Chapter+1

=Chapter 1=

Section 1 9/7/11
What do you see? -In the picture, a car has to stop short because the driver sees a crash happened in front of him. -The man driving's hair is flowing behind him (it wants to stay where it was). There's smoke by the tires showing he's skidding -His car is lifting up in the back a little, probably doesn't have back wheel brakes. There is an oil slick on the road, showing that there is less friction.

What factors affect the time you need to react to an emergency situation while driving? -the amount of time you have to stop -The amount of distance -amount of friction, if your tires are bald -weather -conditions of the road, if the road is wet -How big or small your car is -if you're going up or down

__Investigation Reaction Time: Responding to Road Hazards__
__Moving Foot From Gas to Brake__ Time to move your foot from gas to break 1st -3.63 seconds--> (.363 sec, 1 motion)

__Starting and Stopping stopwatches__ -10cm - 25cm -10cm -21 10cm - 16 Average=10.6 cm

__Stopwatch Reaction:__ 6.61sec and 6.80 (.19) 6.87sec and 7.60 (.19) 7.83 sec and 8.24 sec (.24)

__Catching Ruler(Distractions)__ 10cm --> 25cm (15cm)

__Catching Ruler (Reaction with decisions added)__ 10cm --> 24cm (14cm) 10cm --> 26cm ( 16cm) 10cm --> 20cm (10cm) Average: 13.3 Seconds .15

__Catching Ruler (distractions and descisions)__ 10cm --> 22cm (12cm) 10cm --> 24cm (14cm) 10cm --> 16cm (6cm) Average: 10.6 Seconds .14 sec

Lab Questions: How much experience they've had driving or how many video games they play. -How does your reaction time with needing to make a decision compare with your reaction time without needing to make a decision? Because your brain needs to take a second to think about what you're doing, rather than just doing something on impulse, it takes at least a tenth of a second longer -**What does this difference in reaction time when making a decision apply to your ability to avoid road hazards?** It makes a difference because you may have any seconds to think about something -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? The reaction time without the distraction of texting is much quicker -What does this difference in reaction time when distracted apply to your ability to avoid road hazards while texting? When you're not texting, you have a much quicker reaction time.
 * - ** What factors may affect reaction time for people of the same age, like your classmates?

Wednesday 9/7 Homework:
1. How do distractions affect reaction time? -They slow your reaction by time by at least a tenth of a second or two. 2. Why is driving under the influence of alcohol or drugs illegal? -Because driving normally is hard enough, with phones and people and music, but if you have your reflexes slowed even more by drugs or alcohol, there is an even bigger chance of having an accident. 3. Name three factors in addition to distractions and drugs or alcohol that can affect reaction time. -1. if you have a lot of things on your mind -2. Not getting enough sleep -3. Health problems or mental problems

__ Reaction TIme: Science of the Fastball __
1. How long does it take a fastball to reach home plate? .5 seconds 2. Why if your reaction time is under .5 seconds do you still sometimes not hit the ball? because your bat may not be at the right angle 3. Explan what else might take some time before you hit a pitch besides simply reacting to it? Likewise when you are diriving what else do you need to do besides react to an obstacle before you come to a stop? Moving the bat to hit the ball, and while driving you must move your foot from the gas to the brake. 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? How many times you've played baseball, and the wind that day 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? To constantly watch everything so you don't have an accident. To be ready at all times just in case the pitcher throws the ball. 6. Why might someone yell from the dugout at a batter, "hey batter batter swing batter batter!"? To distract the batter so he misses the ball, this is similar to texting while driving because both are distractions that take your attention away from the task at hand

9/9/11 Active Physics Plus: Section 2
d=1/2at^2

d: distance an object is dropped (measured in cm) t:The time it takes the object to drop "d" a: acceleration of the object (__980 cm__ s^2

units of each: d= cm a=? t= sec^2

cm=cm/s^2 X s^2


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

Reaction time ruler: for homework, put masking tape on the ruler, start at .3 cm, write on the masking tape where .025 seconds is and so on.
 * Distance || Time ||
 * .3 || .025 ||
 * 1.23 || .05 ||
 * 2.76 || .075 ||
 * 4.9 || .1 ||
 * 7.66 || .125 ||
 * 11.03 || .150 ||
 * 15 || .175 ||
 * 19.6 || .2 ||
 * 24.81 || .225 ||

DO NOW: Both need to have really good reaction times because a race car will take longer to stop since racecars are going at a faster speed than someone driving in a school zone, but also, if you're driving in a place where kids are running around, you always have to be on alert just in case a kid comes running out in front of you. Going faster just means you cover more distance in a same amount of distance.

__Reaction Time Ruler Note__

Can you catch a dollar bill? Dollar bill: d=15.4cm Reaction time needed to catch a dollar bill: d=1/2at^2 15.4=1/2(980cm)(t^2) 15.4=(490cm)t^2 15.4/490=t^2 .0314s^2 =t^2 t= .177sec^2 .177s

Homework 9/12 23.5=1/2(980)(t^2) t=.69 seconds

6. If someone's reaction time is very slow, driving could be hazardous. They could hit someone because they didn't step on the brake fast enough or they could crash into another car. Someone with a slow reaction time is going to go a greater distance than someone with a fast reaction time who is going the same speed 7. Teenagers aren't used to rhythm of driving, they've only been behind the wheel for a few months so of course insurance is going to be a lot more expensive for teenagers than for people who have been driving for many years. Teenagers also talk on their phones a lot more than adults, they're also more likely to drive drunk.

What are the top two causes of accidents on the road? -Rubbernecking and driver fatigue What is rubbernecking? -Rubbernecking is when there's a crash on the side of the road and everyone slows down to take a look at it and causes traffic. Its a distraction

9/14 Active Physics Plus
Standard Form ax^2+bx+c=0 ax^2+0x+d=0 (x,y) (t,d) (.02,.196)

d= 1/2 (980)(.02)^2 __.196cm =.196cm__ .784cm=(1/2)(980)(.04)^2

.784cm=.784cm

Two factors effect falling: The heavier it is, the less it is affected by the air Air resistance, surface area. with no air, everything falls at the same rate.

Section 1 Quiz: -Know how to measure time to move your foot -know how to measure reaction time with stop watches -know how to measure reaction time dropping ruler -know how to measure reaction time with descions and distractions -Alcohol/drugs, driver fatigue and rubbernecking -Talking on cell phone---> distractions -Texting If you're going faster, you're covering more distance in shorter amount of time. -How fast you're going, doesn't change your reaction time :) Know how to do this: d=1/2at^2

=Section 2= Grade:

Objectives
-Calibrate the length of a stride -Measure a distance by pacing it off and by using a meter stick -Idenity sources of error in measurement -Evaluate estimates of meausrements as reasonable or unreasonable

In the picture, one kid is measuring the stride of a teenager and a little girl. So they can figure out how long it takes each to get somewhere?

One of them could have used the wrong units WDYThink questions are incomplete. Please include the question in your answers.

Investigate:
Stride 1: 19 Stride 2: 19 Stride 3: 17

Stride Average: 18.3

Stride length average: .56 meters

Estimated Length of area: 10.25 meters

13.5 meters

9/16

12.09 meters || 13.61 || 10.8m || 13.2 || 13.34 || 13.3 || 11m || 14 || 15.54 || 13 || 10.25 || 13.5 || Average Stride Measurement 12.17 Average Meter-stick Measurement 13.33
 * Group || Strides || Metersticks ||
 * 1 || (13 strides)(.93meters
 * 2 || (20strides)(.54m)
 * 3 || (18)strides (.74m)
 * 4 || (22strides) (.5m)
 * 5 || (21 strides)(.74m)
 * 6 || (18.3strides) (.56m)

WIKI: 1) Do the measurements listed on your class table agree? -for the strides no, but for the meter stick measurements yes.  2) By how many meters do the results vary? -For the strides by 2 or 3 meters, for the metersticks by a few centimeters 3) Why are there differences in the measurements made by different groups? List as many reasons as you can think of -There are differences because each person has a different stride, and strides change as you walk so there numbers are not exact. Most strides are not consistent  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. -Everybody uses the same method, and make sure that everybody is very exact 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? -I think the measurements were be a lot more exact, but there are always errors

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 -Strides will always have random error, and no measurement can ever be completely exact every time

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) -Strides were not systematic errors because you can't fix them with a calculation, like using a yardstick instead of a meter stick.

Demo- How long is the tube? 14 sided ruler
.75 meters .75 .82 .75 .72 .71 .78 .76 .73 || .1m .84 .85 .84 .82 .82 .84 .82 .81 .83 || .01m .81 .81 .81 .81 .81 .81 .8 .82 .81 .815 .81 || .001m .815 .814 .8145 .813 .815 .814 .812 .812 .813 .812 ||
 * Interval || 1meter

Homework for 9/19

Physics: The study of matter and energy and the world around us.

Explain the difference between systematic errors and random errors. -Systematic errors are errors that you can fix, like if you measured a length in terms of centimeters and really had to measure it in meters. It would be easy to convert the units. A random error would be if someone measured a length by a stride and found that the real length was different than what they measured, there is no way to know how you got the wrong measurement since strides can be very random.

Explain why there will always be an uncertainty in measurement? There will always be uncertainty because each person who measures something does it different. One might start measuring a centimeter before or after another person. It is impossible to overcome human error in these situations.

What would the positions of arrows on a target need to be to illustrate measurements that are neither accurate nor precise? The arrows would have to be spread out over the target in a very random way, with none of the arrows hitting the bullseye.

SI system
 * Quantity || BaseUnit || symbol ||
 * Distance || Meters || m ||
 * Mass || gram || g ||
 * Time || seconds || s ||

1m= .001km || X .01 || 1cm= .01m 1m= 100 cm || X .001 || 1 mm=.001m 1m = 1000m ||
 * prefix || Symbol || MultofTen || Exp ||
 * kilo || k || 10^3 X 1000 || 1 km=1000m
 * Centi || c || X 10^-2
 * Mili || m || X 10^-3

Physics To Go: 3. A value that my friend and I would agree on would be that there are 4 reese's pieces in a package. An estimated value that my friend and I would not agree on is that there are about 30 m'nms in one package 4. An oil tanker holding five million barrels of oil sounds like a great estimate. There might be a hundred more or a hundred less than five million barrels of oil, which would change the value of the oil on the tanker by a lot. 6a. I do not think this is a very accurate estimate. 2Ls only make 8 cups. If there are 12 people, some may not get any 6b.Yes, this is an accurate statement because a normal car holds about 20 -30 gallons of gas, and gets 25 miles to the gallon, so a car could easily travel this distance. 7. No, if you're off by one meter in your room, then everything is going to be out of proporation, because a person's bedroom is such a small space. If one is off by a meter when measuring the miles between the school and one's house, then that wouldn't make that much of a difference because one meter doesn't amount to much in such a long way. 8a. If one wants to make sure that they do not exceed the speed limit, then they would have to go 60 as shown on the speedometer to make sure that they don't go over 65, since the error can be up to 5 mph. 8b. If the passenger wants to see how off the speedometer really is, then they can use their watch to see how like it takes them to go a mile. 60miles an hour = 1 mile a minute 9.I believe that no one should get a ticket for going 1 mph over the speed limit. First of all, going 1 mph over the speed limit doesn't really change how fast one is going, its too close to actually have a palpable difference. Second of all, all speedometers have a margin of error, most are between 1 mph and 5 mph, a driver may think that they are going the exact speed limit, as it says on their speedometer, but the police officer's speedometer may say differently. In my opinion, a driver should not be punished for having a faulty speedometer.

-If I was buying gas by the gallon, i'd want it to be very accurate because, if they're gonna carry out their price to the second decimal, then their measurements should be precise to the second decimal place. If I was buying carpeting by the yard, i would need it to be very accurate also, otherwise it wouldn't fit in my house
 * 1) 1 If i was buying vegetables by the pound I wouldn't expect it to be that accurate. I'd want it to be at least as accurate to the second decimal place.

-there was as systematic error for all the odd groups because they didn't start measuring from the right place on the ruler.
 * Measuring a Copper Tube:**
 * Groups || Measurement of copper tube ||
 * group 1 || 66 cm ||
 * Group 2 || 64.15 cm ||
 * Group 3 || 66.1 cm ||
 * group 4 || 64 cm ||
 * group 5 || 66 cm ||
 * group 6 || 64.1 cm ||

1. What is the range of lengths for 50-m pools that have an uncertainty of + -10 cm? + -1cm? +-1mm? 10 cm X.01m + - .1 m--> 50.1m-49.9 m
 * Active Physics Plus:**

+ - 1cm--> 50.01 m-49.99 m

+- .001m-->5.001 m-49.000 m

+- 1 cm 50.01-49.99 5m/2s=2m/5

1500 m/15 min(60 sec)=1500m/900 seconds= .5 m/Ts 1 person goes: 50.01 X 30 =1500.3 meters other person goes: 49.99m X 30= 1499.7 m 1500.3 m/15.36 s= 97.68m/1s (speed per second for person swimming longer pool) 1499.7 m/15.35= 97.7m /1s (speed per second for the second person swimming shorter pool)

What does it mean? -This error is systematic because one can fix the measurement with an equation. systematic effects accuracy, random affects precision How do you know? -A jeweler cannot be absolutely sure of this because no weighing device is that accurate to know that it is exactly 1 oz

Answer Q: This wouldn't work because there is no way for a radar gun to have that much of a margin of error.

__Estimation Activity:__ -true -cannot be true -24 hours a day, not true -poodles can't be 120lbs, not reasonable -LxWxH= 45,000ft. The book's estimate is not true not reasonable -This is not reasonable. -no, you would not be able to pass it time __50mi__=__(SO)4(1/4mi)__= __200(1/4)__ 1 hr 60min 60min -You would have to estimate how fast its going to see if you would make it across the bridge -Yes, you would be able to make it under the bridge.

=Section 3= missing learning outcomes || show calculations in #8 || missing following jack do now ||
 * ** Section3 ** || **Points** ||
 * WDYSee/Think: || 7/10
 * Investigate: || 19/20
 * PhysicsTalk: || 20/20 ||
 * PhysicsPlus: || 17/20
 * PhysicsToGo: || 20/20 ||
 * Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**92** ||

Average Speed: Following Distance and Models of Motion. What do you see? -A rear-end collision, because on car is distracted. What is a safe following distance? -About a car for every ten miles an hour. or the 3 second rule; 3 miles between you and the car in front How do you decide what a safe following distance is? -Your speed or the car in front of you's speed.

INVESTIGATION-
-Strobe Photos-sends out a flash at equal amounts of time, a strobe photo is a combined photo of pictures taken at equal intervals of time. -hook up motion detector to USB port -Plug in USB port -Open Datastore

2b. The image had the same amount of space inbetween it, but were farther apart than when it was going 40 mph. 2c. They're farther apart because the car is going faster and covering more distance. 3a. In diagram A, its the fastest one, in diagram C, its the slowest one. 3b. They're all traveling at a constant speed.

Walking to it at a normal speed walking away at a slow speed Walking away at a normal speed



Walking away and then walking toward Walking away and toward at a high speed. 4e. When someone walked toward the motion sensor, the line goes up and when someone walks away, the line goes down because the signal the motion sensor sends out doesn't have to go that far if it bounces off someone closer. 5a. If someone walks toward the motion sensor at a slow speed the line will gradually go up and then the line will shoot back down when they walk toward it at a fast speed 5b. Walking away slowly and then walking toward very fast Walking away fast Walking away slow -The faster you walk, the steeper the slope, a gradual slope is slow speed, a steep slope is a fast speed 6b. If someone forgot to label the lines you could tell which one was the one someone walked fast by how sharply the line curves downward 7a. we walked about 2.5 meters according to the graph 7b. It takes about 4 seconds to walk the distance in most of the graphs 7c. .625 = average speed 7d. If someone doubled their distance and time, then you could estimate how far they walked 8a.The automobile travels 30 feet in half a second 8b. The car will travel 90 more feet if the reaction time increases 8c. the car will travel 25 feet in half a second and 100 more feet in 1.5 seconds 8d. 35 feet per .5 seconds and 75 feet more when the reaction time increases 8e. The car should be 20 feet at least behind the other car. 8f. about four car lengths is how much one car travels at 60ft/s

__Physics Words:__ -Speed- the distance traveled per unit (time 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 the distance -Instantaneous Speed- the speed at a given moment -Reaction distance- the distance that a vehicle travels in the time it takes a driver to react -Velocity- the speed in a given direction

__Checking up Q's :__ 1. Explain how the average speed of a vehicle is different from instantaneous speed -The average speed is the distance divided by the time, this will give an estimate on how fast someone is going while intantaneous speed changes and could be different at different times. 2.How are the speed and velocity of an object different? Speed is just a general term for how much distance someone has covered in an amount of time while velocity is the speed in a specified direction 3. If the distance-time graph shows a straight, inclined line, what does the line represent? The line represents This person is traveling at a fast speed 4. How does reaction time affect reaction distance? If someone's reaction time is low then they will have a shorter reaction distance 5. What is the difference between constant speed and average speed?

9/27/11

V(av)= 90ft/s delta(t)= 6s delta(d)= x

90ft/s=6/x 90(.6) = 54ft

90(1.5) 135ft

Delta(d)=200m Delta(t)=1.27-1.11= 16 sec Vav= 200m/16s= 12.5m/s = 25mph = = 20 mph || 40 mi || ∆t 1 =? || Trip V av = ? || 80 mi || ∆t || Vav= __20mph+ 40mph__= 30 mph 2
 * Part || Distance || Time ||
 * Part1
 * Part 2 40mph || 40mi || ∆t 2 =? ||
 * Whole

Part 1: Vav= __∆d__ ∆t 20mph= __40mi__ ∆t1 ∆t1= 2 hours

Part 2: __∆d__ ∆t 40mph= __40mi__ ∆t2 ∆t2= 1 hour

Vav= __80mi__ = __80mi__ = 26.67 mph ∆t(total) 3 hours

Make a Table of the Trip 1mph || 50mi || 50 hours || Trip || 100 mi || 51 hours || Vav= __100mi__= 1.96mph 51
 * Part || Distance || Time ||
 * part1
 * Part 2 50mph || 50mi || 1 hour ||
 * Whole

Part 1 - - - - -- - - - -- - - - - - - - - - - - - - - <> 40 mi Part 2 - - - - - - -- - <-> 40mi Whole Trip: - - - - -- - - - -- - - - - - - - - - - - - - - - - - - - - -- -

Part 1 - - - - -- - - - - -- - - - - - -- - - -- - - - 50 miles (1mph)

Part 2 - 50 miles (50 mph) Whole Trip: - - - -- - - - - -- - - - - - -- - - -- - - - -

about 5 mph average speed 6.25 average speed
 * 1) 3 8 hour trip

__Physics to Go__ 1a. The car is moving at a slow and steady speed 1b. The car sped up and then slowed down, changing its speed. 2a. --- --- --- --- --- --- (car starting from rest and then reaching its final constant speed) 2b. --- --- --- --- --- --- --- --- --- (car coming to a stop from a constant speed) 3. The man drove 7,000 feet in 20sec, (350 X 20) 4a. 47.78 mph 4b. Its impossible to know how fast the woman was going when she went through Baltimore because we can only find out her average speed 5. One would have to bike 20 mph 6a. The vehicle went fast (it drove away) and then stopped 6b. The vehicle went very fast, driving away, stopped and then drove back very fast 6c. The vehicle drove away slowly, and then increased its speed and drove very fast 6d. The vehicle is driving slowly away 7a. My reaction time is .19s so the vehicle would go 4.75 meters in my reaction time 7b. The vehicle travels 3.04 meters in my reaction time, 1.71 meters less 7c. .38 would be my reaction time, so I would go 9.5 meters. 8a. Experts can use seconds as a way to tell people how close they can be to the car in front of them because they know that normal reaction times are below three seconds so its safe to use this unit. 8b. No, because on a highway you are going much faster than on a small road, so people would have to have more distance between them and the car in front of them. 9a. The vehicle will go 33.33 feet in the time of a sneeze 9b. yes, it is longer than my classroom. 10a. 44 feet 10b. about 3 automobile spaces 10c.22 feet, and about 1 and a half car spaces 10d. 66 feet, about 5 automobile spaces, and it would be about 1/8 of a football field 10e. in one second, the car goes 44 feet at 30mph, the car goes 88 feet at 60mph and 134 feet at 90mph 11.As the reaction time goes up, so does the distance traveled ( i dont know how to make a graph on here)

10/3/11 Following Jack: Part1 -I believe that the drips show that Jack started walking slowly, then walked faster and continued to walk at that same speed because the drips are closer together at the beginning and then get farther apart, but the drips are equidistant at the end. Following Jack: Part 2 -The graph D corresponds to the drips because the equal spacing of the dots at the end says that we will have a straight line in the graph. __Distance VS Time Graph__ 4a. How does the motion of the cyclist A in this graph compare to that of A in question 3 -They're almost exactly the same. 4b. How does the motion of cyclist B in this graph compare to that of B in question 3? -In this graph B starts at a much farther and higher distance than A and then goes back 4c. Which cyclist has the greater speed? How do you know? -A has the greatest, constant speed 4d. Describe what is happening at the intersection of lines A and B. -A and B are going the exact same speed at 5s 4e. Which cyclist has traveled further during the first 5 seconds? How do you know? -A has traveled further

10/3/11 Walking the Graph For this graph, we were close and stayed still for the first few seconds and then jumped backward and stayed still for three more seconds

For the graph, we were far away and stayed still for 3 seconds and then jumped forward and stayed still for 3 seconds For this graph, we stayed close, for a few seconds, moved far away really fast, stayed still for a second, and then moved back really fast and stayed there. For this graph we were far away, then moved close really fast, stayed still for a second, and then moved back really fast and stayed still.

=SECTION 4= <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing a run || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing #9 || What do you see, what do you think? -The car's front is up because he's speeding up. A man is running out of the way, maybe the man is accelerating to make the light
 * <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;">10/10 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">15/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">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;">19/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">20/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**94** ||

-An automobile and a bus are stopped at a red light. What are some differences and similatries? A bus takes longer to accelerate, because it is heavier, The quicker it takes to get up to speed, the quicker the acceleration. Acceleration- change of velocity in a period of time Tangent Line- A line that only touches a curve at one point -Each tangent line slope tells us the instantaneous velocity at a specific point in time

Learning Outcomes: Measure- a change in velocity(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

__Investigation__: 1. No, because as it rolls down the track, gravity pulls on it and it accelerates.



The graph on the bottom left moves at a constant velocity The graph on the bottom right does not move The graph on the top right starts off fast and then slows down The graph on the top left starts off slow and then speeds up __Run1 Prediction:__

(it wouldn't let me put the tangent at 1.5 seconds, so this is at 1 second) Calculate Acceleration of Run 1: a= ∆v/∆t a= 1.1-1/ 2.5-1 a=1/1.5 a=.67



10/6/11 Acceleration- the change in velocity with respect to a change in time Vector- a quanity that has both magnitude and direction Negative acceleration- a decrease in velocity with respect to time. The object can slow down (20m/s to 10m/s) or speed up ( -20m/s to -30m/s) or slow down (-20m/s to -10m/s) Tangent Line- a straight line that touches a curve in only one point 1. Give the defining equation for acceleration in words and by using symbols? -__change in velocity__ change in time the sym= a= ∆v/∆t 2. What is an SI unit for measuring acceleration? Use words and unit symbols to describe. <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">a length with a measure of time= m/s <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">3.What is the difference between a vector and a scalar quanity? <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">-something has to do with direction and size is a vector quantity and something that has size but not direction is scalar quanity <span style="font-family: Tahoma,Geneva,sans-serif; font-size: 12px;">4.Sketch a distance-time graph for constant Velocity and constant acceleration 5.What does the slope of a velocity-time graph represent? How something's velocity changes in relation to time

10/7/11 1.a= 30-0/5-0 a=6 2. Vectors: velocity, a force (push or pull) Scalar: speed (doesn't have direction), calories, anything you can count

__Car VS Pick-up Truck VS bus__ Instantaneous Velocity at 2, 3.5 and 2.5 seconds^ Speeding up Instantaneous Velocity at 2,3.5, and 2.5 seconds^ Slowing down.

Slowing Down a=__∆V=__ __V3.5sec- V2sec__ = __4m/s-.25m/s__ = .__15m/s__ = .1/ m/s^2 ∆t 3.5s-2s 3.5s-2s 1.5 s Speeding Up a= __∆V__= __.35-.85__= __-0.5m/s__ = -0.3m/s^2 ∆t 3.5-2 1.5s

a=slope a=is constant Gravity does not shut off d= 1/2at^2 + Vit

ACTIVE PHYSICS PLUS
1. a= __vf-vi__ ∆t Vf= 20m/s Vi=7m/s a=3m/s ∆t=?

3m/s= __20m/s-7m/s__ ∆t __3m/s(∆t)__= __13m/s__ 3m/s 3m/s

∆t= 4.33 seconds

2. a= __vf-vi__ ∆t 1.5m/s^2= __vf-7__ 10s 15=vf -7 vf= 22

3.a= __∆v__ ∆t a= __20-0__ a= 4m/s 5 d= 1/2at^2+Vi(t) d= 1/2(4)(5)^2+0(5) d= 1/2(4)(25) d=50m

40-0 10 d=1/2at^2+Vi(t) d=1/2(4)(10)^2 d=1/2(4) (100) d=200

__Physics to Go__ 1. Yes, but this can only happen when a vehicle is traveling at a constant speed. 2. No, because acceleration is what happens to velocity, it is impossible to have acceleration without velocity 3. No, the vehicles could be accelerating at the same rate, but one might be going slower than the other 4. Yes, because if two vehicles are going the same velocity, that means that their speeds will be constant and there is no way for one to accelerate faster than the other. 5. Yes, no matter how fast the constant velocity car is going, the other car can always go faster 6.Yes, because the speed limits are set just according to how fast the cars are going, it doesn't matter what direction and MPH 7a. a= (24s)2 a=48 mph 7b..8 miles 8a. a= __∆v__ ∆t a=__75-0__= 25 9 3 a= 8.33 8b. Vav= ∆d ∆t Vav= 4 8c. 675 meters 8d. a= 9.37 Vav=9.37 d= 600 meters

9a. a= __∆v__ ∆t a= __.6-4.5__ 1.3 a=-3 9b. d=1/2at^2+Vi(t) d=1/2(-3)(1.3)^2+.6(1.3) d= -2.535+.78 d= -1.76 9c.a=__.6-4.5__ 1.1 a=-3.55 9d.the one described in 9c 10a. 12mps 10b. a= __12-0__ 9 a= 1.3 10c.The final velocity would be a lot faster 11a. Graph B 11b. Graph D 11c. Graph E 11d. Graph C 11e. Graph F 11f. Graph C 12a. BC 12b. BC 12c. DE 12d. FG 12e. 1,000 meters 12f. The car was back at the place where it started 13a. __250-0__ 30sec 8.33 m/h 13b. 8.33= __vf-0__ 45sec 13c. in another 30s the plane will be at 500mph 13d. d=1/2at^2+vi(t) d=1/2(8.33)(30)^2 d= 12455 14a. Vi=4.5 9.8(4.5) a=44.26=Vf 14b.100 seconds 14c. vav=980 10 vav= 98 14d. 980m 14e. 1.6= __∆v__ ∆d 1.6= __vf-0__ 100 16=vf 15a. a= __632-0__ 1.4 a= 451.43 15b. __9.8__=201.615 = 20.573 15c. d=1/2at^2+Vi(t) d=1/2(451.43)(1.4)^2 d= 442.40m

16a. 4m 16b. d=1/2at^2+Vi(t) d=1/2(4)(2)^2 d=8 16c. d=1/2(4)(3)^2 d=18 16d. d=1/2(4)(4)^2 d=32 16e and 16f 16g. it is very similar

ACTIVE PHYSICS PLUS

1) 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? d=1/2(-7)9+63 d=31.5 no you will not hit the deer.

SECTION 5
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing WDYthink || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing some units in many of your calculated answers || 10/18/11 What do you see? -The front wheels are down and the back wheels are up. -How fast you're going will determine if you can stop in time -Friction, the weight of the car -An old car or a new car
 * <span style="font-family: 'Courier New',Courier,monospace;">** Section4 ** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 20px;">**Points** ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">WDYSee/Think: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">7/10
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">20/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">20/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">16/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">20/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**93** ||

Learning Outcome: -Plan and carry out an experiment to relate braking distance to initial speed -Determine braking distance -Examine accelerated motion

Objective: To determine the effect of initial speed on braking distance.

Copy into excel, make a graph of different trials. Vi on the bottom, braking distance y axiz title it braking distance VS Vi and make it a scatter plot IS the increase exponential? To get the initial velocity: all cars have a flag that will touch the photo gate, so the laser will tell us how long it was broken for. if you measure the flag, and how long it breaks the laser, divide the two and that's your initial Vi=l__ength of flag__ time in gate
 * Vi || Brakingdistance ||
 * One Textbook .59 || 2.8m ||
 * Two Textbooks .94 || 5.15m ||
 * Three Textbooks 1.27 || 11.32m ||
 * Three Textbooks 1.26 || 11.3m ||
 * Three Textbooks .8 || 4 ||
 * Three Textbooks .97 || 5 m ||
 * Three Textbooks 1.23 || 11 ||

vi= .10cm .0017 cm

.10 .00106 .94 cm .10 .00079 .10 .0079 .10 .00094 .10 .00081

D= 1/2at^2 t^2 ~ d x2~x4 x3~x9

Vi^2~ Braking Dist X2~x4 etc .59 X2 1) If initial velocity is doubled by what factor does braking distance increase. It increases by 4 2) If initial velocity is tripled by what factor does braking distance increase. It increases by 9 3) If initial velocity is quadrupled (x4) by what factor does braking distance increase. It increases by 16 4) Do #8 in book on page77 b.209/118 1.77 or 1.33^2 5) If the sports car changed its speed to 30 mph what do you expect its braking distance to be? (//Hint: if you half the speed by what factor will the braking distance change?)// 60/30= 2 30x1/4 7.5

6) If the touring sedan changed its speed to 30 mph what do you expect its braking distance to be?

7) If each car (sports car and touring sedan) decreased its speed to 15 mph what would their braking distances be? (//Hint: if you quarter speed by what factor will the braking distance change?)// 8) 1.33 80-60 118-208 209/118 1.77 1.33^2 1.77

Negative Acceleration- a change in velocity in respect to time of an object by decreasing speed in the positive direction or increasing speed in the negative direction. __Checking Up Questions-__ 1.It has undergone negative acceleration because for a few seconds, it was going slower than it was before 2. If the vehicle is going very fast it has a lot of force behind it so it would take longer for it to stop than a car going slower 3.Because deceleration is when a car is running out of energy, negative acceleration is when something is happening to the car to make it go slower,
 * __//H//omework__** 10/19/11

10/20/11

__Vi(2__= ratio of inital velocities Vi(1)

__Braking Dist (2)__= ratio of braking distances Braking Dist (1)

__(Vi(2))__^2= __Braking dist(2)__ (Vi(1)) Braking dist(1)

__60mph__^2= 1__23ft__ 30mph BDci

3600D=3690

D=30.75

3600 123 100 BD 12300=3600BD

3600= 123 400 BD 49200=3600BD BD=13.67

3600=123 1600 BD 3600BD=196800 BD=54.67

3600= 123 2500 BD 3600BD=307500

3600=123 4900 BD 3600BD=602700 167.42

3600=123 6400 BD 3600BD=787200 218.67

3600=123 8100 BD 3600BD=996300 BD=276.75

3600=123 10000 BD 3600BD= 1230000

4) If Initial Velocity is doubled how does stopping distance change? increases by 4x 5) If the Initial Velocity is multiplied 4 times how does the stopping distance change? increases by 16 6) If the Initial Velocity is halved how does the Stopping Distance Change? It would be divided by 4 7) If the initial Velocity is quartered how does the stopping distance change? it would be cut by 1/16th 8) What speed would you need to have a stopping distance of a mile 393.1 mph 60mph= 123ft ? mph 5280ft 3600= 123 X^2 5280 19008000=123X^2 X^2=154536.59 X=393.11
 * Initial Velocity || Stopping Distance ||
 * 10 mph || 3.42ft ||
 * 20 mph || 13.67 ||
 * 30 mph || 30.75 ||
 * 40 mph || 54.67 ||
 * 50 mph || 85.42 ||
 * 60 mph || 123 feet ||
 * 70 mph || 167.41 ||
 * 80 mph || 218.67 ||
 * 90 mph || 276.75 ||
 * 100 mph || 341.67 ||

__Direction of Acceleration__ <- -> positive negative

Same direction= speeding up Different directions= slowing down __Equations of Motion__ if: Vav= __∆d__ t Then: Vav( t) = ∆d

Vav= __vi+Vf__ 2 Vav=1/2(vi+vf)
 * IF constant acceleration is present

∆d=1/2(Vi+Vf)t -> 1st equation of motion

If: a= ∆v ∆t a=vf-vi ∆t Vf=vi+at > 2nd equation of motion

If: ∆d= 1/2(Vi+Vf)t vf=vi+at ∆d=1/2(vi+vi+at)t

d= 1/2(2vi+at)t d=1/2(2vit+at^2) d=Vi(t)+1/2at^2 -> 3rd equation of motion

d=1/2(vi+vf)t (a) Vf=vi+at (d) d=Vi(t) +1/2at^2 (Vf) Vf^2=Vi^2+2ad (t) Vav=∆d (only used in constant velocity) t EQUATIONS OF MOTION ^

__Physics to Go__ 1. As Initial speed goes up the distance to stop goes up, they are proportionally related 2. Automobile A is safer because it has a shorter stopping distance. 3a. Vf^2=2ad+vi^2 0=2(20)a+30^2 0=40a+900 900=40a a=22.5 0=2(22.5)d+15^2 0=45d+225 225=45d d=5 3b. 0=45d+60^2 3600=45d d=80 3c. 0=45d+45^2 2025=45d d=45 3d. 0=45d+75^2 5625=45d d=125 4. speed=10m/s Reaction time=.9s d=30m

V=d/t vt=d (10m/s)(.9)= d 9m=d

5. don't have the info for this one 6. don't have the info for this one 7. don't have the info for this one

8. What do you now know about stopping that will make you a safer driver? -total stopping distance=d(reaction)+ dBraking V=__dreaction__ treaction Total stopping dist=vtreaction+dbraking Vf^2=Vi^2+2ad(braking) 0m/s=vi^2+2(negative) (braking) -vi^2=2(-aneg)(dbraking)

Total Stopping Distance=Vt(react+vi^2/2a

-That reaction time is a big factor on if you will get into an accident or not and how as speed goes up so does the amount of distance need to stop.

__Active Physics Plus__
10/24 PTG #5,6,7 d=1/2(Vi+Vf)t Vf=Vi+at d=Vit+1/2at^2 Vf^2=Vi+2ad

__Average Acceleration__ 1. a=-4.1m/s^2 t=? d vi=9 Vf=0

Vf=Vi+at 0=9+-4.1t -9=-4.1t t= 2.2s

2) a=2.5 t=? d vi=7 vf=12

Vf=vi+at 12=7+(2.5)t 5=2.5t t=2

3) a=-0.5 t=? d vi=13.5 Vf=0

Vf=vi+at 0=13.5+(-0.5)t -13.5=-0.5t t=27

4) a=? t=1500min d vi=-1.2 vf=-6.5

Vf=vi+at -6.5=-1.2+1500a -5.3=1500a a=-0.004

5a) 5(60) 300(.0047) 1.41m/s^2 5b) Vf=vi+at Vf=1.7+(.0047)(300) Vf=3.11m/s^2

__Displacement with constant uniform acceleration__ 1) a t=3.65 d=? vi=23.7km/h=6.58 vf=9.89

__23.7km__ = 1,000m 1hr 1min = 6.58 h 1 km 60min 60sec 23.7=1,000 m/s = 6.58

d=1/2at^2+vit d=1/2(.92)(3.6s)^2+(6.58m/s)(3.6s) d=29.7m

2) d=1/2(vi+vf)t d= 1/2(15-0)2.5 d=1/2(15)2.5 d=18.75

3) d=1/2(-5+0)(100) d=1/2(-5)(.8) d=-2

4) d=1/2(vi+vf)t 101=1/2(78+0)t t=2.59

5) d=1/2(vi+vf)t 3.2=1/2(6.4+vf)210 3.2=3.2 + vf1/2 3.2=672+105vf -68.8=105vf vf=-0.66

__Velocity and displacement with uniform acceleration__ 2. a=3m/s^2 t=5s d=? vi=4.3 Vf=? Vf=4.3+(3)(5) vf=19.3 d=1/2at^2+vit d=1/2(3)(25)+4.3(5) d=37.5+21.5 d=59

3. a=-1.5m/s t=5s d=? Vi=0 Vf=? Vf=vi+at Vf=0+(-1.5)(5) Vf=-7.5

__Final Velocity after any Displacement__ 2. a=.8m/s^2 t= d=245 Vi=7m/s Vf=?

Vf^2=Vi^2+2ad Vf^2=7^2+2(.8)(245m) Vf^2=49+392 vf^2=441 vf=21m/s

2b. Vf^2=vi^2+2ad Vf^2=(49)+2(.8)(125) Vf^2=249 vf=15.78

2c. Vf^2=vi^2+2ad vf^2=49+2(.8)(67) vf^2=156.2 vf=12.498

3. Vf^2=0+2(2.3)(55) vf^2=253 vf=15.91

4. d=86.5 5. a=2.27

6. Vi=6.5m/s W+ E- Vf=1.5 a=-2.7m/s^2 d=? Vf^2=vi^2+2ad 1.5^2=6.5^2+2(-2.7)d 2.25=42.3-5.4d -40.1=-5.4d 7.42=d

Practice Conversions: km/h to m/s 1__00km__ x __1000m__ x __1k__ x 1min h 1km 60min 60s

or just divide 100km by 3.6 to get to m/s

25km/h= 6.94m/s 220km/h= 61.1m/s

TOTAL STOPPING DISTANCE; raction distance + braking distance Vi= intial velocity before braking treaction=reaction time a= acceleration A=good brakes a=bad brakes

if Vi^ then Total Stopping Distance ^ Vi~TSD if treaction^ then TSD ^ treact~TSD if a^ then TSD (goes down) TSD~1/a

__Total Stopping Distance Activity__ __#1__ 2) There is negative acceleration in the diagram above because we're slowing down. Our velocity is positive, so that means that the acceleration is negative and we are slowing down.

__Calculating Reaction Distance__ __1)__ No acceleration during dreaction so: (t)v=d/t (t) vt=d vit=dreaction 2) An older person is driving a buick up to a yellow light. How far does the person go while deciding if they should break (reaction distance) They are traveling at 10m/s and have a reaction time of 1 sec. 10(1)=dreaction dreaction=10m/s 3) 20(1) dreaction=20 4) An increase in speed changes the reaction distance, because the faster you go, the longer your reaction distance will be When you're going slower, it doesnt take as long to stop

Calculating Braking Distance Vf^2=vi^2+2ad 0=vi^2+2adbraking -vi^2=2adbraking d=-vi^2/2a

5) Vf^2=vi^2+2ad 0=51^2+2(-11)(d) 0=2601-22d -2601=-22 -118.2

Mini Chapter Challenge -Paragraph explaining your storyline -3 ways you will use equations in the storyline -3 ways you'll use graphs in your storyline
 * Email text to Kibala for 1,2,3

**Mini Challenge Lab October 27, 2011** To convince our parents to let us drive the family car, we have to explain to them the safety of the car and how great of a driver we are. We will explain how the car accelerates, the velocity when speeding up and slowing down and the stopping distance. The younger we are, the faster reaction time we have in case of emergencies. Our story line will consist of examples of situations when we had to have fast reaction time or really know how well the car accelerates or how we are able to think on the spot.

Equations:
 * 1) We can see how fast the car accelerates while speeding up; we can see the cars stopping distance while slowing down.
 * 2) We can test our reaction time when having to make emergency decisions.
 * 3) We can make a final destination and calculate the distance of driving.

Graphs:
 * 1) Reaction time
 * 2) How the car is safer then others
 * 3) Teen accidents compared to Adult accidents

Story Line:
 * Taking a trip to the Jersey Shore
 * I want to drive the car because its safer then other cars
 * I can’t be drinking if driving
 * I should drive over friends because I am a safer driver
 * You should trust me over my friends (you don’t know how they drive)
 * I have a fast reaction time and I don’t get distracted easily

__Calculating Braking Distance__ a=-11m/s vi= 51m/s vf=0 d=? vf^2=vi^2+2ad 0^=51^2+2(-11)d 0=2601+ (-22)d -2601=-22d d=118.23m

6. a=-24 vi=51 vf=0 d=? vf^2=vi^2+2ad 0=51^2+2(-24)d 0=2601 -2601=-48d d=54.1

Preparing for the Chapter Challenge 1) Tr=.5sec vi=11m/s a=-4m/s dr=?

v=dr/tr

11=dr/.5 dr=5.5m Vf^2=vi^2+2ad 0=11^2+2(-4)db 0=121+(-8)db -121=-8 db=15.13

TSD=dr+db = 20.63

2) v=dr/tr tr=.5 vi=27 a=-4 dr=? v=dr/tr 27=dr/.5 dr=13.5

Vf^2=vi^2+2ad 0=27^2+2(-4)db -729=-8db 91.13=db TSD=dr+db TSD=104.6

3) tr=1 vi=27 a=-4 dr=? v=dr/tr 27=dr/1 dr=27 Vf^2=vi^2+2ad 0=27^2+2(-4)db -729=-8 91.13=db TSD= 118.13

4) tr=.5 vi=27 a=-2 dr? v=dr/tr 27=dr/.5 dr=13.5 Vf^2=vi^2+2ad 0=27^2+2(-2)d -729=-4d db=182.3 TSD=182.3+13.5 TSD=195.75

Create A Problem -Solve for dr, db,TSD -Embed givens in the problem

My Problem:

You are driving a car on the highway at 100 m/s. Its late and you're tired and your reaction time is 2 seconds. Your car is a '95 so your deceleration is -1. What is your reaction distance, braking distance and total stopping distance?

V=dr/tr 100=dr/2 200=dr Vf^2=vi^2+2ad 0=100^2+2(-1)d -200=-2d db=100 TSD=300db

-Your pet unicorn, Fluffy, is flying to the yearly magical creatures convention in Las Vegas. She must respect airway traffic rules by stopping at red lights. If she was flying at a speed of 500 m/s and has a reaction time of .3 seconds, with a maximum decceleration of -2 m/s2, what was her reaction distance, braking distance, and Total Stopping Distance?

V=dr/tr a=-2 tr=.3 vi=500 vf=0 dr=? 500=dr/.3 dr=150 Vf^2=vi^2+2ad 0=500^2+2(-2)db -250000=-4db db=62500

__Section 6__
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">+12 EC || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">can't understand what givens you chose for your hw, stop and go zone of a real intersection <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">they are not listed || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">some things are not titled ||
 * <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 ||
 * <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;">10/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">7/10
 * <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*.75 = 74.25/75** ||

10/26/11 What do you see? -Red car is stopping short, the dog is flying out of the car. Its stopping at a red light -The Green car is speeding up trying to make it through. Learning Outcomes: -Investigate the factors that affect the STOP and GO zones at intersections with traffic lights -Investigate the factors that result in an Overlap zone or a dilemma zone at intersections with traffic lights -Use a computer simulation to mathmatically model that situations that can occur at intersections with traffic lights



Investigation 11/4 p. 91 #3,4

3a. Automobile b is going to be able to make it through the light 3b. Yes because so is A and A is behind B 3c. yes 3d. It might turn red before C gets there and cause an accident 4a. Yes because it is behind D which is in the Stop Zone 4b. No because it will not be able to stop in time, it could cause an accident

__Yellow Light Model Variables__ Vi= initial velocity before braking ty=time of yellow light. w: width of the intersection a= negative acceleration (brakes) tr: reaction time

GZ=Go Zone SZ=Stop Zone

Group Members: _ _ // Directions: List Expand, Shrink, or no effect in each box // Group Members: _ _ // Directions: List Expand, Shrink, or no effect in each box //
 * Go Zone Prediction Sheet **
 * Variable |||| Change || **Predicted** shrink or expansion of Go ZONE ||
 * ty || yellow-light time || Increase ty || expand ||
 * ^  ||^   || Decrease ty || Shrink ||
 * tr || reaction time || Increase tr || No effect ||
 * ^  ||^   || Decrease tr || No effect ||
 * v || speed limit || Increase v || expand ||
 * ^  ||^   || Decrease v || shrink ||
 * a || negative acceleration || Increase a || no effect ||
 * ^  ||^   || Decrease a || no effect ||
 * w || width of intersection || Increase w || shrink ||
 * ^  ||^   || Decrease w || expand ||
 * Stop Zone Prediction Sheet **
 * Variable |||| Change || **Predicted** shrink or expansion of Stop ZONE ||
 * ty || yellow-light time || Increase ty || No Effect ||
 * ^  ||^   || Decrease ty || No Effect ||
 * tr || reaction time || Increase tr || moves it back ||
 * ^  ||^   || Decrease tr || moves it forward ||
 * v || speed limit || Increase v || Shrink ||
 * ^  ||^   || Decrease v || Expand ||
 * a || negative acceleration || Increase a || no effect ||
 * ^  ||^   || Decrease a || no effect ||
 * w || width of intersection || Increase w || expand ||
 * ^  ||^   || Decrease w || move it back ||

Part B: Yellow Light Dilemma -For A I would stop because we would not make the light -For B I would go through the light because if we stopped too short it could cause an accident. -For C I would go through the light -For D I would stop

2. For E and F I would stop. And for H and G I would go through the light. 3.For J and L I would stop, for M and K I would go. 4a. They were different because in each the cars are going different speeds and the yellow light changes at different speeds. 4b.The best choice is look at what is around you, if you think you can make it, its best to just get through the light as fast as possible. 4c.The best choice is to stop just to be safe 4d.Intersection II has a overlap zone and Intersection III has a dilemma zone

1. The fact that when you plug numbers in, it changes according to how it would be in real life 2. It is the spot in an intersection where it is still safe to go when there is a yellow light 3.The spot in an intersection where it is safe to stop when there is a yellow light 4. This is the spot where it is safe to stop and safe to go 5. This is the spot where it is not safe to stop and not safe to go

__Go Zone and Stop Zone Equations__ Go zone v=d/t v=d/ty v=(w+GZ)/ty

(ty)v=w+GZ Go Zone equation= GZ=vty-w ^The furthest distance at which you can go safely, the "edge" of GZ

STOP ZONE: SZ =TSD SZ=dr+db SZ=VTr +-(vi)^2/-2a) The closest position to the intersection in which you can brake safely "edge" of SZ

1) w=25m ty=35sec v=20 a=-5 tr=.7 GZ=vty-w GZ=20(3)-25 GZ=35

i) use excel sheet to make overlap zone of 10-30 ii) calculate stop zone and Go zone iii) diagram stop zone and go zone

DIAGRAMS ARE ON THE BOTTOM!!!

Intersection1 Yellow Light Time- 6sec Reaction Time-0.8sec Speed of Vehicle- 20 m/s Negative Acceleration- -5m/s^2 Width of intersection - 40m Go Zone- 80 meters

Stop Zone- 56 meters Dilemma- 24 meters

GO ZONE w=40m ty=6sec v=20 a=-5 tr=.8 GZ=vty-w GZ=(40)(6)-40 GZ=80

STOP ZONE SZ=vi(tr)-vi/2a SZ= (20)(.8)-(20)/2(-5) SZ=16-20/-10 SZ=18

Intersection 2: Go Zone GZ=vty-w GZ=(12)(7)-40 GZ=44

Stop Zone SZ=vi(tr)+vi^2/2a SZ=12(.8)+(12)^2/2(-5) SZ=9.6+144/-10 SZ=24

Intersection 3: Go Zone: GZ=vty-w GZ=(12)(7)-40 GZ=44

Stop Zone SZ=vi(tr)+vi^2/2(a) SZ=(12)(.4)+(12^2)/2(-5) SZ=4.8+14.4 SZ=19.2

Intersection for a cognizant driver: Intersection for a fatigued driver:

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 shrinks

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)---> by making the yellow light longer tr (reaction time), w(width of intersection), a(negative acceleration), vi(initial speed before braking).

3) A driver with worn out brakes approaches an intersection while a driver with superior brakes approaches an intersection next to him. Which driver will have to brake first? Which drive has a larger total stopping distance? Which driver has a stop zone that is pushed further back from the intersection? -The driver with worn out brakes will have to stop sooner and have a larger stopping distance because bad brakes take a long time to stop. the one with better brakes

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 one that is moving faster has a larger stopping distance and has a stop zone that is pushed 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? -They have the same stopping distance, but the sober one has a stop zone that 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.

Intersection 1: 2. 25 meters 3.Reaction time 0.6sec and -5.8sec deceleration 4. GZ=vty-w GZ=11.18(3.5)-25 GZ=14.13

SZ=vi(tr)+vi^2/2(a) SZ=(11.18)(.6)+(25^2)/(-5.8) SZ=6.71+625/33.64 SZ=25.29 5. GZ= (22.35)(3.5)-25 GZ=53.23 SZ= vi(tr)+vi^2/2(a) SZ=(22.35)(.6)+22.35^2/2(-5.8) SZ=13.41+499.52/134.56 SZ=17.122 Does this make a more or less safe intersection if you travel at double the speed limit? Why do you think a ticket is such a large fine for traveling at double the speed limit? Does the Stop Zone or Go Zone Change more when the speed is doubled? Explain why. -it makes it a more dangerous intersection which is why tickets are a lot of money if you go double the speed limit the go zone gets bigger if you double the speed limit because when you are going faster, there is more time to go through the light.

SECTION 7:
Centripetal Force: Driving on Curves

<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing centripetal acceleration notes ||
 * <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;">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;">**67.5/75** ||

Learning Outcomes: __Physics Talk__ Force- A push or a pull Centripetal Force- a force directed toward the center to keep an object in a circular path Centripetal acceleration- A change in the direction of the velocity with respect to time

1.positive 2.Centripetal force 3.friction 4. An object can start going backward 5. Going down a hill, pressing the gas and being pushed by another car 6.Gravity


 * Investigation**

HYPOTHESIS: a car can go faster around a wide turn than a sharp turn

Part one: -inner radius: 5.87sec Time for 1 revolution: .59 -Outer radius: 11.6 Time for 1 revolution:1.2

C=2(pi)r C=2(3.14)(18.5) C=116.2

7/.59 18/1.2 inner radius:11.86 Outer radius: V=d/t = 2(pi)r/t
 * Radius || Max Speed ||
 * 7 || 11.86 ||
 * 18 || 15 ||

1) 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? You can go faster on a wider turn

Part two: HYPOTHESIS: a car can go faster on a dry surface than an icy surface 1.Turning lazy susan: Normal: 16.9---> 1 revolution: 1.7 Sandpaper: 11.06---> 1 revolution:1.11

Normal: C=2(pi)r C=2(3.14)15 C=94.2

Sandpaper: C=2(3.14)18 C=113


 * Radius || MaxSpeed ||
 * Normal15 || 55.3 ||
 * Sandpaper18 || 101.8 ||

2)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? -Normal aspalt you go faster than on an icy surface

Part 3: HYPOTHESIS: a less massive vehicle can go faster around a turn than a bigger vehicle Bigger Mass: Radius-12 Time-18.7 One Revolution: 1.87 Small Mass: Radius-12 Time-11.7 One Revolution:1.17

C=2(pi)r

Bigger Mass C=2(3.14)(12) C=75.4

Small Mass C=2(3.14)(12) C=75.4

1) Can you achieve a larger maximum speed before skidding out in a heavier vehicle or a lighter vehicle? How does your data show this? -The lighter vehicle can go faster, because theres less mass, the small mass had a higher max speed than the large mass
 * Mass || MaxSpeed ||
 * Small || 64.4 ||
 * Large || 40.3 ||

NOTES: Centripetal acceleration: a change in velocity caused not by a change in speed but just by a change in direction -When turning the lazy susan, the cork leans toward the center

Acceleration and force are unidirectional. They point in the same direction.

HOMEWORK 11/16

Active Physics Plus

1) A car makes a left turn at a small intersection with turning radius 10 m. The mass of their car is 2000 kg. The pavement provides a frictional force of 13,720 N. What maximum speed can the car make the turn with? 8.3 m/s

2) Now the pavement is wet and made on the same turning radius and with the same car. The wet pavement causes the frictional force to cut in half. What maximum speed can the car make the turn with? 5.6 m/s

Fc=mv^2/r 6,860=2000(v^2) 10 68600=2000(v^2) 34.3=v^2 5.9=v

3) What minimum radius can a car make a turn with if it is traveling at 10 m/s, has a mass of 3000 kg and has a provided frictional force of 20,580 N? What can the car do to make the turn at an even sharper radius? 14.6 m

20,580=(3000)(100) r 20580r=300000 r=14.6

4) What minimum radius can a car make a turn with if it is traveling at 5 m/s, has a mass of 2200 kg and has a provided frictional force of 6,000 N? What can the car do to make the turn at an faster speed? 9.2 m

Fc=mv^2/r 6,000=(2200)(25) r 6,000r=55000 r=9.2m

5) Calculate the centripetal acceleration of a car making a 12 m radius turn at 10 m/s. 8.33 m/s2

Ac=V^2/r Ac=10^2/12 Ac=8.33

6) Now calculate their centripetal acceleration at 20 m/s. Twice the speed. How many times more centripetal acceleration is produced? 33.3 m/s2

Ac=400/12 Ac=33.3

7) Use the Givens in number 5 but now change the radius to twice the radius. What is the centripetal acceleration? What fraction is the centripetal acceleration of the centripetal acceleration in number 5 when the radius was still 12m? 4.17 m/s2

Physics to Go 1. V=__2(pi)r__ = 2(pi)r t V=__2(3.14)(6400)__ 24 hr V=1675km/h

2. V=__2(pi)r__ T V=__2(3.14)(1.5x10^8)__= 107515km/h 8766

3. 30/60 .5 revolutions per second 4. 30/20 1.5 m/s a.the speed lessens b. you would spin out c.you have an even better chance of going into the ditch 5. car around a race-track, spokes on a wheel 6. 7. Centripetal force turns the automobile anyway it wants no matter how one turns the wheels, friction keeps the car on task. 8. 270/2000 Acceleration:0.14 9. The second person is right, because they feel that it is normal to stay in a straight line, there is not force doing it. 10.Friction 11. These turns are more dangerous because there is more and more force building up on your car 12. if it were bending to the right you would end up in oncoming traffic and if you were bending to the left you'd end up in the ditch

8. 72.9 m/s2
 * 1) 465.4 m/s, 129.3 km/h
 * 2) 4753 m/s, 1320.3 km/h
 * 3) 56.5 m/s

ACTIVE PHYSICS PLUS
 * 1-4

Fc=__m v^2__ r Ac=V^2/r

Force: measured in Newtons(N) Fi=__mv^2__ r Fc=mv^2/r

=v^2 8.3m/s=v