Chapter+1

=Chapter 1 - Driving the Roads= toc

Section 1
Grade: 100 - I see a car crash and then a driver on the road coming up behind the crash trying to stop in time so that he doesn't collide. It looks like the car on the bottom of the crash swerved and the yellow car behind it was going to fast (or was too close to the first car) to stop in time, and ended up hitting the car in front and somehow flying over and landing on top of it. The oil from the first car means there is less friction for the next car. The blue car coming up behind looks like its going uphill which means that it would be easier to stop because it was working against gravity. The rabbit heard the commotion and ran to the side of the road to check out the scene. What factors affect the time you need to react to an emergency situation while driving? - Some factors include: how tired/aware you are, distracting views (mountains, animals), distracting noises (radio), size of the vehicle, hydroplaning (no friction to stop you), what you were previously doing as you noticed the emergency situation, the speed you were going, if you were drunk or not, texting or talking on the phone, decision (what side to go around the accident)
 * What do you see, what do you think? (9/7)**


 * Investigation: Reaction Time (9/7)**

Measuring Time To Move Your Foot 1) 26 movements 2) 26 movements 3) 27 movements 4) 28 movements 5) 28 movements average: 10 seconds/ 27 movements = .37 seconds/movement

Method A: Stopwatch (Decision Time) 1) .21 s 2) .26 s 3) .26 s average: .24 s

Method B: Ruler (Without Decisions) Time (s) || 1) fastest: Harvey at .093 seconds slowest: Giovanni at .27 seconds average reaction time: .19 seconds
 * Trial || Distance(cm) || Reaction
 * 1 || 7 || .13 ||
 * 2 || 10 || .14 ||
 * 3 || 14 || .17 ||
 * average || 10.3 || .15 ||

2) The factors that can affect reaction time are gender, athletic ability, and their attentiveness.

With Red/Green 1) 83 cm time: .62 s 2) 65 cm time: .45 3) 37 time: .27 4) 43 time: .3 5) 80 time: .6 average: 61.6 cm time: .41 s

2) My reaction time with a distraction is much greater than without a distraction.

3) This difference in reaction time means that it would be a lot harder to avoid road hazards if faced with a distraction while driving.

With Texting 1) 73 cm time: .52 s 2) 52 cm time: .37 s 3) 42 cm time: .28 s 4) 90 cm time: .71 s 5) 45 cm time: .3 s average:60.4 cm time .40 s

2) My reaction time with distractions and texting is slightly lower than my reaction time with only distractions.

3) This difference in reaction time would affect my ability to avoid road hazards because my reaction time with only distractions is greater, but it would only affect my ability slightly.

1. How do distractions affect reaction time? - Distractions slow reaction time.
 * Checking Up Questions:**

2. Why is driving under the influence of alcohol or drugs illegal? - Alcohol and drugs can significantly slow a person's reaction time.

3. Name three factors in addition to distractions and drugs/alcohol that can affect reaction time. - Three other factors are gender, practice, and attentiveness.

Do Now: 1) It takes a fastball about half a second to reach home plate. 2) It takes extra time after reacting to actually swing the bat. 3) Before you hit a pitch it may also take some time to actually swing the bat. Likewise, while driving it could also take time to decide to move your foot from the gas to the break when reacting to an obstacle. 4) Some factors not in the program are how heavy the bat is for your specific body weight and strength, deciding whether the throw is a ball or a strike, and also having to go through the entire motion of swinging the bat. 5) To reduce reaction time while driving once you are alert to an obstacle, you could already move your foot off the gas and towards the break. A batter might start positioned in their backswing in order to reduce the reaction time and hit the ball faster. 6) Yelling out things to the batter before a pitch is a distraction that increases their reaction time, which is similar to texting while driving. Texting is another distraction that would generally increase one's reaction time.
 * Reaction Time: Science of the Fastball (9/9)**

d=(1/2)at^2
 * Notes Active Physics Plus (9/9)**
 * d: distance any object falls in a time "t" (seconds)
 * t: time any object falls a distance "d" (centimeters)
 * a: acceleration of any object falling (980cm/s^2) -- cancels the seconds so the answer is in centimeters

d=(1/2)at^2 d=(1/2)(980)(.025s)^2 d=.31cm
 * Time || Distance ||
 * .025s || .31cm ||
 * .05s || 1.2cm ||
 * .075s || 2.76cm ||
 * .1s || 4.9cm ||
 * .125s || 7.66cm ||
 * .15s || 11.03cm ||
 * .175s || 15cm ||
 * .2s || 19.6cm ||
 * .225s || 24.81cm ||

d=(1/2)(980)(.05s)^2 d=1.2cm

d=(1/2)(980)(.075s)^2 d=2.76cm

d=(1/2)(980)(.1s)^2 d=4.9cm

d=(1/2)(980)(.125s)^2 d=7.66cm

d=(1/2)(980)(.15s)^2 d=11.03cm

d=(1/2)(980)(.175s)^2 d=15cm

d=(1/2)(980)(.2s)^2 d=19.6cm

d=(1/2)(980)(.225s)^2 d=24.81cm

Does a race car driver need a faster reaction time than someone driving in a school zone? Explain your answer, giving examples of the dangers each driver encounters. - A race car driver does not necessarily need a faster reaction time, because they both have their dangers that would be better to be able to react fast to. For example, a race car goes faster so it would need to react very fast to obstacles that come in its way, such as an overturned car. However, a normal car driving in a school zone would also need a fast reaction time because children could be running all around and school buses coming and going from all over.
 * Do Now (9/12)**
 * A faster car could cover more distance in the same amount of time as a slower car. The faster you go, the more distance will pick up per second.

Max distance the dollar bill can drop: 15.55cm d=(1/2)at^2 reaction time needed to catch the dollar bill: 15.55cm=(1/2)(980)**t**^2 31.1cm=980**t**^2 .031s^2=**t**^2 .18s=**t**
 * Can You Catch a Dollar Bill? (9/12)**


 * Reaction Time Ruler (9/12)**

average reaction time: .195s
 * Testing My Mom's Reaction Time (9/12)**
 * Trial || ReactionTime ||
 * 1 || .250s ||
 * 2 || .190s ||
 * 3 || .190s ||
 * 4 || .179s ||
 * 5 || .170s ||

1) The consequences of driving if one's reaction time is slow rather than quick is that when faced with an obstacle, it will be harder to avoid and could be more dangerous. 2) Although teenagers often have good reaction times, auto insurance is more expensive for teenage drivers because they are less experienced and more likely to get into an accident. Teenagers are more distracted.
 * Physics To Go (9/14)**

What are the top two causes of accidents on the road? - Driver fatigue, and rubbernecking.
 * Reflecting on the Chapter Challenge (9/14)**

What is rubbernecking? Does rubbernecking constitute a decision or distractions? - Rubbernecking is is when you are driving and being distracted by looking at an accident on the side of the road. This would constitute distractions.

1)
 * Active Physics Plus (9/14)**

Section 2
Grade: 100 Learning Outcomes: - Calibrate the length of a stride. - Measure a distance by pacing it off and by using a meter stick. - Identify sources of error in measurement. - Evaluate estimates of measurements as reasonable or unreasonable.

**What do you see, What do you Think? (9/15)** - I see a little boy recording something on a clipboard looking at a measuring tape stretched across the floor. A little girl is walking next to the tape, and so is an older boy. It looks like the little boy is recording the lengths of their strides while they walk. - With the 3 meter vs. 10 meter situation, one of them has definitely made a mistake because they are so far off from each other. However, in the situation where the differences were 3 meters and 3.01 meters, I do not think one of them has made a mistake because the numbers are so close it doesn't make much of a difference.


 * Investigation (9/15)**

My Number of paces: 21 My measurement of stride: .70 meters My calculations: (21 paces)(.70 meters) = 14.7 meters long

1) The measurements on the class table generally do not in all cases agree between their stride method versus meter stick method. 2) Most of the differences are only about 1 meter off. 3) There are differences in measurements made by different groups because different groups have different stride lengths. 4) A method to make the class's measurements more precise would be to just measure the distance using meter sticks. If all groups used this method this would reduce the range of measurements collected because every meter stick is exactly the same distance, rather than based on strides.

Measurement of length with meter stick: 13.12 meters long 1) Not all the measurements on the class table agree. 2) The length of meters in strides varies from 9 meters to 14 meters. In meter sticks it varies between 13.7 meters and 12.2 meters. 3) There are differences in the measurements because each group has different length strides. Some groups could have accidentally misread the meter stick. When you measure your stride, its not necessarily true that each stride will be that measurement. 4) A method to make the class's measurements more precise would be to just measure the distance using meter sticks. If all groups used this method this would reduce the range of measurements collected because every meter stick is exactly the same distance, rather than based on strides. 5) If each group was given a really long tape measure, the results of each group would be the same because it is just one ruler they are using for the whole distance, and if some groups were off by a little bit, it could just mean that they misread the ruler. 6) There will always be differences in measurements because different people would think that the end might be closer to one number or another. 7) a: We may have had some systematic errors if we started at the wrong number or didn't move the ruler of correctly. b: In the methods, the random errors would be off by 1.08 meters. 8) Three ways a systematic error may have been introduced are: by starting measuring at the wrong number, by looking at the wrong number while measuring, by moving the meter stick ahead to far while measuring.
 * Group || Strides || Metersticks || Group || Strides || Metersticks ||
 * Zack,Jenna,Christina,Steven || 14.06m || 13.17m || Kelsey,Carrie,Ryan || 14.2m || 13.12m ||
 * Joy, Aaron, Steve, Rolando || 9.24m || 12.28m || Michael, Carly, Jared || 13.44m || 12.73m ||
 * Ashley, Ben, Alyssa || 9m || 13.1m || Dana, Jason, Chelsea, Brandon || 13.11m || 13.165m ||

random error - an error that cannot be corrected by calculation systematic error - an error produced by using the wrong tool or using the tool incorrectly for measurement and can be corrected by calculation accuracy - an indication of how close a series of measurements are to an accepted value precision - an indication of the frequency with which a measurement produces the same results
 * Physics Words (9/17)**

1) Random errors cannot be corrected by calculation because there will always be uncertainty in measurement, while systematic errors can be corrected by calculation. 2) There will always be uncertainty in measurement because the tool nor the person can ever be completely exact. 3) With arrows, not having precision or accuracy would mean that the arrows are randomly spread out outside of the bulls-eye.
 * Checking Up Questions (9/17)**


 * Do Now (9/19)**

.8 .85 .832 .8 .85 .84 .83 .8 || .85 .83 .82 .84 .83 .83 .82 .817 .81 .825 || .81 .812 .813 .809 .81 .82 .815 .82 .812 ||  || Adding lines on the ruler makes the measurement more precise.
 * DEMO - How Long is the Tube? / 4-sided Ruler**
 * Interval || 1m || .1m || .01m || .001m ||
 * || .8


 * Physics Talk Notes -- SI System (9/19)**
 * Quantity || Unit || Symbol ||
 * distance || meters || m ||
 * mass || grams || g ||
 * time || seconds || s ||

1m = .001 km || 1 m = 100 cm || 1 m = 1000 mm ||
 * Prefix || Symbol || Multiple often by Which Base unit is Multiplied || Example ||
 * kilo || k || 10^3=1000 || 1km = 1000m
 * centi || c || 10^-2 = .01 || 1 cm = .01m
 * milli || m || 10^-3 = .001 || 1 mm = .001m

3. Me and my friend both agree that the price of a good lacrosse stick would be $90. Me and my friend disagree on the amount of people a medium-sized pizza could serve. I said it could serve 8 people, and she said it could serve 10 people. 4. The measurement is pretty accurate. If each barrel is worth $100, the possible uncertainty in value of the oil tanker's oil is that if there are not exactly 5 million barrels, then it would not be worth as much. How big is the oil tanker and the barrel? 6. a) No, 2 liters is not enough of a soft drink to serve 12 people. b) Yes, a mid-sized car with a full tank of gas could travel from Boston to NYC without stopping. 225 miles of driving. 30mi/1 gallon *12 gallons = 360 mi 7. Being off by 1 m in measuring a room is a lot bigger of an error as being off by 1 m in measuring the distance between my home and school, because between my home and school there is more room for error so 1 m does not seem like so much. 8. a) You should drive at 65 mph. b) The passenger could count how many miles you go in a certain number of seconds. 9. Driving 31 mph in a 30 mph zone should not earn the driver a ticket because there could be error in the car's speedometer so it would be unfair. If the speed exceeds above 35 mph in a 30 mph zone, then the driver should receive a ticket because it is a more noticeable difference.
 * Physics To Go (9/19)**

If I was buying vegetables by the pound, gas by the gallon, or carpeting by the yard, I would expect very exact and certain measurements of each because I would actually have to pay for the amount of each I want. So if the measurement was off, I could end up paying more than I need to and that would be unfair. Also, especially with carpeting, I wouldn't want to end up having extra material because it needs to fit just right. The government, through different organizations, ensures the accuracy of commercial weighing and measuring devices at the local level. The government regulates measurement standards in quantity and quality of marketplace, and in industry measurements in mass, density, and volume.
 * Inquiring Further #1 (9/19)**

1) uncertainty __+__ 10cm --> __+__ 10 (.01)m --> __+__ .1m (50.1-49.9m) __+__ 1cm --> __+__ 1(.01)m --> __+__ .01m (50.01-49.99m) __+__ 1mm --> __+__ 1(.001)m --> __+__ .001m (50.001m-49.999m) 2) speed = distance/time speed = (50m)/25s speed = 2 meters/second .02m/s = 2t t = .01 seconds 3) (30 laps)(.02m) = .6 meters <--- range 1500 = dt 1500 minutes = 1.66 m/s t=d/s t = .6/1.66 = .36s
 * Measuring a Copper Tube (9/20)**
 * Groups || Measurements ||
 * G1 || 66cm ||
 * G2 || 64cm ||
 * G3 || 66cm ||
 * G4 || 64.1cm ||
 * G5 || 66cm ||
 * G6 || 64cm ||
 * Active Physics Plus p.29 (9/20)**

What does it mean? - This would be a systematic error because it can be corrected through calculation. Systematic errors would affect accuracy because since the measuring tool is incorrect, the final measurement will not be close to the correct value but it could still be precise.
 * Do Now p.31 (9/21)**

Why do you believe? - You can trust experiments even if all measurements have uncertainties because certain measurements that are very important to physics knowledge will have been tested many times to get as close to the accurate value as possible.

Reflecting on the section and the challenge: You go 75 mph in a 30 mph zone. You explain to the police officer that his reading on his radar is due to the uncertainty in the instrument. Is this valid? - This is not valid because even though there is probably some uncertainty in the reading of the radar, the uncertainty is not enough to be a difference of 45 mph.

If 49.9//9//m is distance of a pool, the last decimal place is the uncertain one = __+__ .01m = range 50.00 - 49.98m 50.//5//m = __+__ .1m = range 50.6 - 50.4, 5//1//m = __+__ 1m = range 52 - 50m //5//0m <-- 0s don't count for uncertainty = __+__ 10m = range 60 - 40m
 * Notes (9/22)**

a) reasonable, because football players tend to be heavier than the average person b) unreasonable, because basketball players are usually 6-7ft c) unreasonable, because teachers work about 8 hours per day, not 24 hours d) unreasonable, because even large dogs only way about 50 lbs e) unreasonable, it is way too small, our classroom is about 2 times that size f) unreasonable, the estimation is too big because g) yes, because about the full length of the school track would be enough distance away from the other car to pass h) you must estimate the speed the dump truck is going and distance to the bridge for you and your speed i) unreasonable, because an average RV by itself is about 25 feet
 * Estimation Activity p.24 #9 (9/22)**

Section 3
Grade: 100 I see a two-lane road. One one side, there is a car jam bumper to bumper, but on the other side there is no traffic. It looks like the crowded side has a three-car collision. A safe following distance would be a two car distance between you and the vehicle in front of you. You can decide what a safe following distance is because you should stay 1 car away for every 10mph both cars are going.
 * What do You see? What do you Think? (2/23) **

1) a)
 * Investigation (9/23)**

2) a) This is the car at 45mph. They are more spread apart. b) The automobile is not the same distance apart between successive photos. The rectangles were closer together when traveling at 30mph. In one minute, the car travels .5 miles. c) This is the car at 60mph. I decided how far to place them because since the vehicle is moving faster, the rectangles are spaced farther apart.

3) a) The automobile is traveling the slowest in C, where they are closer together. It's traveling the fastest in A, because they are spaced farther apart. b) Each vehicle is traveling at a constant speed because the distances between them are the same.

4) a) person walking towards detector at a normal steady speed: b) person walking away from the detector at a normal speed: c) person walking away from detector, then toward it at very slow speed: d) person walking in both directions at fast speed: e) In the graphs, the cases where the person walked fast had more condensed lines, while the slow walking made the line more spread out because it took more time. In the graphs walking away from the graph, the line goes from low to high, but walking to the detector the line goes from high to low.

5) a) If you walk toward the detector at a slow speed but then walk away fast, the graph will have a a long spread out line going from high to low and then switches to a condense line going from low to high.

b) This is the actual graph: My prediction was very accurate.

6) a) trial 1 (red): person walks away slowly; trial 2 (green): person walks away quickly b) You could determine which graph goes with which lines because the more condensed line would be the one that the person walked away quickly because it takes less time.

7) a) In the first trial, I walked about 2.5m. In the second trial I walked about the same, just at a faster speed. b) In the first trial, it took about 5.8s. In the second trial it took about 2.4s. c) trial 1: 2.5m/5.8s = .43m/s trial 2: 2.5m/2.4s = 1.04m/s

8) a) v=d/t d= 30ft b) 60=d/1.5 90ft, 60 more feet c) 25ft, 75ft which is 50 more feet d) 35ft, 105ft which is 70 more feet e) a driver should be 20 ft behind f) 60=15/t it travels .25 ft/sec

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 any given moment velocity- the speed in a given direction Doppler effect- the change in the pitch, or frequency of a sound (or the frequency of a wave) for an observer that is moving relative to the source of the sound (or source of the wave) reaction distance- the distance that a vehicle travels in the time it takes the driver to react
 * Physics Words (9/26) **

1) The average speed is the total distance divided by time, while instantaneous speed is the random speed at any given moment. 2) The speed of an object is the distance traveled per unit of time and has no direction, while velocity is the speed of an object in a given direction. 3) It represents that as the time increases, the distance steadily increases as well in relation creating a constant speed. 4) Reaction time affects reaction distance because reaction distance is the distance the driver travels in the time it takes the driver to react. If the driver has a greater reaction time, then the distance will be greater.
 * Checking Up Questions (9/26)**

How does velocity related to reaction time and reaction distance?
 * My Own Checking Up Question (9/26)**

An automobile is traveling at 90ft/s (60mph). If the driver's reaction time is .6s, how far does the automobile travel during this time? - v= d/t 90ft/s = d/.6s d = 54ft How much further will the car travel if the driver is distracted by texting, so that reaction time is increased to 1.5s? - 90ft/s = d/1.5s d = 135ft
 * Do Now (9/27)**

20mi/hr || 40 mi || ∆t=2 hours || 40mi/hr || 40 mi || ∆t=1 hour || avg velocity=? || 80 mi || ∆t wholetrip=3 hours || v = d/t average velocity = ∆ distance/∆ time v = 80mi/∆t whole trip Part 1: 20mi/hr = 40mi/∆t ∆t = 2 hours
 * Active Physics Plus (9/28)**
 * Part || Distance || Time ||
 * Part1:
 * Part 2:
 * Whole Trip:

Part 2: 40mi = 40mi/∆t ∆t = 1 hour

Part 3: 1 hour + 2 hours = 3 hours

v = 80 mi/3 hrs v = 26.67mi/hour It isn't an even 30 because you are going 20 for a longer amount of time so it will be closer to 20.

1mi/hr || 50mi || 50 hours || 50mi/hr || 50mi || 1 hour || v = 100mi/51hrs v = 1.96mi/hr It is closer to 1mi/hr because it was going that speed for 50 hours, so it dominates.
 * Part || Distance || Time ||
 * Part 1:
 * Part 2:
 * total || 100mi || 51 hours ||

2) 3) 50mi/hr || 50mi || 1 hour || 25mi/hr || 50mi || 2 hours || 10mi/hr || 50mi || 5 hours || Part 1: 50 = 50/t t = 1 hr Part 2: 25 = 50/t t = 2 hrs Part 3: 10 = 50/t t = 5 hrs
 * Part || Distance || Time ||
 * Part1:
 * Part 2:
 * Part 3:
 * total || 150mi || 8 hours ||

a) I estimate the average speed to be 15 mi/hr. b) v = 150/8 v = 18.75 mi/hr This was very close to my estimate because I knew the average speed was going to be closer to the 10 because that's the part of the trip that took the longest.

1) a) In this diagram, the automobiles are traveling at a constant speed and go the same distance every 3 seconds because the spaces between them are exactly the same. b) In this diagram, the automobiles are not traveling at a constant speed because at some parts the cars (as shown every 3 seconds) are close together, but then there is a big gap which means that the automobile sped up during that segment of time. They go slow, speed up, then go slow again.
 * Physics To Go (9/27)**

2) a) An automobile starts from rest and reaches a constant speed.

b) An automobile moves at a constant speed and then comes to a stop.

3) v=d/t 350ft/s = d/20s d = 7000 ft

4) a) average speed = total d/t s = 215m/4.5hr s = 47.8 mph b) We cannot tell how fast she was going when the driver passed through Baltimore because our calculation is just the average speed after the whole trip, so we do not know the instantaneous speed at any random point.

5) s = 5m/.25hr s = 20mph

6) a) This graph shows that the automobile traveled at an exact constant speed but then stopped because the distance stopped increasing as time progressed. b) This graph shows that the automobile traveled at an exact constant speed at a very fast rate but then stopped, and then suddenly began going backwards at a constant speed that is not as fast as the speed in the beginning. c) This graph shows that the automobile traveled at an exact constant speed at a slow rate but then suddenly switched to a faster speed while remaining constant. d) This graph shows that the automobile's distance increased as time increased, but it was not and exact constant speed.

7) a) v =d/t 25m/s = d/.37 d = 9.25 meters b) 16m/s = d/.37 d = 5.92 meters This distance is shorter than when I was going at a faster speed, which means that when I am going faster the car goes further when I react and try to stop the car. c) 25m/s = d/.74 d = 18.5 meters

8) a) Traffic experts can be sure that 3 seconds is a safe following distance if the experts tested using the average reaction time of a a lot of different reaction times to get this number. b) Three seconds following distance would not be equally safe on both types of roads because a rural road could have more obstacles that a driver should watch out for like children or people walking around. However, on an interstate road the cars may be going faster because it's a highway so they might want to stay farther apart than three seconds.

9) a) v = d/t 100ft/s = d/.3333333s d = 33.33 ft b) 33.33ft is longer than the length of our physics classroom going from the front board to the back wall because that would be approximately 19.68ft.

10) a) 88ft/s = d/.5s d = 44ft b) This would be about 3 automobile spaces if the automobile was 15ft long. c) 30mi/hr = 44ft/sec 44ft/s = d/.5s d = 22 ft This would be about 1.5 automobile spaces long. d) 90mi/hr = 132 ft/sec 132 ft/s = d/.5s d = 66 ft This would be about 4.4 automobile spaces long. e) 30mi/hr 44ft/s = d/3 d = 132 ft, about 8.8 spaces long 60 mi/hr 88ft/s = d/3 d = 264 ft, about 17.6 spaces long 90mi/hr 132ft/s = d/3 d = 396 ft, about 26.4 spaces long

11) 88ft/sec 88 = d/ .25 d = 22ft 88 = d/.5 d = 44ft 88 = d/.75 d = 66ft 88 = d/1 d = 88ft Explain your thinking. Provide an explanation for your answer. - Since the dots are close together in the beginning, it means Jack was barely moving. The dots get farther apart, which means he kept on walking faster until eventually the dots become an equal distance apart because Jack was walking at a constant speed.
 * Following Jack: Part 1 and Part 2**

Explain your thinking. - This graph best matches Jack's motion because he starts off slow and then gets faster until he reaches a constant speed.

3) a) No, the cyclists start at different points because B started at a greater distance. b) At t=7seconds, cyclist A is ahead. I know this because A is higher on the graph at that point. c) Cyclist A is traveling faster at 3s because it is a steeper line. d) At any time, their velocities are not equal because their slopes are different, which means that their speeds are different. e) At the intersection of the lines, cyclists A and B have reached the same distance at the same time.
 * Distance vs. Time Graphs Cyclists**

4) a) In this graph, the motion of cyclist A is about the same, because it is the same line with the same slope. b) In this graph, the motion of cyclist B is reversed. It started at a greater distance and is coming closer, because it has a negative slope. c) Cyclist A has the greater speed because the line is steeper. d) At the intersection, both cyclists have reached the same distance at the same time. e) Cyclist A has traveled further during the first 5 seconds because it started father away from the intersection line on the graph.

1) For this graph, we stood still close to the detector, took a jump away from the detector and then stood still again. 2) For this graph, we stood still farther away from the detector, took a jump towards the detector and then stood still again. 3) It is impossible to create a complete circle on the graph because you cannot travel back in time.
 * Walk the Graph (10/4)**

4) For this graph, we stood still farther away then sped up walking towards the detector and back then stood still again. 5) For this graph we stood still in front of the detector then quickly walked away at a decreasing speed and then cam back with increasing speed, and then stood still.

Section 4

 * ** Section4 ** || **Points** ||
 * WDYSee/Think: || /10 ||
 * Investigate: || /20 ||
 * PhysicsTalk: || /20 ||
 * PhysicsPlus: || /20 ||
 * PhysicsToGo: || /20 ||
 * Wiki || /10 ||
 * **TOTAL POINTS** || **100** ||

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.

- It looks like one car tried to accelerate very fast but it looks burnt out. It looks like it tried to accelerate too fast after a red light turned into a green light. The other car is going at a normal speed. The red car gets to top speed quicker. - If an automobile and a bus are going from stop to 30mi/hr, some similarities are that they both start from the same velocity and aim to reach the same velocity. The car will reach top speed quicker because it is not as heavy as the bus. Therefore, the car has the greater acceleration. __Tangent Line:__ line that touches a curve at only one point
 * What do you see, What do you think? (10/5)**
 * Notes (10/5)**

__Slope of a tangent line on a distance vs. time graph:__ the instantaneous velocity at any point in time

1) If I were to place the cart at the top of the ramp and release it to freely move down the ramp, it would not travel the same distance in the second half of the ramp because it would be traveling faster since it accelerates.
 * Investigation (10/5)**

2)



__Run 1:__ prediction: tangent line: The velocity is changing with respect to time because in the beginning there is a lower velocity but it becomes higher as time goes on. As the velocity goes up, the time goes up velocity time graph: the velocity increases as time increases The slope and the tangent line are the same thing. calculate acceleration: a=v/t a=(.2/1.5) a = .13333 m/s^2

__Run 3:__ prediction: tangent line: the velocity decreases as time increases

velocity time graph: As time increases, the velocity looks like it decreases. The slope and the tangent line are the same thing. calculate acceleration: a = v/t a = .39/.4 a = .975m/s^2

__Run 4:__ prediction: tangent line: the velocity changes because at first it decreases as it goes up the ramp, and then it increases as time goes down because it comes back down the ramp velocity vs. time graph: As time increases, velocity decreases at first, but then reaches a point and begins to increase.In this velocity versus time graph, the velocity makes its turn at about .45 seconds on the graph. I know that this is the time it makes the turn because there is a sharp "v" in the graph and the vertex is the point it makes the change.The acceleration is pretty constant because when it decreases, there is a straight line. When it increases, the line is mostly constant and straight although there are a few little segments that mess it up. It is possible that the cart could have a constant velocity even though it makes the turn if the slopes on either side of the change are the same. calculate acceleration: a = v/t part 1 -- a = .3/.1 a = 3 m/s^2 part 2 -- a = .95/.3 a = 3.167 m/s^2

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 (it can slow down 20 to 10 or speed up -20 to -30) positive acceleration: an increase in velocity with respect to time (can speed up 20 to 30 or slow down -20 to -10) tangent line: a straight line that touches a curve in only one point
 * Physics Words (10/6)**

1) Acceleration is the change in velocity with respect to time. The equation is a=change in v/ change in t. 2) The SI unit for acceleration is meters per second squared. m/s^2 3) A vector quantity has size and direction while a scalar quantity only has size. 4) 5) The slope of a velocity time graph represents the change in velocity with respect to time.
 * Checking Up Questions (10/6)**

1) a = 30/5 a = 6 m/s^2 2) 2 vectors are pressure, force, or velocity. 2 scalars are speed, calories, or The difference between a vector and a scalar is that a vector quantity has direction and size, but a scalar quantity only has size.
 * Do Now Physics Talk Review (10/7)**


 * Bus versus Car (10/7)**

1) Yes, a situation can exist where there is zero acceleration but a nonzero velocity because as long as the velocity is constant there will be zero acceleration because there is no increase or decrease. 2) Yes, if there is a zero velocity there can be acceleration is it is at the apex point when an object is thrown in the air where velocity is 0 but acceleration is still going because the velocity is changing direction. 3) If two objects have the same acceleration they would have the same velocity because acceleration is velocity divided by time, so if they have the same time and acceleration, they would end up having the same velocity. The only difference could be that one is a negative acceleration. 4) If two automobiles have the same velocity, they do not necessarily have the same acceleration because the change in velocity is different than velocity can be different for each of them. 5) Yes, an accelerating automobile could be overtaken by a car with a constant velocity if it's constant velocity is higher than what the accelerating automobile's velocity reaches. 6) It is correct to refer to speed-limit signs because speed is the same thing as velocity, just without the negative or positive direction, which doesn't matter as long as the car doesn't go too fast or too slow. Speed-limit signs in the U.S. refer to miles per hour. 8) a. a=75/9 c. d = (1/2)(8.33)(81) d. a = 75/8 9.38 = v/8 75 = d/8 9) a. a = 3.9/1.3 b. d = (1/2)(-3)(1.21) + (1.1)(4.5) c. 3 = v/1.1 d. The second trial would get her to third base faster because it has a faster acceleration. 10) a. The top speed would be about 11m/s b. a = 11/9 c. If the object fell from a greater height, then the acceleration would be constant, but the velocity would increase. 11) a. Graph B would show the boy's velocity verses time as he coasts up the hill. b. Graph D would show the boy's distance versus time as he coasts up the hill. d. Graph A would show the boy's speed as he coasts down the hill. e. Graph F would show the boy's speed for the whole hill. f. Graph C would show the boy's distance as he goes down the hill. 12) a. segments d and e-f show constant speed b. segment a-c show increasing speed c. segment c-e shows at rest d. segment e-g shows decreasing speed e. the car traveled about 1100m during the whole test f. the car was at the starting point when the test was completed 13) a. a = 250/.0083 b. 30120.48 = v/.00416 c. 30120.48 = 500/t d. 500 = d/.017 14) a.a = v/t 9.8 = (vf-0)/4.5 b. a=9.8, vi=0, 11m = (1/2)(9.8)t^2 + 0 c. 9.8 = v/10 d. d = (1/2)(9.8)(100) + 0 e. 1.6 = v/10 v = 16m/s 16 = d/10 d = 160m If the acceleration was 1.6m/s^2, then the velocities, distances, and times in the answers to these questions would be much smaller. 16) a. 2m b. 8m c. 18cm d. 32m
 * Physics to Go (10/11)**
 * a = 8.33m/s^2**
 * d = 337m**
 * a = 9.38m/s^2**
 * v = 75m/s**
 * d = 600m X --> 300m**
 * a = -3m/s^2**
 * d = 5.85m**
 * v = 1.2m**
 * a = 1.22 m/s^2**
 * a = 8.33**
 * v = 375mi/hr**
 * t = 60 sec**
 * d = 15000m**
 * vf = 44.1 m/s**
 * t = 4.5sec**
 * v = 98 m/s**
 * d = 490m**

(see worksheet)
 * Predicting v vs t graphs from d vs t graphs (10/12)**

Equations: a = v/t
 * Active Physics Plus (10/12)**

d = (1/2)at^2 + vt (v = initial velocity)

1) A car accelerates from a stop light with acceleration 3 m/s2 from an initial rolling speed of 7 m/s to a speed of 20 m/s. How long does it take the car to do this? a = 3m/s^2, vi = 7m/s, vf = 20m/s, t = ? 3 = (20-7)/ t 2) If a car accelerates from 7 m/s to with an acceleration of 1.5 m/s2 for 10 seconds what speed is it now going? a = 1.5, vi = 7, t = 10, vf = ? 1.5 = (v-7)/10 3) A car accelerates from rest at a stop light to a speed of 20 m/s in 5 sec. 1. v = 20, t = 5, a = ? a = 20/5 2. d = (1/2)(4)(25) 4) A car accelerates from rest at a stop light to a speed of 40 m/s in 10 sec. That is an acceleration of 4 m/s2. What distance has the car gone? v = 40, t = 10, a = 4, d=? d = (1/2)(4)(100) 5) A car rolls up to a stop light with initial speed 3 m/s and accelerates to 23 m/s in 5 sec.
 * t = 4.33s**
 * vf = 22 m/s**
 * 1) What is the car’s acceleration?
 * 2) What distance has the car gone?
 * a = 4 m/s^2**
 * d = 50m**
 * d = 200m**
 * 1) What is the car’s acceleration?
 * 2) What distance has the car gone?

6) A car slows down coming to a red light from a speed of 25 m/s to 0 m/s in 4 sec. 1. vi = 25, vi = 0, t = 4 a = -6.25 2. because it is slowing down 3. d = (1/2)(6.25)(16) + 100
 * 1) What is the car’s acceleration?
 * 2) Why is it negative?
 * 3) How far did it take the car to stop?
 * d = 150m**

8) A car accelerates at a rate of -7 m/s2 up to a deer in the road which is 40 m away in 3 seconds from a speed of 21 m/s to a stop. Will the car hit the deer? d = (1/2)(-7)(9) d = -31.5 No, you do not hit the deer.

9) Now the car is moving twice as fast, 42 m/s and slows down with the same acceleration of -7 m/s2. (-7) =
 * 1) How much time will it take the car to come to a stop?
 * 1) If the car is attempting to avoid a deer which is 80 m away will the car hit the deer?

10) A ferrari whizzes by a cop at a red light at constant velocity of 40 m/s (80 mph), clearly speeding. The cop tries to catch our red light running speeding friend who never notices the cop behind him so continues at simply 80 mph. 7 = 40/t 2. How far will the cop have gone to catch the speeder? d = (1/2)(7)(32.49)
 * 1) If the cop car accelerates toward the speeder at a rate of 7 m/s2 how long will it take the cop to catch the speeder.
 * t = 5.7s**
 * d = 113.72 m**

- find instantaneous velocity on distance time graph using slope of tangent line - use instantaneous velocity to find acceleration - use the two formulas - know how to find acceleration of a velocity time graph (slope)
 * Quiz Topics (10/13)**

Section 5
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">+16 EC ||
 * <span style="font-family: 'Courier New',Courier,monospace;">** Section5 ** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 20px;">**Points** ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">WDYSee/Think: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**116*.75 = 87/75** ||

Learning Objectives: - **Plan** and carry out an experiment to relate breaking distance to initial speed. - **Determine** braking distance. - **Examine** accelerated motion.

- I see a car about to hit a moose in the road and it looks like the car is trying to come to a stop pressing hard on the breaks but is still moving. - To determine how much distance there is for you to stop to avoid hitting the animal, you must consider some factors including reaction time, under the influence, how far the moose is away, the weather, and if the driver is distracted.
 * What do you see? What do you think? (10/18) **

120 cm || .246 m/s || 15.75m || 110 cm || .268m/s || 13.5m || 100 cm || .263m/s || 13.0m || 90 cm || .598m/s || 9.75m || 1st Period's Data (see graph) (1.23m/s)/(.59m/s) = 2.08 <-- ratio of Vi --> (Vi 1/Vi 2)^2 11m/2.8m = 3.92 < ratio of braking distance ---> brake distance 1/ brake distance 2
 * Investigation (10/19)**
 * || Vi(m/s) || Breaking Distance (m) ||
 * Trial 1
 * Trial 2
 * Trial 3
 * Trial 4
 * Vi (m/s) || Breaking Distance (m) ||
 * .59 m/s || 2.8m ||
 * 1.27 m/s || 11.32m ||
 * .94 m/s || 5.15m ||
 * 1.06 m/s || 6.5m ||
 * .8 m/s || 4m ||
 * .97 m/s || 5m ||
 * 1.23 m/s || 11m ||

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

Investigation Questions: 1. If initial velocity is doubled, braking distance increases by a factor of 4. 2. If initial velocity is tripled, braking distance increases by a factor of 9. 3. If initial velocity is quadrupled, braking distance increases by a factor of 16. 4. a) The braking data is located on the right side of p. 177. b) No, the ratio of the initial velocities and braking distances would not be the same because the braking distances would increase with the square of the intial velocities. The ratio of the braking distances is 1.769.The actual ratio of the braking distances is 1.769, which is the initial velocity squared. c) This data follows the same pattern as the data in our experiment because the ration of initial velocity squared is the ratio of braking distances. 5. The braking distance would be less than the original braking distance ... 6. The braking distance would be 7.

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
 * Physics Words (10/19)**

1. If a vehicle is traveling at constant velocity and then comes to a sudden stop, it has still undergone a negative acceleration because although it may be a sudden stop, the car is decreasing speed in a positive direction. 2. Increasing the velocity of an automobile increases the breaking distance because if it is going at a faster speed, it takes longer to come to a complete stop and therefore travels a further distance while braking. 3. The term negative acceleration is used instead of deceleration because when something is slowing down, it could have a negative acceleration by decreasing its speed in the positive direction or increasing its speed in the negative direction.
 * Checking Up Questions (10/19)**

(Vi2/Vi1)^2 = BD2/BD1 (60mph/30mph)^2 = 123ft/BD1 60^2/30^2 = 123/BD1 60^2(BD1) = 123(30^2) BD1 = 123(30^2)/(60^2) BD1 = 30.75ft
 * Honda Civic Stopping Distances (10/20)**
 * Initial Velocity || Stopping Distance ||
 * 10 mph || 3.42 ft ||
 * 20 mph || 13.67 ft ||
 * 30 mph || 30.75 ft ||
 * 40 mph || 54.67 ft ||
 * 50 mph || 85.42 ft ||
 * 60 mph || 123 feet ||
 * 70 mph || 167.42 ft ||
 * 80 mph || 218.67 ft ||
 * 90 mph || 276.75 ft ||
 * 100 mph || 341.67 ft ||

4. If Initial Velocity is doubled how does stopping distance change? The stopping distance is x4.

5. If the Initial Velocity is multiplied 4 times how does the stopping distance change? The stopping distance is x16.

6. If the Initial Velocity is halved how does the Stopping Distance Change? The stopping distance is x(1/4).

7. If the initial Velocity is quartered how does the stopping distance change? The stopping distance is x(1/16).

8. What speed would you need to have a stopping distance of a mile? You would need to go a speed of 393.11 mph.


 * Direction of Acceleration (10/21)**


 * Equations of Motion (10/21)**

1. as the initial speed increases, so does the braking distance. 2. The braking distance for automobile B would have a larger braking distance because its slope is steeper, which means that over a shorter period of time it is going farther than A. This means that automobile B has a greater braking distance, so automobile A is "safer" because it wont take as long for the car to come to a stop. 3. a) 5 m b) 80 m c) 45 m d) 125 m 4. total stopping distance = reaction distance + braking distance TSD = (reaction time)(velocity) + 30m TSD = (.9s)(10m/s) + 30m 8. Some factors that affect stopping distance are reaction time, under the influence, the weather, the speed of the car, and if the driver is distracted.Making me a safer driver, I now know that it takes more than just reaction and pressing the brakes for the car to come to a stop. I will be more aware on the roads so that I have enough time to come to a complete stop before hitting something.
 * Physics To Go (10/20)**
 * TSD = 39 m**

TSD = reaction distance + breaking distance reaction distance: reaction distance =vt → TSD = Vt r + (V i ²/2a) So how did we get (V i ²/2a) anyway? Derivation:
 * Total Stopping Distance (10/26)**
 * < If V i goes up (↑) ||< then TSD ↑ ||< V i ~ TSD ||
 * < If t r ↑ ||< then TSD ↑ ||< t r ~ TSD ||
 * < If a ↑ ||< then TSD ↓ ||< (1/a) ~ TSD ||
 * a = negative acceleration slowing down (brakes); big acceleration means good brakes, small acceleration means bad brakes

Average Acceleration 1. a = -4.1, vi = 9, vf = 0, t =? 0 = 9+(-4.1)t 2. vi = 7, vf=12, a = 2.5, t=? 12=7 +(2.5)t 3. a=-.5, t=?, vi=13.5, vf =0 0=13.5+(-.5)t 4. vi=-1.2, t=1500s, vf=-6.5, a=? -6.5=-1.2 +a(1500) 5. a = 4.7x10^(-3) a) 4.7x10^(-3) = v/300s b) vf = 1.7 + 4.7x10^(-3)x(300)
 * Worksheet (10/24)**
 * t = 2.2 s**
 * t = 2 s**
 * t = 27 s**
 * t = -.004 m/s^2**
 * v = 1.41 m/s**
 * vf = 3.11 m/s**

Displacement with Constant Uniform Acceleration 1. vi=0, vf=6.58m/s, t=6.5s, d=? d=(.5)(6.58)(6.5) 2. vi=15, vf=0, t=2.5s, d=? d=(.5)(15)(2.5) 3. vi=100, vf=0, a=-5, d=? 0=(10000)+2(-.5)d d=1000 m ---> **no it cannot because that would only be 800m** 4. v=21.67m/s, d=99m, t=? 99=21.67t 5. vi=6.4, d=3200m, t=210s, vf=? 3200=(3.2 + .5vf)210 3200=672 +105vf
 * d = 21.4 m**
 * d = 18.75 m**
 * t = 4.57 s**
 * vf = 24.08 m/s**

Velocity and Displacement with Uniform Acceleration 1. vi=6.58m/s, d=?, a=.92, t=3.6s, vf =? d=(6.58)(3.6) + (1/2)(.92)(3.6^2) vf=(6.58) + (.92)(3.6) 2. vi=4.3, a=3, vf=?, d=?, t=5 vf=4.3 + (3)(5) d=(4.3)(5) + (1/2)(3)(25) 3. t=5, a=-1.5, vf=?, d=?, vi=0 vf=(-1.5)(9) d=(1/2)(-1.5)(25) 4. vi=15, vf=10, a=-2, d=?, t=? 10=15+-2t d=(37.5) + (1/2)(-2)(6.25) d = **31.26 m**
 * d = 29.65 m**
 * vf = 9.89 m/s**
 * vf = 19.3m/s**
 * d = 59 m**
 * vf = -7.5 m/s**
 * d = -18.75 m**
 * t = 2.5 s**

Find Velocity After Any Displacement 2. a) vf^2=49 + (2)(.8)(245) b) vf^2 = 49 +(2)(.8)(125) c) vf = 49 +(2)(.8)(67) 3. a)vf^2 = 0 +(2)(2.3)(55) b) 15.9 = 2.3t 4. a=.85, vi=23.05, vf = 26.1, d=? 681.21=531.3 + (2)(.85)d 5. vi=0. vf=33.33, d=240, a=? 1111.11=2a240 6. vi=6.5, vf=1.5, a=2.7, d=? 2.25=42.25 + 2(2.7)d
 * vf = 21m/s**
 * vf = 15.78 m/s**
 * vf = 12.5 m/s**
 * vf = 15.9 m/s**
 * t = 6.9 s**
 * d= 88.2 m**
 * a = 2.31 m/s^2**
 * d= 7.41 m**

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Key Concepts: <span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">reaction time <span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">distance versus time <span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">systematic/random error <span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">velocity versus time <span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">acceleration versus time <span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">reaction distance <span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">measuring parts of a trip (velocity and time)
 * Mini Challenge Plan (10/27)**

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Storyline: <span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Esther’s parents are allowing her and her friends to meet them and her little brother at the family lake-house in upstate New York. Her parents said that if she drove safely the entire way and obeyed the speed limit she would be able to use the family car more often. She must be very careful because there are many distractions and roadway obstacles in order to reach the mountainous destination. She will face treacherous hills and crazy deer wandering along the roads.

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Equations: <span style="background-color: transparent; color: #ff0000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">v = d/t <span style="background-color: transparent; color: #ff0000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">To calculate the distance and time from Esther’s house to the family lake-house in upstate New York, so that Esther’s parents can tell that she went the speed limit the entire way if they know her starting time and ending time. <span style="background-color: transparent; color: #674ea7; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">(vi2/vi1)^2 = bd2/bd1 <span style="background-color: transparent; color: #674ea7; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">To calculate Esther’s braking distance when she sees a deer in the middle of the road on her way to the lake-house. <span style="background-color: transparent; color: #4a86e8; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">vf = vi + at <span style="background-color: transparent; color: #4a86e8; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Calculate Esther’s vf as she’s going up Baljeet Hill

<span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Graphs: <span style="background-color: transparent; color: #00ff00; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">d vs t <span style="background-color: transparent; color: #00ff00; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Shows Esther’s distance and time as she goes up and down Baljeet Hill.- CARLY IS DOING THIS! no <span style="background-color: transparent; color: #ff00ff; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">v vs t <span style="background-color: transparent; color: #ff00ff; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">How her velocity decreased when she sees a deer in the road. <span style="background-color: transparent; color: #4a86e8; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">a vs t <span style="background-color: transparent; color: #4a86e8; display: block; font-family: Arial; font-size: 11pt; text-decoration: none; vertical-align: baseline;">Shows Esther’s acceleration as she goes up and down Baljeet Hill on her way to the lake-house. 1) Draw a diagram of a car moving down a road and braking for an object. 2) Assuming we are traveling forward with a positive velocity, then we would have a negative acceleration when we are braking.
 * Total Stopping Distance Activity (10/28)**

__Calculating Reaction Distance__ 1) Finding reaction distance using diagram above v=d/t (<-- this is the one without acceleration because during this time there is no acceleration) reaction distance=v/reaction time 2) v=10m/s, t=1s d=10m 3) v=20m/s, t=1s d=20m 4) An increase in speed changes the distance because the vehicle would travel farther during that time. A decrease in speed would mean a shorter distance. One second is a pretty short reaction time, and it may be this large because of decisions (whether to stop or not at the yellow light), age, or intoxication.

__Calculating Braking Distance__ 1)vf^2=vi^2 +2a(braking distance) 0 = Vi^2 + 2(a)(braking distance) d = -(vi)^2/2a 5) d = -(51^2)/2(-11) d = 118.23 m 6) d = -(51^2)/2(-24) d = 54.19 7) Better brakes have a larger acceleration. Better brakes make for a shorter braking distance.

__Preparing for the Chapter Challenge:__ 1) tr=.5s, vi=11, a=-4, dr=?, db=?, TSD = ? 11 = d/.5 (v=reaction distance/reaction time) 0 = 121 + 2(-4)d 2) vi=27, dr=?, db =?, TDP =?, a=-4, t=.5 27=d/.5 0 = 729 + 2(-4)d 3) vi=27, t=1, dr=?,db=?, TSD =?, a=-4 0=729 + 2(-4)d 4) t=.5, vi=27, dr=?, db=?, TSD=?, a=-2 27=d/.5 0 = 729 + 2(-2)d
 * reaction distance = 5.5 m**
 * braking distance = 15.13 m**
 * total distance = 20.63 m**
 * reaction distance = 13.5 m**
 * braking distance = 91.13 m**
 * total distance = 104.63 m**
 * reaction distance = 27 m**
 * braking distance = 91.13 m**
 * total distance = 118.13 m**
 * reaction distance = 13.5 m**
 * braking distance = -182.25 m**
 * total distance = 195.75 m**

You are driving on a highway at 50 m/s, but you see a fallen down tree across the highway in front of you. Your reaction time is .65 s, and your car has a deceleration of -.5 m/s^2. What is your reaction distance, braking distance, and total stopping distance? 50=d/.65 0 = 2500 + 2(-.5)d <span style="font-family: Arial,Helvetica,sans-serif;">Your grandma is driving you home from school one afternoon at a speed of 13.41 m/s (30 mi/h) when a rabbit jumps out into the road up ahead and she puts her foot on the brake. Her reaction time isn't the best - 4 seconds - and your car has a maximum deceleration of -3 m/s². What is the reaction distance, braking distance, and total stopping distance? 13.41=d/4 0 = (179.83) + 2(-3)d
 * Create Your Own Problem (11/2)**
 * reaction distance = 32.5 m**
 * braking distance = 2500 m**
 * total stopping distance = 2532.5 m**
 * reaction distance = 53.64 m**
 * braking distance = 29.97 m**
 * total distance = 83.61 m**

**Section 6**
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">+16 EC ||
 * <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;">/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">/10 ||
 * <span style="display: block; font-family: 'Courier New',Courier,monospace; font-size: 15px; text-align: right;">**TOTAL POINTS** || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">**100** ||

**Learning Objectives:** In this section, you will - **Investigate** the factors that affect the STOP and GO Zones at intersections with traffic lights - **Investigate** the factors that result in an Overlap Zone or a Dilemma Zone at intersections with traffic lights - **Use** a computer simulation to mathematically model the situations that can occur at an intersection with traffic lights

I see a woman on the side of the road with her hands on her hips angrily because a man in a car in front of her tries to come to a stop as the light turns red. The person in the green car goes right through the right. It looks like the first car was unable to stop inside to stop at the intersection so he goes into the middle a little bit because he was going faster. Is green car doing something legal or not? - No, because he is speeding through a red light. But he didn't think it was wrong because he was in the intersection when the light turned red.
 * What Do You See? What Do You Think? (11/3)**


 * Stop Zone/ Go Zone Notes (11/3)**

3. a) Yes, automobile B will be able to make it through during the yellow light. b) Yes, automobile B is in the Go Zone because that is the zone in which you can legally make it through the intersection when the light turns yellow. c) Yes, any automobile closer to the intersection than automobile A will be in the Go Zone because automobile A is, so they will be too. d) No, automobile C is not in the go zone because if automobile A is at the end of the Go Zone and automobile C is behind it, then automobile C is in the stop zone. If it tries to make it through the yellow light, then automobile C will have to come to a sharp stop when the light turns red, or it will continue into the intersection and be illegal. 4. a) Yes, automobile E is in the Stop Zone because it would be able to come to a safe stop when the light turns red. b) No, automobile F is not in the Stop Zone because it is too close to the intersection, so if it decides to stop it will not stop before the intersection. So it might end up in the middle of the intersection.
 * Investigation #3,4 (11/4)**

ty (yellow light time) w (width of intersection) a (negative acceleration) || Vi (speed limit) ty (yellow light time) w (width of intersection) a (negative acceleration) tr (reaction time) ||
 * Go Zone Stop Zone Notes (11/7)**
 * GoZone || StopZone ||
 * Vi (speed limit)

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

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

1) The spreadsheet is referred to as a model because it is a mathematical equation. 2) The Go Zone is the space before the intersection where you can safely make it through a yellow light while following the speed limit. 3) The Stop Zone is the space before the intersection where you can safely come to a stop in the time the yellow light turns red while traveling the speed limit. 4) The Overlap Zone is the space where when a light runs yellow, you can choose to either stop or go through the intersection safely before the yellow light turns red. 5) The Dilemma Zone is the space between the Stop and Go Zones where even if you are going the speed limit, you cannot safely stop before the intersection or pass through it completely before the light turns red.
 * Checking Up Questions (11/8)**

1. a) Automobile A would stop, automobile B would have a choice but if it were me then I would speed up and go through the yellow because there's not enough space to stop, automobile C would go, and automobile D would stop. 2. a) Automobile E would stop, automobile F would stop, automobile G would go, and automobile H could do either because there is an overlap but I would just go because there is enough time. 3. a) Automobile K would go, automobile L would stop, automobile M would speed up and go through, automobile J would stop. 4. a) the intersections are different because the second one has an overlap zone, and the third one has a dilemma zone. b) Your choices would be to stop or go, and it would be safe to do either because there is enough space and time. c) Your choices would be to stop or go, but they would not be legal choices because there are no legal choices in that situation theat would be safe. d) Intersection II has an overlap zone and intersection III has a dilemma zone.
 * Part B: Yellow Light Dilemma p. 94 #1-4 (11/8)**

__Go Zone:__ v=d/ty v=(w + GZ)/ty GZ = vty - w __Stop Zone:__ SZ = TSD SZ = dr + db SZ = Vi(tr) - (Vi)^2/2a
 * Deriving Go Zone/ Stop Zone Equations Notes (11/8)**

1. 19 m dilemma zone; w=25m ty=3s, vi=20m/s, a=-5 GZ = (20)(3) - 25 GZ = 35 m SZ = (20)(.7) - (400)/2(-5) SZ = 54 m 2. GZ = (15)(4) - (15) GZ = 45 m SZ = (15)(.7) - (15^2)/2(-8) SZ = 24.56 m GZ = (15)(3) - 10 GZ = 35 m SZ = (15)(.7) - (15^2)/2(-5) SZ = 33 m GZ = (15)(4) - 10 GZ = 50 m SZ = (15)(.7) - (15^2)/2(-5) SZ = 33 m 3. ty=4s, tr=.5s, v=15m/s, a=-6, w=10m GZ = (4)(15) - 10 GZ = 50 m SZ = (15)(.5) - (15^2)/2(-6) SZ = 26.35 m 4. ty=4s, tr=2s, v=15m/s, a=-4, w=10m GZ = (15)(4) - 10 GZ = 65 m SZ = (15)(2) - (15^2)/2(-4) SZ = 58.13 m
 * Active Physics Plus Section 6 (11/9)**

1. Approaching an intersection in the Midwest, I realize that it is wider than I expected. Because of this, the Go Zone shrinks. 2. The driver could have increase velocity to compensate for the width of the intersection and increase the Go Zone. 3. The driver with the bad brakes will have to brake first, and they are the one will the larger total stopping distance. They are also the one that has a Stop Zone pushed back farther away from the intersection. 4. The faster car has the larger total stopping distance, and they are the one with the stop zone pushed back farther away from the intersection. 5. The drunk driver has a larger stopping distance and a stop zone pushed farther back away from the intersection. 6. The stop zone is affected by speed more drastically because in the formula, speed is used twice, and one time it is even squared so it makes more of a distance. Since stop zone is made up of braking distance and reaction distance, it is used twice, once in each mini-equation.
 * Active Physics Plus Homework (11/9)**

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

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

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

Calculations for when I double the speed limit: GZ = (17.88)(3) - (20.2) SZ = (17.88)(.5) - (17.88)^2/2(-4) If you travel at double the speed limit, the intersection is even less safe because the dilemma zone is much greater. I think that a ticket for traveling at double the speed limit is such a large fine because it puts you and the cars around you in a lot more danger than just if you were speeding by only a few more m/s. Since the dilemma zone is greater, there is more of a chance of an accident.The stop zone or go zone changes more when the speed is doubled because the stop and go zones are quadrupled.
 * GZ = 33.44 meters long**
 * SZ = 48.90 meters long**

Section 7
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing centripetal force 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;">/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** ||

**What do You see? What do you think? (11/15)** I see a car driving on a mountain going really fast trying to turn. It looks like since it is going to fast for the sharp turn and is kind of driving along the edge of the mountain. Why is the sign indicating to slow down? - It is telling you to slow down because if you go too fast you will not be able to complete the turn and will go off the road. How is the amount you should slow down determined? - You should slow down to half of your speed.

In this section you will: - Recognize the need for a centripetal force when rounding a curve. - Predict the effect of an inadequate centripetal force. - Relate speed to centripetal force.
 * Learning Objectives (11/15)**

__force__ - a push or a pull __centripetal force__ - a force directed towards the center to keep an object in a circular path __centripetal acceleration__ - a change in the direction of the velocity with respect to time
 * Physics Words (11/15)**

1. The force is directed to the center, perpendicular to the object. 2. Centripetal force is the name of the force. 3. Friction is the centripetal force for an automobile. 4. Velocity can change because direction is changing. 5. Acceleration can take place when an object speeds up, slows down, or changes direction. 6. Gravity is the Earth's centripetal force.
 * Checking Up Questions (11/15)**

PART 1 hypothesis: A car can go faster around a wide turn because you do not have to completely turn the wheel and you have more space to drive. 4. inner: c=2pi6
 * Investigation (11/5)**
 * Radius(cm) || MaxSpeed(cm/s) ||
 * 6 cm || 12.56/10 =1.26s ||
 * 14 cm || 23.98/10 = 2.4s ||
 * c= 37.69 cm**/1.26 = s = 29.91 cm/s

outer: c=2pi14 6. You can achieve a larger speed while going around a sharp turn. My data shows that when the circular mass was place closer to the center of the circle, it achieved a larger speed.
 * c= 87.96 cm**/2.4 = s = 36.65 cm/s

PART 2 hypothesis: A car can go faster turning on a dry surface because when you are on an icy surface, there is little friction and the driver can not turn the car. 3. no sandpaper: c=2pi11 sandpaper: c=2pi11 6. You can achieve a larger maximum speed on normal asphalt than on an icy road. According to my data, the circular mass was able to go faster when it was on the sandpaper.
 * Radius(cm) || MaxSpeed(cm/s) ||
 * 11 cm (no sandpaper) || 26.1/10 = 2.61s ||
 * 11 cm (sandpaper) || 14.56/10 = 1.46s ||
 * c= 69.12 cm**/ 2.61 = s = 26.48 cm/s
 * c= 69.12 cm**/ 1.46 = s = 47.34 cm/s

PART 3 hypothesis: A less massive vehicle can go faster turning because it has less mass to move with it. heavier: c=2pi6 = 37.7 cm/1.96 = 19.23 cm/s lighter: c=2pi6 = 37.7 cm/1.63 = 23.13 cm/s You can achieve a larger maximum speed with a lighter vehicle. According to my data the smaller mass was able to go faster.
 * Mass || MaxSpeed(cm/s) ||
 * heavier || 19.62/10 =1.96s ||
 * lighter || 16.33/10 =1.63s ||

centripetal acceleration - change in the direction of an object's velocity but not a change in its speed
 * Centripetal Acceleration Notes (11/16)**

centripetal acceleration and centripetal force are unidirectional (towards center) a=v/t v=c/T(time it takes to go around once) a=v^2/r F=mv^2/r a=acceleration v=velocity m=mass r=radius f=force (N - Newtons) (n) = (kg)(m/s^2)/m --> kgm/s^s = (N)

centripetal acceleration = v^2/t

1. r=10, m=2000, f=13720, v=8.3 13720 = (2000)v^2/10 2. f=6860 6860 = (2000)v^2/10 3. v=10, m=3000, f=20580 20580 = (3000)100/r 4. v=5, m=2200, f=6000 6000 = (2200)(25)/r 5. r=12, v=10 a=100/12 6. a = 400/12 7. r=24, v=10 a=100/24
 * Active Physics Plus (11/16)**
 * v = 8.3 m/s**
 * v = 5.86 m/s**
 * r = 14.6m**
 * r = 9.2 m**
 * a = 8.3 m/s^2**
 * a = 33.3 m/s^2**
 * a = 4.12 m/s^2**

1. t=24h, r=6400km s=c/t c=2pir c=2pi6400 c=40212.39 km s=40212.39/24 2. r=1.5x10^8 km, c=2pir = 9424777961 km c=2pi 3. s = 7200pi, r = 15cm s=v^2/t 4. a) if the curve is tighter, then the car will spin off b) if the road surface becomes slippery, the car will spin off and go straight because friction is no longer keeping it in control c) if the road is slippery AND the curve is tighter, the the car will spin off because there is little friction 5. A baseball bat travels in a circular path when being swung, and the muscles of the batter produce the force that keeps the object towards the center of the curve. 6. 7. "The driver may turn the wheels but it is the road that turns the automobile." This means that although the driver of a car chooses either to turn left or right through the steering wheel, without the friction of the road that is a centripetal force, the car would not be able to turn either way at all. 8. s=270m/s, r=1000m, a=? a = (270^2)/1000 9. Both explanations are correct because they are describing the same thing how since there was not enough friction between them and the seat, they kept going in a straight line because nothing was holding them back. The car including the seat did turn though because of friction from the road. 10. The force is friction. 11. These types of turns that get tighter and tighter are especially dangerous because you do not always see it coming, and in the beginning when it is gentle you may be going faster, but when it gets tighter you don't have enough time to significantly decrease your speed, and you could end up driving off the road. 12. If you are curving to the right, you will end up on the other side of the road because the car would want to keep moving straight. If you are curving to the left, you would end up on the right side.
 * Physics To Go (11/16)**
 * s = 1675.52 km/hr** (1hr/3600s)(1000m/1km) **= 465.42 m/s**
 * s = 4753 m/s = 1329.3 km/h**
 * s = 56.5 m/s**
 * a = 72.9 m/s^2**