch a Dollar Bill?//**

Parent Average: 0.64 sec (8.508cm)
 * //HOMEWORK 9/12//**

//**PHYSICS TO GO 9/14**// 6,7 6. There could be many consequences of driving if one's reaction time is slow rather than quick. If ones reaction time is slower than you could cause accidents or crash into objects or people. It would take longer for a person to react to the obstacle. If a person has a slow reaction time, the distance is greater than a person who has a faster reaction time. 7. Teenagers may have good reaction times but the insurance is more expensive because teenagers are new to the road and driving. Teenagers are most likely more distracted on the road whether its music or talking on their cell phones or driving under the influence.

//**Reflecting on the Section and the Challenge 9/14**// 1. The top two causes of accidents on the roads are rubbernecking and people driving while tired. 2. Rubbernecking is when there is an accident on the side of the road that is unusual and are wondering what happened. Rubbernecking constitutes of as a distraction because everyone is trying to slow down and watch.

-How to measure time gas to break(foot) -Know how to measure reaction time for stop watch -Measure reaction time of Dropping Ruler, with decisions, with distractions -Under the influence/Drugs/Driver Fatigue/Rubbernecking (slow down nervous system) -Talking on cell phone/texting (distractions) -Yellow light, stop or go (decision) -How fast=How Far (doesn't effect how long it takes to react) -Equation -Ordered Pairs (t,d) -Covering more distance when driving faster -Rise gets greater
 * //Section 1 Quiz//**
 * Investigation**
 * Effects of Reaction Time**
 * Falling Ruler**

//SECTION 2-Measurement//
//Grade: //

**Calibrate** the length of a stride.
 * Learning Objectives**
 * Measure ** a distance by pacing it off and by using a meter stick.
 * Identify** sources of error in measurement.
 * Evaluate** estimates of measurements as reasonable or unreasonable.

I see students walking through the hallway. I see a boy recording numbers down for the little girl walking next to the ruler. The taller boy is much taller so the younger girl is being compared with their strides while walking. Maybe to figure out why it takes a person a longer time to get to a place because of small strides. Measuring how long their stride is in a certain distance. -I think that the student who measured 10 m because 3 m and 10 m have a big distance between them. Therefore the one with the stride of 10 m made a mistake by using the wrong type of measuring device. -If the student measure 3 m and 3.01 m, it depends how it looks on the measuring device.
 * 9/14: What do you see/What do you think?**

//**Investigation 9/15**// 22 **Strides**: Tape to Tape 11m 50 cm: Stride length 14 Metersticks-14m Average Stride Measurement: 12.17m Average Meter Stick: 13.36m Longer Strides=Less Strides Strides have the most randomness 1) Do the measurements listed on your class table agree? The class average is 13.36m which is 2) By how many meters do the results vary? They vary from 5 meters. 3) Why are there differences in the measurements made by different groups? List as many reasons as you can think of. Everyone has a different stride length and each stride was probably different for each time with add a percentage to an error. 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. Suggest that each person use the same method, take the same length measurement, and for them not round the answer. 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? If each group was given a really long tape measure each group would have the same measurement. Starting and ending on which side of the tape. 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 I think that there will always be differences in the experiment. People are different. Shorter and taller people can't produce the same stride length as each other. 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 meter stick?” technique) a. When measuring the hallway, we didn't have systematic errors because they cant be corrected by a calculation. b. The random percentage is about .36m 8) If you did not have any systematic errors then name 3 ways a systematic error may have been introduced. You could have used a yard stick, which then you would convert into meters.
 * Group || Strides || Meter Sticks ||
 * 1 || 12.09m || 13.16m ||
 * 2 || **10.8m** || 13.2m ||
 * 3 || 13.34m || 13.31m ||
 * 4 || 11m || //**14m**// ||
 * 5 || **15.54m** || **//13m//** ||
 * 6 || 10.25m || 13.5m ||

.82 .75 .72 .71 .78 .76 .73 || .84 .85 .84 .84 .82 .82 .84 .82 .81 .83 || .81 .81 .81 .81 .81 .81 .80 .81 .815 .81 || .815 .814 .815 .8145 .813 .815 .814 .812 .812 .813 .812 ||
 * 9/16-DeMo: How long is the Tube? (4 sided ruler)**
 * Interval || 1m || .1m || .01m || .001m ||
 * || .75


 * 9/17 Physics Talk**
 * random error:** an error that cannot be corrected by calculation.
 * Measuring tool/person measuring responsible for uncertainty.
 * systematic error:** an error produced by using the wrong tool or using the tool incorrectly for measurement and can be corrected by calutlation.
 * Can be avoided or corrected by calculating.
 * 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.
 * Measurements can vary with accuracy and/or precision.

1**.** Systematic errors can be corrected by calculations. Random errors cannot be corrected by calculations. 2. There are always uncertainty in measurements because people could measure different ways because there is no perfect tool. 3. The positions of arrows on a target need to be to illustrate measurements that are neither accurate nor precise, they would need to be somewhat close together or close to the target. The placement would be completely random to make it neither accurate nor precise.




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

x1000 || 1km=1000m 1m=.001km || x.01 || 1 cm=.01m 1m=100cm || x.001 || 1mm=.001m 1m=1000mm ||
 * Prefix || Symbol || Multof10s || Example ||
 * kilo || k || x10^3
 * centi || c || x10^-2
 * milli || m || x10^-3

This represents a systematic error because the odd groups had a measuring device that started at 2 centimeters and the other groups started at 0 centimeters.
 * 9/20-Measuring a Copper Tube**
 * Group || Measurement ||
 * 1 || 66cm ||
 * 2 || 64.15cm ||
 * 3 || 66.1cm ||
 * 4 || 64cm ||
 * 5 || 65cm ||
 * 6 || 64.1cm ||

1. +/-10cm, +/-.1m 50.1m->49.9m +/-1cm, +/-.01m 50.01->49.99m +/-1mm, +/-.001m 50.001m->49.999m (Random Error) 2. +/- 1cm 50.01m->49.99m 50m/25s=2m/1s(speed) 2m/1sec=.02m/xsec (time to swim extra distance caused by random error) 2x=.02 x=.01seconds added to 50.01 compared to the 49.99 pool. 3. .02m(difference in range:due to random error for the 50.01) 30lapsx.02m=.6 m (60cm) 1500m/900seconds(60secx15min) 1500m/900=.6m/xsecs x=.36seconds 4. 15.36s in 50.01pool 15.35s in 49.99 pool The second person broke the record in the smaller pool. But the first swimmer swam faster than the second swimmer 50.01x30=1500.3m 49.99x30=1499.7m 1500.3/15.36=97.68m/1s 1499.7m/15.35s=97.70m/1s
 * 9/20-Active Physics Plus**
 * ___1500m___|

3. My friend and I both agree that an iPad was around $450 based on the qualities the iPad has available. On the other hand, we both disagree on how much we think the price of the Apple iPhone costs. I think that it costs about $200 but my friend thinks it costs around $100. 4. I would say that four million barrels would be a good estimate which means that the measurement is fairly accurate. On the other hand, the tank could told a little less or a little more in the tanks because the workers rounded the prices to the closest full number. The possible uncertainty in value ($100 per tank) of the oil tanker’s oil based on if it’s more or less than five million. The 5 in the 5 million is the only certainty in this problem therefore it is +/-1,000,000. Need to know how much is in the barrel of oil and how big the ship is. 6.a. A 2-L bottle of soft drink is not enough to serve 12 people at a meeting because 2 L is 8 cups therefore the 4 other people wouldn't receive the soft drink. b. A mid-sized automobile with a full tank of gas can travel from Boston to New York City without having to refuel. This is not a correct estimate. 34miles/gallons x 15 gallons=510 miles Therefore you could go to Boston and back to New York on a full tank of guess. 7. Being off by 1 meter in measuring the width of a room is a huge mistake compared to a longer distance from my house to my school because if you were putting in a carpet, the carpet would be too big or too small for the room. It wouldn't make a difference as comparing the distance from home to school. 8. a. I should drive around 60 mi/h because the speedometer is accurate within 5 mi/h. b. They could count count how many miles you do in 60 minutes and count the time between the mile markers but to find out, you would have to drive 60 mph for the experiment to work. 9. I think that driving 31 mi/h in a 30 mi/h zone does not make a difference. I believe that going 5 mi/h over the speed limit should deserve a ticket. In cars, most cars don't show the exact speed that you are going, therefore your estimating your speed at a glance. People may judge their speed differently therefore 5 mi/h is a reasonable zone. //Inquiring Further// I would expect full certainty in measurement when I am purchasing vegetables because I wouldn't want to buy too much or too little of it depending on how much I could fit. The types of measurement standards regulated by the government in industries are density, mass, volume, while food markets are more quality, quantity, and distribution. I would want to the amount of food for the right price to the second decimal place to make the purchase precise.
 * //Homework-Physics to Go/Inquiring Further 9/20//**

//**DoNow Section 2: 9/21**// What does it mean? This is a systematic error because she is using a yardstick for a meter stick which is correctable. Systematic effects the accuracy. (bigger number) Why do you believe? You can trust experiments if all measurements have uncertainties because if there is an uncertainties, the measurements would be around the same number therefore you have an idea of what the answer is going to be. No its not reasonable for a police officer who pulls a person over for going 75 mph in a 35 mph because a radar gun shouldn't be off 45 mph.

//**Estimation Activity: 9/21**// pg 24 a-i a. This a reasonable estimate because college football players are bigger and stronger and weigh about 220 lb. b. This is unreasonable because no one is 13 feet tall. c. This is unreasonable because 1440 min is 24 hours, therefore teachers don't work 24 hours a day. d. This is unreasonable because poodles can't be 132 lb, a standard poodle is around 60 lb. e. This is unreasonable because when calculated, we got 45,000 ft^3. f. This is unreasonable because some schools are bigger than others. g. It is not safe because you are going the same speed as the other car and it takes a while to get around the tractor. h. How far away they are and how fast they are going. i. It would make it safely through because the bike and the motor home is about 15 feet high.

SECTION 3: Average Speed
//**Learning Outcomes**// Define and contrast average speed and instantaneous speed. Use strobe photos, graphs, and an equation to describe speed. Use a motion detector to measure speed. Construct graphs of your motion. Interpret distance-time graphs. Calculate speed, distance, and time using the equation for average speed.
 * ** Section3 ** || **Points** ||
 * WDYSee/Think: || /10 ||
 * Investigate: || /20 ||
 * PhysicsTalk: || /20 ||
 * PhysicsPlus: || /20 ||
 * PhysicsToGo: || /20 ||
 * Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**100** ||

//**What do you see? What do you think? 9/21**// I see a rear-end collision happening. It looks like the blue car was going to slow and watching the bunny and the yellow car is going too fast, then the red car was stuck in the middle. The safest distance is a car length behind the other car or use the 3-second rule My speed and how fast the car in front of you is going and should be going the same speed as the car in front of you.

Strobe Photo: combined photo of pictures taken at equal intervals of time.
 * //Investigation 9/22//**

Instructions: Hook up motion detector to USB Port Plug in the USB drive Open Data Studio


 * //9/23//**

4a. Walking Toward Slowly 4b. Walking Away at Normal Speed 4c. Walking Away and Toward at a Slow Speed 4d. Walking Fast back and forth 5a. .

5b. 6b. The steeper slope is the faster walking. 7a. 2.4 meters 7b. 2.5-3.0 seconds each distance 7c. 2.3 meters/2.7 meters=.816meters/second 7d. Multiply the distance and time by 2 if your walking twice the distance. The assumption would be the walker has a constant pace throughout the experiment. 8a. (60ft/1sec)(x/.5 sec) x=30 ft 8b. (60ft/1sec)(x/1.5sec) x=90 ft 8c. (50ft/1sec)(x/.5) x=25 ft (50ft/1sec)(x/1.5) x= 75 ft 8d. (70ft/1sec)(x/.5) x=35 ft (70ft/1sec)(x/1.5) x=105 ft 8e. (40ft/1sec)(x/.5sec) x=20 ft They would need to stop 20 feet behind. 8f. (60ft/1sec)(x/15sec) x=4 There would 4 car lengths away from each other.

speed: the distance traveled per unit time; speed is a scalar quantity, it has no direction. constant speed: speed that does not change over a period of time. average speed: the total distance traveled divided by the time it took to travel that distance instantaneous speed: the speed at a given moment. velocity: the speed in a given direction. reaction distance: the distance a vehicle travels in the time it takes a driver to react.
 * //Homework-9/26//**
 * Physics Words**:
 * Checking up questions**:
 * 1.** The average speed is the total distance divided by the time whereas instantaneous speed is the speed at a one point in time.
 * 2.** Speed is the distance traveled per time and it has no direction but velocity is the speed in a given direction.
 * 3.** It shows that as more time passes more distance has been covered.
 * 4.** Reaction time affects reaction distance because depending on your reaction time will make a difference in your reaction distance, so if you have a slower reaction time you're going to have a greater reaction distance and the other way around.

//**Warm-Up 9/28**// //**(Picture)**//

//**9/28- Active Physics Plus**//

20 mi/h || 40 mi || ∆t 1=? 2 hours || 1 hour || 3 hours || V=80mi/∆t
 * Part || Distance || Time ||
 * 1st half
 * 2nd half 40mi/h || 40 mi || ∆t 2 =?
 * Total Trip || 80 mi || ∆t total =?
 * V AV =∆d/∆t**

Part 1: 20mi=40mi/∆t 1 2 hours Part 2: 40mi=40mi/∆t 2 1 hour Total Trip: x=80mi/3 hours
 * Total Trip=26.67 mi/h**

-This answer does not make sense because the answer is closer to 27 mi/h instead of 30 mi/h.

50 mi/h || 50 mi || ∆t 1=? 1 h || 1 mi/h || 50 mi || ∆t 2 =? 50 h || 51 h || 100mi/51hr
 * Part || Distance || Time ||
 * 1
 * 2
 * Total || 100 mi || ∆t total =?
 * Total Trip: 1.96 mi/hr**

-The answer would be closer to 1 mi/h because this driver drove at only 1 mi/h for many hours and only got a chance to drive 50 mi/hr for 1 h. The total trip would be 1.96 mi/h.

1. 80 mi Trip 100 mi Trip 2. 80 mi Trip Part 1: _--_--_--_--_--_--=40 mi Part 2: _---_-_-=40 mi Whole Trip: _--_--_--_--_--_--_---_-_--=80 mi

100 mi Trip Part 1:_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_--=50 mi Part 2:_--_--_=50 mi = 50 mi Whole Trip: _-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_- _--_--=100 mi

3. 50 mi at 50mph 50mi at 25 mph 50 mi at 10mph

50+50+50 = 150 1hr + 2 hr + 5hr = 8 Vav= 150/8= 18.75 mi/hr is the average speed.

//**Physics to Go-10/1/11**// 1a. The cars are going at a constant speed. 1b. The cars have a constant speed but then the cars speed up and then back to a constant speed. 2a. --- -- - - - - - [rest to constant speed] 2b. - - - - - - [constant speed to a complete stop] 3. 350 ft/s=x ft/20 s 4a. 215 mi/4.5 s 4b. You don't know the speed at that instant. She could have been going slower than 48 which means she probably was in traffic. 5. 15 min/60hr=.25 min/hour 5 mi/.25 hours= 6a. The person's speed is gradually increasing then stops. 6b. The person's speeds up gradually then stops then slows down at a constant rate. 6c. The person's speed is constantly increasing, then the speed increased. Constant velocity. 6d. The automobile's velocity both speed and direction. Fast velocity. 7a. 25m/s=xm/.71s 7b. 16m/s=x/.71 7c. 11.35m/s (2) 8a. Traffic experts can be sure the 3-second rule is a safe following distance because they could use equations to relate the distances. 8b. Sometimes the 3-second rule is not enough for the highway as on a rural road. There is a less amount of space on a rural road than on a highway. On the highway, a larger following distance is needed to be safer. 9a. 70mi/h=x/1/3 x=23.3 feet 9b. This is not larger than the length of my classroom. The classroom is 30 feet. 10a. 88 ft/s=x/.5s x=30 feet 10b. If the car was 15 feet long, than one car length and one space can fit. 10c. 30 mi/h=x/.5s x=15 1 car length can fit 10d. 90 mi/h=x/.5s x=45 mi/h 1 and a half car length space and 1 and a half spaces. 10e. If talking on the phone doubles your reaction time, then the distance becomes smaller. 11.
 * x=7000 ft/s**
 * average speed= 47.78 mi/hr**
 * x=20 mi/hour**
 * x=17.75 m/s**
 * x=11.36m/s**
 * x=22.72 m/s ( miles)**

C-The drops are closer together then the space increases and continues at a constant speed the rest of the way and are equally spaced.
 * //10/3/11 Do Now//**

3. 4.
 * //10/3/11 Distance vs. Time Graphs: Cyclists//**


 * //10/4/11: Walk the Graphs//**

1. Graph 1 was possible. We had 2 people stand single file in front of the motion detector and then the first person jumped out of the way. 2. Graph 2 was also possible. We had one person stand in front of the motion detector and the other person jumped in front of the person. 3. Graph 3 was not possible. It was impossible to make a perfect circle.

4. Graph 4 was possible. We took our laptop and moved it toward the motion detector to create the graph.

5. Graph 5 is possible. We took the laptop and moved it away from the motion detector to create the graph.

//**SECTION 4: Graphing Motion**//
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing #8,9 ||
 * <span style="font-family: 'Courier New',Courier,monospace;">** Section4 ** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 20px;">**Points** ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">WDYSee/Think: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/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;">18/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <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;">**98** ||

//**Objectives:**//
 * Measure** a change in veloctiy (acceleration) of a cart on a ramp using a motion detector.
 * Construct** graphs of the motion of a cart on a ramp.
 * Define** acceleration using words and an equation.
 * Calculate** speed, distance, and time using the equation for acceleration.
 * Interpret** distance-time and velocity-time graphs for different types of motion.

//**10/5/11: What do you see?**// The red car has rear wheel drive. If the car was stopping, the back of the car would be up if he slammed on the brakes. He was probably accelerating. He was trying to get the red light. The differences and similarities of the motion of an automobile and bus from 0 mi/h to 30 mi/h that a bus takes longer to acceleration to 30 mi/h than a car can. The car has the greater acceleration. Same change in velocity in a period of time.
 * //What do you think?//**

Straight line that only touches a curve at one point. Slope of the tangent: The instantaneous velocity at a specific point in time.
 * //10/5/11: Tangent Line//**


 * //10/5/11: Investigation: Graphing Motion//**

a. The car travels slower at the beginning and slows toward the end. b. The car travels faster at the beginning and slows toward the end. c. The car moves at a constant velocity. d. Graph d is where the car doesn't move.
 * //1.//** It would go faster the second half of the distance because when letting go the car gains speed on the second half of the distance.
 * //2. a b.//**
 * //c d//**


 * //Run 1://**

Prediction

Tangent Line 0.5m/s Tangent Line 1.5m/s Velocity vs. Time Graph

On Velocity Graph at 0.5s=0.6m/s at 1.5s=1.3m/s

Acceleration: A=1.2-0.8/1.45-0.85 A=0.4/0.6
 * A= .67m/s^2**


 * //Run 3//**

Prediction Tangent Line 0.664m/s Tangent Line 0.043m/s Velocity vs. Time Graph

On Velocity Graph

Acceleration

//**Run 4**//

Prediction Tangent Line 0.678 Tangent Line 0.033 Tangent Line -0.481 Velocity vs. Time Graph

On Velocity Graph

Acceleration Acceleration is constant.
 * The top of the graph of d vs t has a positive tangent line. Then as the car comes back down the ramp, velocity goes closer to 0.
 * Then the graph goes into negatives because on the way back it decreases which causes the velocity to be a negative.**
 * The gravity pulls the car back. Acceleration never stops because gravity never turns off.
 * The car spends an instant amount of time having a velocity of 0. Keeps changing.**

//**10/6/11: Checking-Up Questions**//

Acceleration-The change in velocity with respect to a change in time Vector-A quantity that has both magnitude and direction. Negative Acceleration-a decrease in velocity with respect to time. The object can slow down or speed up. Positive Acceleration-an increase in velocity with respect to time. The object can speed up or slow down. Tangent Line-a straight line that touches a curve in only one point.
 * //Physic Words//**

1. Give the defining equation in words, and by using symbols. Acceleration is the change in velocity over the change in time. a=∆v/∆t 2. What is an SI unit for measuring acceleration? Use words and unit symbols to describe the unit. Any measure of length with a measure of time. (for example: m/s^2) 3. What is the difference between a vector and a scalar quantity? A vector has both magnitude and direction, whereas a scalar quantity has magnitude and no direction. 4. Sketch a distance-time graph for a. constant velocity: b. constant acceleration: 5. What does the slope of a velocity-time graph represent? It would equal the to the change in velocity with respect to time. This ultimately represents velocity.
 * //Checking-Up Questions//**

//**10/7/11 DoNow: Physics Talk Review**// 1. 30-0/5 6m/s^2 2. A vector has both magnitude and direction. A scalar quantity has magnitude but no direction. Vectors include: Velocity, Force (push or pull) Scalar Quantity include: Speed, Calories, Anything than can be Count

//**10/7/11 Car vs. Pick-Up Truck vs. Bus**//

Speeding Up

Slowing Down

//**10/9/11 Physics To Go**//

No, an object can never have 0 acceleration with a nonzero velocity whenever you are moving, that is your velocity. You would divide the velocity the amount of time it took to travel that speed and would get acceleration. No one would ever get 0 unless you had 0 m/s for your velocity. No, an object can never have 0 velocity because for it to have 0 velocity it would need to also have 0 acceleration because anything divided by 0 is zero. They don't need to have the same velocity because they could have the same acceleration but could not have achieved it in the same velocity. For example, they could have both 5 m/s for acceleration but one velocity can be 25 m and the other one could be 15 m. They don't need to have the same acceleration to have the same velocity because they could have achieved that velocity in a different amount of time. 20 m could have been achieved in 5 s or 2 s leaving it with different accelerations. It depends whether or not there is a constant velocity of the automobile is but most likely it can overtake an accelerating vehicle. Yes it is more correct to refer to speed-limit signs instead of velocity signs because velocity refers to direction and there would be too many signs on the road if there were velocity limit signs. Speed limits refer to just a speed which refers to drivers are well known by people nowadays. **a. How fast would it be going at t=2min?** 120s/5=24 s 24(2)=48 At 2min the automobile would be going 48 mi/h. **b. How far would it be from the starting line?** d=vt d=48(30) 1.6 mi from the starting line. a=∆v/∆t a=(75-0)/(9-0) a=75/9 a=8.33 m/s^2 **b. What was the race car's average speed during the acceleration?** ∆v=a(∆t) ∆v=8.33(9) ∆v=74.97 m/s d=vt d=74.97(9) d=674.73 m a=∆v/∆t a=(75-0)/(8-0) a=75/8 a=9.4 m/s^2 ∆v=a(∆t) ∆v=9.4(8) ∆v=75.2 m/s d=vt d=75.2(8) d=601.6 m a=∆v/∆t a=(4.5-0.6)/(1.1-0) a=3.9/1.1 a=4.29 m/s d=vt d=3.9(1.3) d=5.07 m a=∆v/∆t a=(4.5-0.6)/(1.1-0) a=3.9/1.1 a=3.55 m/s The second trial would get her from second base to third base faster because there is a faster acceleration. The top speed recorded by the falling objects was 12s. a=∆v/∆t a=(9-0)/(7.5-0) a=9/7.5 a=1.2 m/s The object's acceleration as it falls will increase and the objects final velocity before striking the ground would increase. Causing constant speed and then increase linearly. graph B graph D **c. Bike's acceleration as it coasts uphill?** graph E graph A graph F graph C D-E, F B, E D C-F back to the beginning. a=(250-0)/(30-0) a=8.33 mi/h a=(250-0)/(45-0) a=5.56 mi/h 500=250/∆t ∆t=2 hours d=500(2) d=15,000 mi 44.1 m/s 4.5 sec ∆v=a/∆t ∆v=9.8/10 ∆v=98 m/s d=vt d=(.98)(10) d=9.8 m **15. In 1954, in a study of human endurance prior to the manned space program, Colonel John Paul Stapp rode a rocket-powered sled that was boosted to a speed of 632 mi/h (1017 km/h). The sled and he were then decelerated to a stop in 1.4 s.** **a. What was the acceleration of this stop?** a=∆v/∆t a=(632-0)/(1.4-0) a=451.43 mi/h **b. What is this acceleration in terms of g's?** 1/9.8m=x/201.6 g=20.6 g **c. In what distance did the speed of the sled travel as its speed changed from 1017 km/h to 0?** d=(1017)(1.4) d=1423.8 km 16. a. 2m b. 8m c. 18m d. 32m
 * 1. Can a situation exist in which an object has zero acceleration and nonzero velocity? Explain your answer.**
 * 2. Can a situation exist in which an object has zero velocity and nonzero acceleration, even for an instant? Explain your answer.**
 * 3. If two automobiles have the same acceleration, do they have he same velocity? Why or why not?**
 * 4. If two automobiles have the same acceleration, do they have the same velocity? Why or why not?**
 * 5. Can an accelerating automobile be overtaken by an automobile moving with constant velocity?**
 * 6. Is it correct to refer to speed-limit signs instead of velocity-limit signs? Why or why not? What units are assumed for speed-limit signs in the United States?**
 * 7. Suppose an automobile were accelerating at 2 mi/h every 5 s and could keep acceleerating for 2 min at the that rate. **
 * 8. At an international auto race, a race car leaves the pit after a refueling stop and accelerates uniformly to a speed of 75 m/s in 9 s to rejoin the race.**
 * a. What is the race car's acceleration during this time?**
 * c. How far does the race car go during the time it it accelerating? (wrong)**
 * 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?**
 * 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.**
 * a. What is the player's acceleration during the slide? (wrong)**
 * b. What was the distance of her slide?**
 * c. If she had slid for only 1.1 s, how fast would she have been moving when she reached third base? (wrong)**
 * d. Which of these two trials would get her from second base to third base faster?**
 * 10. a. From the graph, approximately what was the top speed recorded by the falling objects?**
 * b. What is the acceleration of gravity on this planet?**
 * 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?**
 * 11. a. Velocity vs Time as he coasts up the hill?**
 * b. Distance traveled verses time as he coasts up the hill?**
 * d. Reaching the top of the hill, the speed of the boy as he coasts down the hill on the bike?**
 * e. Speed vs Time?**
 * f. Distance vs Time as going down the hill?**
 * 12. a. traveling with constant speed?**
 * b. increasing speed?**
 * c. at rest?**
 * d. decreasing speed?**
 * e. According to the graph, where was the car when the test was completed?**
 * 13. A jet taking off from an aircraft carrier goes from 0 to 250 mi/h in 30 s.**
 * a. What is the jet's acceleration?**
 * b. If after taking off, the jet continues to accelerate at the same rate for another 15 s, how fast will it be going at that time? (wrong)**
 * c. How much time does it take for the jet to reach 500 mi/h? (wrong)**
 * d. How much distance would it take for that same jet to reach 500 mi/h?**
 * 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.**
 * a. If the object falls for 100 m, how fast is it traveling?**
 * b. How much time is required for it to fall this 100 m?**
 * c. If the object falls for 10 s, how fast is it traveling?**
 * d. How far has it fallen in this 10 s? (wrong)**

//**10/12-Speeding Up & Slowing Down Graphs**//

//**Speeding Up**// a=V(at 3s)-V(at 2s)/3.5-2 a=.4m/s-.24m/s/3.5s-2s a=.15m/s/1.5s **a=0.1m/s^2**
 * Speeding Up Acceleration**
 * //a=∆v/∆t//**


 * //Slowing Down//**

a=∆v/∆t a=0.33-0.8/3.5-2 a=-0.47/1.5
 * Slowing Down Acceleration**
 * a=-0.31m/s^2 **



10/12-**//Velocity vs. Time Graphs//**



**//10/12/11-Active Physics Plus//**

d=1/2at^2+Vit a=∆v/∆t or a=Vf-Vi/∆t

1. Vf=20 m/s Vi=7 m/s a=3 m/s^2 ∆t=? 3m/s^2=20 m/s-7 m/s / ∆t 3x=13m/s ∆t=4.33s 2. Vf=? Vi=7 m/s a=1.5 m/s^2 ∆t=10 s 1.5=(x-7) / 10 15=x-7 Vf=22 m/s 3. a. Vf=20 m/s Vi=0 m/s a=? ∆t= 5 s x=(20-0) / 5 x=20/5 a=4 m/s^2 b. 1/2(4)(5)^2 2(25) 50 m  4. Vf=40 m/s Vi=0m/s ∆t=10 s a=4m/s^2 d=? d=1/2(4)(10)^2 2(100) d=200 m  5. a. Vf=23 m/s Vi=3 m/s a=? ∆t=5 s a=(23-3) / 5 a=4 m/s^2 b. d=1/2(4)(5)^2+3(5) d=2(25)+15 d=65 m

SECTION 5: Negative Acceleration
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">did not title preparing for the chapter challenge section of total stopping distance activity ||
 * <span style="font-family: 'Courier New',Courier,monospace;">** Section5 ** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 20px;">**Points** ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">WDYSee/Think: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">9/10
 * <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;">**99** ||

//**10/18/11 Learning Outcomes**//
 * Plan** and carry out an experiment to relate braking distance to initial speed.
 * Determine** braking distance.
 * Examine** accelerated motion.

//**10/18/11 What Do You See/What Do You Think?**// Theres a moose in the middle of the road. The car is trying to stop so he doesn't hit the moose and the back wheels indicate he's slowing down. You must consider the following factors:the mass of the vehicle, how fast your going (the faster you go the further you go), the road and the cars tires relationship (friction), the condition of the internal vehicle, the weather, and reaction time.

//**Investigation-10/18/11**//
 * Objective**: To determine the effect of initial speed on braking (stopping) distance.

Braking Distance-Counting Tiles(30.5) Vi=11cm/time in gate
 * Braking Distance vs. Initial Velocity**
 * Vi Braking Distance**
 * Initial Velocity(m/s) || Braking Distance(m) ||
 * 0.887 || 10.2 ||
 * 1.17 || 5.64 ||
 * 0.705 || 5.34 ||
 * 0.91 || 10.83 ||

11/0.124=88.7cm=.0887 30.5(33.5)=1021.75=10.2

11/0.0943=1.17 18.5(30.5)=564.25=5.64

11/0.1566=.705 17.5(30.5)=533.75=5.34

11/.1202=.91 30.5(35.5)=1082.75=10.83



209/118=1.77 **c)** **The braking distance is our speed squared.** 1. The braking distance will increase by 4 if the initial velocity is doubled. 2. The braking distance will increase by 9 if the initial velocity is tripled. 3. The braking distance will increase by 16 is the initial velocity is quadrupled. 4b. 209/118= 1.77. The breaking distance is the speed squared. 4c. The breaking distances were squared of our speed. 5. Decrease speed is 1/2 --> square of a 1/2 When you square 30 mph (speed), you get 900 miles for your breaking distance. 6. The breaking distance is 900 miles as well because you square the speed (30mph).
 * 1)** The braking distance will increase by 4 if it is doubled.
 * 2)** The braking distance will increase by 9 if it is tripled.
 * 3)** The braking distance will increase by 16 if it is quadrupled.
 * 4)** The braking distance is the speed squared.
 * 5)** ** a) **The decrease speed is 1/2 square of a 1/2

Speed: Vi=80/60=1.33 Stopping Distance: 209/118=1.77 (1.33^2=1.77)


 * //Physics Talk 10/19/11//**

__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 Q's**__ 1). This is undergoing negative acceleration because it is decreasing speed in a positive direction (going forward). 2). If you are going faster and increasing your velocity, the automobile is going to have a bigger braking distance. Since you are going faster, you are going to need more distance to slow down. WHen you are going faster, the car has more time to reach maximum velocity. 3). For a negative acceleration, you are moving forward but you are slowing down to be negative acceleration. The acceleration is getting slower and slower.
 * Vocab Words**

__**EQUATION FOR CALCULATING BRAKING DISTANCE**__

//**10/20/11-Braking Distance vs Initial Velocity**//

Vi(2) / Vi(1) Ratio of initial velocities Braking Distance(2) / Braking Distance(1)

(Vi2 / Vi1)^2=Braking Distance2 / Braking Distance1

(60mph/30mph)^2=123ft/x 30.75ft
 * Initial Velocity || Stopping Distance ||
 * 10 mph || 3.42ft ||
 * 20 mph || 13.67ft ||
 * 30 mph || 30.75ft ||
 * 40 mph || 54.67ft ||
 * 50 mph || 85.42ft ||
 * 60 mph || 123 feet ||
 * 70 mph || 167.28ft ||
 * 80 mph || 218.94ft ||
 * 90 mph || 276.75ft ||
 * 100 mph || 341.94ft ||

4) If Initial Velocity is doubled how does stopping distance change? Increased by 4 (multiply) 5) If the Initial Velocity is multiplied 4 times how does the stopping distance change? Increased by 16 (Multiply) 6) If the Initial Velocity is halved how does the Stopping Distance Change? Decreased by 4 (divide) 7) If the initial Velocity is quartered how does the stopping distance change? Decreased by 16 (divide) 8) What speed would you need to have a stopping distance of a mile? (60mph / x)^2=123ft / 5280 ft 3600 / x^2= 123 / 5280 123(x^2)=19008000 x^2=154536.58 393.1mph

//**10/20/11-Physics To Go**// //**1.**// 2. Automobile A because at a certain point B and A cover the same distance but B covers it in a less amount of time. This makes automobile A "safer" because a driver would have a longer time to react in respect to the velocity and the braking distance. People want a little braking distance rather than automobile B, which has a greater braking distance. 3. //How much distance will it require to stop when traveling at the following? (20 m at 30 mi/h):// <span style="color: #000000; font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">a. 15 mi/h = <span style="color: #0000ff; font-family: Tahoma,Geneva,sans-serif; font-size: 12px;"> 5 m (1/4) <span style="color: #000000; font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">b. 60 mi/h = <span style="color: #0000ff; font-family: Tahoma,Geneva,sans-serif; font-size: 12px;"> 80 m(2^2 or 4) <span style="color: #000000; font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">c. 45 mi/h = <span style="color: #0000ff; font-family: Tahoma,Geneva,sans-serif; font-size: 12px;"> 45 m(9 times 5) <span style="color: #000000; font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">d. 75 mi/h = <span style="color: #0000ff; font-family: Tahoma,Geneva,sans-serif; font-size: 12px;"> 125 m(5^2 or 25 times 5) 4. Speed=10m/s t-reaction=0.9s v=d/t v(t)=d 10(0.9)=d reaction distance=9m total stopping distance=reaction distance+braking distance 39m=9m+30m 5. SKIP 6. SKIP 7. SKIP 8. When you are driving, you need to be very conscious of the surroundings. You need to know how fast you can react to things as well as the speed of your car. If you have a slow reaction time, then you might not stop in time to avoid an obstacle on the road. Also, if you are moving fast, then your car will have a greater braking distance than if you were going slower. V=dreaction/treaction dreaction=V(treaction) Total Stopping Distance=Vtreaction+dbraking Vf^2=Vi^2+2(a)(dbraking) 0=Vi^2+2a(negative)(dbraking) -Vi^2 / -2a(negative)=dbraking Vi^2 / 2a=dbraking Total Stopping Distance=Vtreaction+Vi^2/2a Total Stopping Distance=Vireaction+Vi^2/2a Total Stopping Distance=reaction direction+braking distance Vi-initial velocity before braking treaction: reaction time a:acceleration greater a=good brakes smaller a=bad brakes

If Vi increase before you stop then TSD will increases. Vi~TSD If treaction increases then TSD will increases(far distance to stop). treaction~TSD If a increases(really good brakes) then TSD will decrease. TSD~1/a


 * //10/21/11-Direction of Acceleration/Equations of Motions//**

//**10/24/11-Active Physics Plus**//

d=1/2(Vi+Vf)t Vf=Vi+at d=Vit+1/2at^2 Vf^2=Vi^2+2ad

//**Average Acceleration**// 1. A=-4.1m/s^2 Vf=0 Vi=9 t=? Vf=Vi+at 0=9+(-4.1)(t) -9=(-4.1)t t=2.2 s

2. A=2.5 Vf=12 Vi=7 t=? Vf=Vi+at 12=7+2.5(t) 5=2.5t t=2 s

3. A=-.5m/s^2 Vi=13.5 Vf=0 t=? Vf=Vi+at 0=13.5+(-0.5)t -13.5=-0.5t t=26 s

4. A=? Vi=-1.2m/s Vf=-6.5m/s t=1500s Vf=Vi+at -6.5=-1.2+a(1500) a=-0.004m/s^2

5. A=0.0047 Vi=0 Vf= t=300s

Vf=Vi+at

1.Vi=0m/s Vf=6.6m/s t=6.5s d=? d=1/2(Vi+Vf)t d=1/2(0+6.6)(6.5) d=1/2(42.9) d=21.45 m
 * //Displacement with Constant Uniform Acceleration//**

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

3. Vi=? Vf=0m/s a=-5m/s^2 d=0.8km-> 800m Vf^2=Vi^2+2ad 0^2=x+2(-5)(800) 0=x^2-8000 8000=x^2 Vi=89.44m/s NO

4. Vi=21.7m/s Vf=0 d=99m t=? d=1/2(Vi+Vf)t 99=1/2(78,000+0)(x) 99=1/2(78,000)x 99=39000x t=9.14s

5. Vi=6.4m/s Vf=x t=210s d=3.2 km/1hr x 1000m/1km x 1hr/60min x 1min/60sec = .889 meters d= 1/2(Vi+Vf)t 0.889=1/2(6.4+x)(210) 0.889=105(6.4+x) 0.0085=6.4+x Vf=-6.39m/s

//**Velocity and Displacement with Uniform Acceleration**// 1. Vi=23.7 km/h-1000m/1km-1hr/60min-1min/60sec=6.58m/s a=0.92m/s^2 s=3.6s Vf=? Vf=Vi+at x=6.58+0.92(3.6) x=6.58+3.456 Vf=9.89m/s d=1/2(Vi+Vf)t d=1/2(6.58+9.89)(3.6) d=29.7m

2. Vi=4.3m/s a=3m/s^2 t=5s Vf=? Vf=Vi+at x=4.3+3(5) x=4.3+15 Vf=19.3 m/s d=1/2(Vi+Vf)t d=1/2(4.3+19.3)(5) d=59m

3. Vi=0m/s t=5s a=-1.5m/s^2 Vf=? Vf=Vi+at x=0+5(-1.5) Vf=-7.5 m/s d=1/2(Vi+Vf)t d=1/2(0-7.5)(5) d=1/2(-7.5)(5) d=-18.75m

4.Vi=15m/s Vf=10m/s a=-2m/s^2 t=? Vf=Vi+at 10=15+(-2)(t) -5=-2t t=2.5s d=Vit+1/2at^2 d=15(2.5)+1/2(-2)(2.5)^2 d=37.5-6.25 d=31.25m

//**Final Velocity After Any Displacement**// 2. a=0.8m/s^2 Vi=7m/s Vf=? d=245m Vf^2=Vi^2+2ad x^2=49+2(.8)(245) x^2=49+392 x^2=441 part a-Vf=21m/s

Vf^2=Vi^2+2ad x^2=49+2(0.8)(125) x^2=49+200 x^2=249 part b-Vf=15.78m/s

Vf^2=Vi^2+2ad x^2=49+2(0.8)(67) x^2=49+107.2 x^2=156.2 part c-Vf=12.49m/s

3. Vi=0m/s a=2.3m/s^2 d=55m Vf=? Vf^2=Vi^2+2ad x^2=2(2.3)(55) x^2=253 part A-Vf=15.91m/s

d=1/2(Vi+Vf)t 55=1/2(15.91)(x) 55=7.95x part B-t=6.92s

4. a=0.85m/s^2 Vi=83km/h-23.06 Vf=94km/h-26.11 d=? Vf^2=Vi^2+2ad 681.21=531.76+2(0.85)(x) 681.21=531.76+1.7x 149.45=1.7x x=86.9m

5. Vi=0m/s Vf=120km/h33.3m/s d=240m a=? Vf^2=Vi^2+2ad 33.3^2=2(240)(d) 1108.89=480d 2.31m/s^2

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

//**10/25/11-STUDY GUIDE**// Stopping Distance effected by Vi Direction of acceleration --->v speeding up >a

>v <-a slowing down Equations of Motion Vi^2=Braking Distance If doubled=Vi^4

100km/h-1000m/1m1h/60min1min/60s= 27.78m/s 25km/h1000m/1km-1h/60min1min/60s= 6.94m/s 220km/h1000m/1km1h/60min1min/60s= 61.11m/s
 * //10/26/11-Practice Conversions km/h---m/s//**

//**10/28/11-Total Stopping Distance Activity**// 1. 2. The sign of the acceleration is negative because the car is slowing down in the picture above. 1. v=d/t (no acceleration during dreaction) d=vt dreaction=Vit(treaction) 2. d=10m/s(1s) d=10m 3. 20m/s(1s) d=20m 4. The faster you are going the longer the stopping distance is. The slower you are going the shorter the stopping distance is. 1 is a really long reaction time. It was an old man driving, if theres yellow light, some people want to go through it but other people want to slow down. Making a decision makes the reaction time longer.
 * Calculating Reaction Distance**


 * //10/28/11-Total Stopping Distance Activity//**

1. Vf^2=Vi^2 + 2ad 0=Vvi^2+2adbraking -(Vi)^2=2adbraking -(Vi)^2/2a=dbraking **a is negative if braking and positive direction**
 * Calculating Braking Distance**

5.-(Vi)^2/2a=dbraking -2601/2(-11)=dbraking dbraking=118.2m

6. a=-24m/s^2 Vi=51m/s Vf=0m/s d? 0=2601+2(-24)(x) -2601=2(-24)x dbraking=54.19m

1. t=0.5 Vi=11m/s a=-4m/s^2 dr=? v=dr/tr 11m/s=dr/0.5 dr=5.5 m Vf^2=Vi^2+2ad 0=(11)^2+2(-4)(d) -121=-8d db=15.13m- TSD=dr+db TSD=20.63m

2. Vi=27m/s t=0.5 a=-4m/s^2 dr=? v=dr/tr 27=dr/0.5 dr=13.5m Vf^2=Vi^2+2ad 0=729+2(-4)(d) -729=-8d db=91.13 TDS=91.13+13.5 TSD=104.63

3. v=dr/dt 27=dr/1 dr=27 0=729+2(-4)(d) -729=-8d db=91.13 TSD=91.13+27 TSD=118.13m

4. v=dr/tr 27=dr/.5 dr=13.5 0=729+2(-2)(d) -729=-4d db=182.25m TSD=13.5+182.25 TSD=195.75m

Reaction Distance- Velocity & Reaction Time Braking Distance- Acceleration & Velocity

//**11/2/11-Create a Problem**// -solve for dr, db, TSD -embed givens in problem -don't embed givens that don't need to be mentioned

My Question A car is moving at a speed of 30 m/s (67mph). Assume that you have a reaction time of 0.7 seconds including moving your foot from gas to brake. You drive on to the highway with a maximum deceleration of -3 m/s^2. What is your reaction distance, braking distance and total stopping distance? Vi=30m/s Vf=0m/s tr=0.7 s a=-3m/s^2 d=? v=dr/tr 30=dr/0.7 dr=21m 0=900+2(-3)(d) -900=-6d db=150m TSD=150+21 TSD=171m

I am driving on a 30m/s highway with an acceleration of -24m/s^2. I have a reaction time of .7 seconds. What is my reaction distance, braking distance, and total stopping distance? Vi=30m/s Vf=0m/s tr=0.7s a=-2.4m/s^2 d=? v=dr/tr 30=dr/0.7 dr=21m 0=900+2(-2.4)(d) -900=-4.8d db=187.5 TSD=187.5+21 TSD=397.5m
 * //Dara's Question//**

**//Mini-Challenge//**

 * //Plan//**

**//<span style="background-color: transparent; color: #000000; text-decoration: none; vertical-align: baseline;">Paragraph //**

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">To represent his African American heritage, Maurel wore the hat, passed down in his family from generation to generation. Maurel came charging in on his unicorn named Bradley to rescue the fair maiden, Nicolette from the dragon. Maurel had to travel up hill toward the castle which was 1000 meters away where Nicolette was being held. His unicorn was traveling 20mph. But, what’s this? Oh no! There is work being done on the road! Gasp! Now Maurel must take a detour! Now having to take a new route, Maruel is suddenly find himself truding through the mountain region of The Land of Awe where the hills are short, but there are several of them. Bradley was going up and down and all around. Then suddenly! Gasp! He trips! Maurel is propelled off of Bradley’s back and thrown into the bushes! Poor Maurel had only been traveling for 9 seconds. He tries to stand, pushing himself up from the ground, brushing the leaves off his tights. “Bradley! Come hither!” Bradley trotted to him as he mounts his unicorn. “Onward and upward!” Finally he reaches the castle. He jumps of his steed. With his hands on his hips, he puffs out his chest and commands that he will rescue the princess.

//** Given **//

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">d = 1000 m from starting point to castle

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">Vi = 20 mi/hr → 8.94 m/s

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">Vf = 0m/s

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">t_falling = 9 s

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">t = ?

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">a = ?

**//<span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">equations: //**

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">d=1/2(Vi+Vf)t-To find the distance of Maurel’s journey.

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">Vf=Vi+at-Find the acceleration of Maurel’s journey to the castle.

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">d=1/2(Vi+Vf)t-To find where Maurel tripped to find how much longer he has to travel to get to the castle.

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">1.

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">1000 = ½ (8.94 + 0) t

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">1000 = 4.47 (t)

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">1000 / 4.47 = t

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">t = 223.71 s

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">2.

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">0 = 8.94 + a(223.71)

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">-8.94 = a (223.71)

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">-8.94 / 223.71 = a

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">a = 0.03996

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">a = 0.04 m/s^2

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">3. d = ? Vi = 8.94 Vf = 0 t = 9

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">d = ½ (8.94 + 0) (9)

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">d = 40.23 m

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">1000 - 40.23 = 959.77 m left in his journey!

//**<span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">graphs: **//

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">-going up the hill (speeding up and slowing down)

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 12pt; text-decoration: none; vertical-align: baseline;">-combined graph

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Story-Christine

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Equations-Allyson & Kim

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Graphs-Jessi & Alex

//SECTION 6: Using Models//
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">+12 EC || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing stop and go zone of a real intersection || -//**Investigate**// the factors that affect the STOP and GO Zones an intersections with traffic lights. -//**Investigate**// the factors that result in an Overlap Zone or a Dilemma Zone at intersections with traffic lights -//**Use**// a computer simulation to mathematically model the situations that can occur at an intersection with traffic lights.
 * <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 ||
 * <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;">20/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">20/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">0/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**92 *.75 = 69/75** ||
 * //Learning Outcomes//**

Red car-slowing down/stopping Green car-speeding up Its a red light, green is trying to run yellow light.
 * //What do you see/What do you think?// **

//** 11/4- Investigation **// pg. 91 #3,4 3a. Yes because if automobile A can make it through, then B can make it through. B is in front of A. 3b. Yes B is in the go zone because if A is in the go zone, then B is also in it. 3c. Yes, if automobile A is in the go zone, then any car thats closer will also be in it too. 3d. Autombile C is not in the go zone because he will not make the yellow light because he is too far back. If he tries to keep going he will run a red light. 4a. E is in the stop zone because if the car in front of it (D) is in the stop zone, then E is in the stop zone as well. 4b. F is in the Go zone and if they tried to stop short than it would stop too sudden and it might cause an accident.


 * //11/7/11-Go Zone & Stop Zone//**
 * //Go Zone//**
 * Variable |||| Change || **Predicted** shrink or expansion of GO ZONE ||
 * ty || yellow-light time || Increase ty || Further Away ||
 * ^  ||^   || Decrease ty || Closer ||
 * tr || reaction time || Increase tr || no effect ||
 * ^  ||^   || Decrease tr || no effect ||
 * v || speed limit || Increase v || Further Away ||
 * ^  ||^   || Decrease v || Closer ||
 * a || negative acceleration || Increase a || no effect ||
 * ^  ||^   || Decrease a || no effect ||
 * w || width of intersection || Increase w || Closer ||
 * ^  ||^   || Decrease w || Further Away ||


 * //Stop Zone//**
 * Variable |||| Change || **Predicted** shrink or expansion of STOP ZONE ||
 * ty || yellow-light time || Increase ty || no effect ||
 * ^  ||^   || Decrease ty || no effect ||
 * tr || reaction time || Increase tr || Further Away ||
 * ^  ||^   || Decrease tr || Closer ||
 * v || speed limit || Increase v || Further away (need more stopping distance) ||
 * ^  ||^   || Decrease v || Closer ||
 * a || negative acceleration || Increase a || Closer ||
 * ^  ||^   || Decrease a || Further Away ||
 * w || width of intersection || Increase w || No effect ||
 * ^  ||^   || Decrease w || No effect ||

//**11/8/11-pg 95-96 Part B #1-4**//

1. A should stop, D should stop, B should go, and C should go. 2. E and F would stop. G would go. H can stop because its in an Overlap Zone. 3. J, L, M would stop. K would go. Car M has no choice with a Dilemma Zone 4. Compare the GO Zone and the Stop Zone for Intersections I, II, III, IV. a. The GO ZONE change in each intersection. Different speed limits The 5 variables. b. You could go or stop. It would be safer to go so the car doesn't smash into you. c. You could either stop or go. I would be safer to stop. d. Intersection 3 is a Dilemma Zone and Intersection 2 is an Overlap Zone.

__GO__ V=d/t V=d/ty V=(w+GZ) / ty V(ty)=w+GZ __**GZ = V (ty) - w-**__ the furthest distance at which you can go safely.
 * //11/8/11-Go Zone & Stop Zone Equations//**

__**STOP**__ 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-"edge" of Stop Zone

GZ- Velocity, Yellow Light Time, and SZ- Reaction, Speed, Velocity, Reaction TIme

1. A spreadsheet is a model because it is used to determine the values for the GO Zone and the STOP Zone. It lets you analyze an intersection much more precisely than just watching cars pass. 2. The GO Zone is the area before the intersection where it is safe to go through when the light is yellow. 3. The STOP Zone is the area before the intersection where it is safe to stop when the light is yellow. 4. The Overlap Zone is when the light turns yellow and the driver has the choice of stopping or going through the intersection safely before the light turns red. 5. The Dilemma Zone is a space between the STOP and GO Zone where the driver cannot safely stop before the intersection or pass through the intersection. The driver has to choose between 2 different dangerous options.
 * //11/8/11-Physics Talk//**

//**11/9/11-Active Physics Plus**// 1.

2. 3. Yellow light time=4.5 seconds Reaction Time=0.5 seconds Speed=20m/s Negative Acceleration=-6m/s^2 Width of intersection=20 m

GO Zone=70m STOP Zone=43.33m Overlap Zone 26.66m

4. Yellow-Light TIme=6 seconds Reaction Time=2seconds Speed of Vehicle=25m/s Negative Acceleration=-4m/s^2 Width of Intersection=10m

GO Zone=140 meters STOP Zone=128.125meters Overlap Zone=11.875meters

//<span style="font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">**__11/13/11-ACTIVE PHYSICS PLUS: Homework Questions__** //

** 1) Approaching an intersection in the Midwest you realize it intersects another wide highway. Since the intersection is wider than you anticipated how has the Go Zone changed compared to what you had anticipated? ** The Go Zone has expanded or is greater than what was anticipatied.

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

** 3) A driver with worn out brakes approaches an intersection while a driver with superior brakes approaches an intersection next to him. Which driver will have to brake first? Which drive has a larger total stopping distance? Which driver has a stop zone that is pushed further back from the intersection? ** The driver with worn out brake 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) Two drivers approach an intersection equal in all respects except one is moving faster than the other. Which one has a larger total stopping distance? Which one has a stop zone that is pushed further back from the intersection? ** The 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) Two drivers, equal in all respects except one drunk and one sober approach an intersection. Which one has a larger stopping distance? Which on has a Stop Zone which is pushed further back from the intersection? ** The sober driver has a larger stopping distance.  The drunk driver has a stop zone which is pushed further back from the intersection.

**6) Why does speed affect the Stop Zone more drastically than the Go Zone? You can use the formulas for Stop Zone and Go Zone to explain. Consider also how speed affects braking distance compared to how it affects the GoZone** . Speed affects the stop zone more drastically than the go zone because it must be considered with the persons reaction time.

//**11/16/11-Go Zone and Stop Zone of a Real Intersection**//

Yellow Light Time Crossing Hillsdale Ave. driving on Kinderkamack: 3 sec Speed Limit: 20 mph

__ 5 variables: __ ty=3 seconds tr=0.2seconds w=20m a=-11m/s^2 vi=20mph-9.1m/s

GZ = V (ty) – w GZ=9.1(3)-20 GZ=7.3m

SZ=Vtr + Vi^2 / 2a SZ=9.1(0.2)-9.1^2/2(-11) SZ=1.18-82.81/-22 SZ=1.18+3.76 SZ=5.58m



Doubled Speed Limit __ 5 variables: __ ty=3 seconds tr=0.2seconds w=20m a=-11m/s^2 vi=17.88m/s (40mph)

GZ=17.88(3)-20 GZ=33.64m

SZ-17.88(0.2)-17.88^2/2(-11) SZ=18.11m



When doubling the speed limit, it creates a safer intersection with a larger overlapping zone. There are 15.53 meters for a driver to safely cross the intersection. When traveling at a faster speed limit, it creates a larger Go Zone to pass through the intersection or stop in the Stop Zone before the light changes red. On the other hand, going faster than the speed limit, could cause accidents. The only reason that we doubled the speed limit in this situation was to create a safer intersection. The reason for a large ticket amount is because there is a bigger area for an accident to occur. A driver's total stopping distance will become larger because you are driving faster. As the speed increases, the total stopping distance increases. The problem is, if there is an emergency, it will take the driver longer to stop. The Stop Zone changes the most when the speed increases because there is a larger total stopping distance, the driver will need more room to stop.

**// SECTION 7-Driving on Curves //**
-**Recognize** the need for a centripetal force when rounding a curve. -**Predict** the effect of an inadequate centripetal force. -**Relate** speed to centripetal force.
 * <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;">**75/75** ||
 * //Learning Outcomes//**

-Car went around the curve too fast -Car tipped over because they turned the wheel around -Making a left hand turn.
 * //11/14/11-What do you see?//**

-The sign says to slow down because the car could spin out if your going too fast. -You determine it by hw fast your going, how sharp the turn is (slower=slower turn), the weather (leaves snow), what size car you have, and how worn out your tires are (no friction).
 * //11/14/11-What do you think?//**


 * //11/15/11-Investigation//**

Hypothesis: Answer the Question posed in the Title of Part 1 and give a reason for this hypothesis. A car can go faster around a wide turn because
 * __ PART 1 Can a Car Go Faster around a __**__ Wide Turn **or Sharp Turn?** __

Procedure: 1) Choose two radii to place your circular mass on, one towards the center of your Lazy Susan and one towards the edge. The radius towards the center will simulate the car on a sharp turn while the radius towards the edge will simulate the car on a wide turn. 2) Spin the Lazy Susan with the mass at the inner radius for 10 revolutions then spin the Lazy Susan with the mass at the outer radius for 10 revolutions, recording the time each takes to make 10 revolutions. 11.98s 3) Calculate the time it took the Lazy Susan to make one revolution for each radii. 0.98s 1.198s 4) Use the formula C=2 πr to solve for the circumference of the path that the weight traveled for each radii. Next, Divide the Circumference for each path by the time for the weight to make one revolution for the inner and outer radius. This will tell you the speed of the weight while it traveled around the Lazy Susan.

5) Create a table similar to below to organize your data for Part 1 6) Can you achieve a larger maximum speed on a wide turn (large radius) or a sharp turn (small radius)? How does your data show this?
 * Radius (cm) || Max Speed (cm/s) ||
 * 3 (9.8) || 15.73cm/s ||
 * 14(11.98) || 73.4cm/s ||

You could achieve a larger maximum speed on a wide turn because our maximum speed is larger for a wide turn.

Hypothesis: Answer the Question posed in the Title of Part 2 and give a reason for this hypothesis. A dry surface because there is more friction.
 * __ PART 2] Can a Car Go Faster Turning on an Icy Surface or __**__ Dry **Surface?** __

Procedure: 1) Spin the Lazy Susan with the mass on the Wood Surface for 10 revolutions then spin the Lazy Susan with the mass at the same radius on the sandpaper for 10 revolutions, recording the time each takes to make 10 revolutions. 2) Calculate the time it took the Lazy Susan to make one revolution for each surface 1.321s 1.105s 3) Use the formula C=2 πr to solve for the circumference of the path that the weight traveled for each radii. Next, Divide the Circumference for each path by the time for the weight to make one revolution for the inner and outer radius. This will tell you the speed of the weight while it traveled around the Lazy Susan.

4) Create a table similar to below to organize your data for Part 2 5) Can you achieve a larger maximum speed on sandpaper (normal asphalt) or on the wood surface (slippery road conditions)? How does your data show this? We can achieve a larger maximum speed on the sandpaper because the coin can grip the sandpaper better which would cause the coin to go faster. The coin would have better friction.
 * Radius (cm) || Max Speed (cm/s) ||
 * 14(13.21) || 66.1cm/s ||
 * 14 (11.05) || 79.56cm/s ||

Hypothesis: Answer the Question posed in the Title of Part 3 and give an reason for this hypothesis.
 * __ PART 3] Can a more massive vehicle go faster turning or a __**__ less massive vehicle go faster turning**?** __

The less massive vehicle go faster turning because with the less weight, the car can travel faster.

Procedure:

1) Spin the Lazy Susan with the lighter mass on the Wood Surface for 10 revolutions, then spin the Lazy Susan with the heavier mass at the same radius on the sandpaper for 10 revolutions, recording the time each takes to make 10 revolutions. 2) Calculate the time it took the Lazy Susan to make one revolution for each mass. 1.321s 1.239s 3) Create a table similar to below to organize your data for Part 2 large || 70.99cm/s || 4) Can you achieve a larger maximum speed before skidding out in a heavier vehicle or a lighter vehicle? How does your data show this? You can achieve a larger maximum speed with the lighter vehicle because in the end, it takes a quicker time for 10 revolutions than for the lighter vehicle. A larger mass vehicle would tip it because of the distribution of weight. It has nothing to do with friction.
 * Mass || Max Speed (cm/s) ||
 * 14(13.21) small || 66.59cm/s ||
 * 14(12.39)


 * //11/15/11-Homework//**

__Centripetal Force:__ a force directed toward the center to keep an object in a circular path. __Centripetal Acceleration:__ a change in the direction of the velocity with respect to time 1. The force is pulling on the object in a straight line when moving in a circle. 2. Centripetal force keeps an object moving in a circle. 3. Friction keeps an automobile moving in a circular path on a road. 4. The velocity can change even if the speed is not changing when the car changes changes direction. 5. Acceleration can change when a car changes velocity, speed, or direction. 6. The centripetal force is gravity that keeps Earth moving in a circle around the Sun.
 * Vocab Words**
 * Checking Up Questions**


 * //11/16/11//**

__**//Change of velocity=Acceleration//**__

__**Centripetal Acceleration**__-a change in velocity not by 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-Active Physics Plus**//

1) (2)(π)(6400km)/(24 h)= 1675.52 km/h -- 465.42 m/s (rotation of axis) 2) (2)(1.5x10^8)(π)/(8766-hours) = 107515.15 km/h -- 10908307.82 m/s 3) (2)(π)(15)/(.016)= 5890.49 cm/s 4a) The car will have to slow down and the radius will be smaller. If not, the car would spin out and cause you to travel into ditch. 4b) The car might lose traction if the road is too slippery. Would spin out to the ditch. 4c) If the turn is too tight and the road is too slippery the car might spin out because there won't be enough friction. Causing the car to travel into the ditch. 5) Frisbee, baseball bat, 6) 7) Depending on the conditions of the road, the road controls your tires. The road has to have enough friction to make sure that your tires aren't sliding all over the road. The wheels of a car wouldn't be able to turn the car. The wheels help slow down or stop with no traction. 8) a=v^2/r a=(270)^2/1000 a=72.9 m/s^2 9) Yes both explanations are correct. In both examples a sharp left turn was made and the person ended up sliding to the right and hitting their right shoulder on the side of the door. 10) Friction on the road and wind resistance keep the car moving in a circular motion and pulled in towards the center. 11) They are especially dangerous because you are going fast around the wide turns and then the turns get tighter and tighter and you have to adjust your speed, but if you don't it can be very dangerous. 12) If the curve bends to the right you would end up in the land of oncoming traffic but if it bends to the left you would end up in the ditch.


 * //11/17/11-Active Physics Plus//**

Forced measured in newton N = kg(m^2/s^2) / m 1/m / 1/m
 * 1) 1-4 **Fc = m (V^2) / r**
 * N= kg(m^2/s^2)**

1. 13720N= 2000 (v^2) / 10---137200 = 2000 (v^2)68.6N = v^2-- 8.28 m/s 2. 6860N= 2000 (v^2) / 10-68600N= 2000 (v^2)34.3=v^2-- 5.86m/s 3. 20580N=3000 (10^2) / r--20580 (r) = 300000 14.6 m 4. 6000N = 2200 (5^2) / r---6000(r)=55000-- --9.17 m

5. a = (10^2) / 12--- a=8.33m/s^2 6. a = (20^2) / 12-- 33.3m/s^2 7. a = (10^2) / 24 4.17m/s^2
 * 1) 5-7 ac = v^2 / r