Chapter+1



= **SECTION 1:** = Grade: toc What do you see? You see an accident up ahead on the road and you see a man trying to stop before he crashes into the rest of the cars.The smoke by the tires shows that he is skidding to a stop. The lines behind the car show that he must have been going to fat to begin with. ... You also see a rabbit.
 * DO NOW 9/7 **

What factors affect the time you need to react to an emergency situation while driving?
 * old tires
 * friction
 * down hill or up hill

Timer Test 1st time Christine: 5.67 Kristoff: 6.04
 * Investigation Reaction Time**

2nd time Kristoff: 5.06 Christine: 5.41

3rd time Kristoff: 4.21 Christine: 4.67

Average: 5.17 seconds


 * SECTION 1 : DO NOW 9/12**

Does a race cardriver need a faster reaction time than someone driving in a school zone? Explain your answer, giving examples of the dangers each driver encounters.

The race car driver due to the fact he is traveling at 200 mph. A school zone driver has to be driving at 25 mph or less because students are crossing the roads and are aware of this concept and are being more cautious


 * Reaction Time Ruler Notes 9/12**
 * Can you catch a dollar bill? 9/12\**


 * d=1/2at^2**

Dollar Bill: 15.4cm

d=15.4 15.4=1/2(980cm/s^2)t^2 t = .177s

Reaction time needed: .177s

What are the consequences if one's reaction time is slow? Why are auto insurance more expensive for teenage drivers than it is for older, more experienced drivers?
 * SECTION 1: DO NOW 9/14**
 * have an accident

What are the top two cases of accidents on the road? What is rubbernecking?
 * 9/14 Reflecting on the Chapter Challeng**e
 * driver fatigue
 * rubbernecking
 * when someone sees an accident or something along the road. it's a distraction

d=1/2at^2 .196=1/2(980)(.02)^2

SECTION 1 QUIZ REVIEW
 * Know how to measure time to move your foot
 * Know how to measure reaction time with stop watches
 * dropping the ruler
 * w/ decisions
 * w/ distractions
 * Alcohol/Drugs
 * Talking on cell phone
 * Texting
 * Driver Fatigue
 * Rubbernecking

= SECTION 2: = Grade: What do you see? Kids with a tape measure on the floor of a hallway. A man is walking along the tape measure and a boy is documenting it.
 * DO NOW 9/14**


 * Measurements: Errors, Accuracy, Precision**

Investigation 9/15

Strides: 21 Stride Length: 54cm

David Katie Megan || (13 strides)(93m) = 12.09m || 13.16m || Allison Anna Allyson Emily || (221)(.5m) = 11m || 14m || Noelle Christine Kristoff || (20 strides)(54m) = 10.8m || 13.20m || Mel Nicky Alex Kim || (21 sticks)(.74m) = 15.56m || 13m || Krista Jessi Zach || (18 strides)(.74m) = 13.34m || 13.31m || Dara Natalie Christine Kaitlin || (18.3 strides)(.56m) = 10.25m || 13.5m || Average Stride Measurement: 12.17m Average Meterstick Measurement: 13.33m
 * Group || Strides || Metersticks || Group || Strides || Metersticks ||
 * Jackson
 * Sam
 * Tiffany

.82 .75 .72 .71 .78 .76 .73 || .84 .85 .84 .84 .82 .82 .84 .82 .81 .83 || .81 .81 .81 .81 .81 .81 .81 .81 .81 .8 .82 .81 .815 .81 || .815 .814 .815 .8145 .813 .8 .815 .814 .812 .812 .813 .812 ||
 * Demo How Long is the Tube? 9/15**
 * Interval || 1meter || .1m || .01m || 00.1 ||
 * || .75


 * DO NOW 9/19**

Which thrower has systematic error? Which has random error?
 * Precise not Accurate
 * but not Precise


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

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

1.) 10 c m 10 (.01) m +/- .1 m > 50.1m - 49.9m +/- 1 cm > +/- .01m > 50.01m - 49.99m +/- .001m > 5.001m - 49.999m
 * 9/20 Measurement of Copper Tub**
 * Groups || Measurement of Copper Tub ||
 * 1 || 66 cm ||
 * 2 || 64.15 cm ||
 * 3 || 66.1 cm ||
 * 4 || 64 cm ||
 * 5 || 66 cm ||
 * 6 || 64.1 cm ||
 * 9/20 Active Physics Plus**

2.) +/- 1 cm 50.01m - 49.99m __50 m__ = __2 m__ 25 s 1 s __2 m__ = __.02 m__ 1 s (t) s (t = time to swim extra distance caused by random error)

3.) 50.01 - 49.99 = .02 m 30 laps x . 02m = .6m or 60 cm t = .36s

4.) 15.36 s __50.01 m x 30 laps = 1500.3 m__ __15.35 s__ 49.99 m x 30 laps = 1499.7 m

What does it mean? Why do you believe? Reflecting on the section and the challenge? Do you think it is reasonable to tell a police officer who pulls you over for going 75mph in a 30mph zone that your speed is due to the uncertainty present in her radar gun?
 * DO NOW 9/21**
 * systematic because it is a problem that can be fixed and the precision
 * incomplete


 * PHYSICS TO GO 9/20**
 * Pages 32/33**

3.)

4.)

7.)

8.)

9.)

Inquiring Further 1.)

S**ECTION 2 PAGE 24**
 * a.)** reasonable 3
 * b.)** unreasonable (no one is 13 ft)
 * c.)** unreasonable
 * d.)** unreasonable
 * e.)** unreasonable
 * f.)** unreasonable
 * g.)** reasonable

= SECTION 3: = Grade:

missing wDYthink and learning outcomes || unorganized, not numbered and graphs not described, #8 is missing calculations shown || did not answer checking up Qs || calculations not shown, in cyclist worksheet please write answers in complete sentencnes including the question in your answer missing following jack do now || incomplete ||
 * ** Section3 ** || **Points** ||
 * WDYSee/Think: || 4/10
 * Investigate: || 10/20
 * PhysicsTalk: || 15/20
 * PhysicsPlus: || 11/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">0/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/10 ||
 * <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;">**60** ||

What do you see?
 * The yellow car is speeding and seems to be rear-ending the red car and as a result of that, the red car is now rear-ending the blue car. Behind that, you see another red car that also appears to be speeding and it looks as though it is going to rear-end the car up ahead.

walking away walking away (slow) walking toward Number 5 (fast)
 * 9/22 Investigation**


 * Physics Talk 9/26**


 * Constant Speed:** speed that does not change over a period of time
 * Speed:** the distance traveled per unit
 * Average Speed:** the ratio of the total distance traveled to the total elapsed time
 * Instantaneous Speed:** the speed at a given moment
 * Velocity:** the speed given in a direction
 * Reaction Distance:** the distance that your automobile travels until you respond

How do you find the distance an automobile traveled if only given the reaction time?
 * Checking Up Question:**

a) 30ft b) 90ft c) 25 ft d) 75 ft e) 20ft f) 4 l/s
 * 8# on page 36**


 * DO NOW 9/27**

Vav = ∆d / ∆t Vav= 90ft/s ∆t = .6s ∆d = ? 90ft/s = ∆d / .6s 54ft = ∆d
 * An automobile is traveling at 90 ft/s (60mph). If the driver's reaction time is .6s, how far does the automobile travel during this time?**


 * 9/28 GO-CART TEST RUN**


 * 9/28 Active Physics Plus**


 * Part || Distance || Time ||
 * 20mi/hr || 40mi || ∆t = 2 hours ||
 * 40mi/hr || 40mi || ∆t = 1 hour ||
 * Vav = ? || 80mi || ∆t = 3 hours ||


 * Paragraph**
 * Part || Distance || Time ||
 * 1mi/hr || 50mi || 50h ||
 * 50mi/hr || 50mi || 1h ||
 * Vav = ? || 100mi || ∆t = 51hr ||


 * Mile Trip**

STROB PROBLEM


 * 10/3 Page 50**

15min/60hours = .25 hours ∆t = .25 hours ∆d = 5 miles Vav = ∆d/∆t = 5miles/.25hours = 20mi/hr


 * Distance vs Time Graphs**

PART 1 a.) no because cyclist B starts higher up the graph (ahead of the other) b.) cyclist A is traveling at a faster rate then B because it is above B in the graph meaning is covering more ground c.) cyclist B because it's above A in the graph and is covering more ground d.) yes at 5 seconds because they meet each other in the graph e.) they are at equal velocities PART 2 a.) No difference really b.) Cyclist B is going down in the graph meaning it is decreasing in speed c.) Cyclist A has the greater speed because it's slope is steeper d.) They are at equal velocities e.) cyclist A because the slope is steeper

= SECTION 4: =

<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing learning outcomes and what do you think || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing calculation of acceleration, missing two runs || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing alot check my wiki to see. ||
 * <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;">3/10
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">5/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">0/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">5/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">0/20 ||
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/10 ||
 * <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;">**23** ||


 * What do you see?**
 * red car speeding
 * green light
 * dog and guy running


 * 10/5 Tangent Line:** a line that only touches a curve at one point

V vs. T (velocity vs. time)
 * LAB**

1.) it would go faster the second half of the distance because when letting go, the car gains more speed towards the bottom 2.) graph D is where the car doesn't move



PREDICTION


 * DO NOW 10/7**

1.) From a stop light a car accelerates when the light turns green from 0 m/s to 30 m/s (60mph) in 5 seconds. What is the acceleration of the car? 2.) Name 3 vectors and 3 scalars. What is the difference between a vector and a scalar.
 * Review**
 * From a stop light a car accelerates when the light turns green from O m/s to 30 m/s (60 mph) in 5 seconds. What is the acceleration of the car?
 * A= ∆v / ∆t
 * A= (30-0) / (5)
 * A= 6 m/s^2
 * Name 3 vectors and 3 scalars. What is the difference between a vector and a scalar.
 * Vectors: Velocity, Force (push or pull)
 * Scalars: Speed, Calories, Anything you can count


 * SPEEDING UP GRAPH 10/12**

Slowing down velocity vs. time graph decreases from positive to negative (negative slope) a= slope acceleration is constant; gravity does not shut off a= V(f) - V(i) / ∆t
 * 10/12**

a = ∆v / ∆t = V(3s) - V(2s) / 3.5s - 2s = .4m/s - .25m/s / 3.5s - 2s = .15m/s / 1.5s = .1m/s^2
 * SPEEDING UP / ACCELERATION**

a = ∆v / ∆t = V
 * SLOWING DOWN**


 * ACTIVE PHYSICS PLUS**

3.) Vf = 20m/s Vi = 0m/s ∆t = 5 sec a = ? a = Vf - Vi / ∆t a = 20m/s - 0m/s / 5sec a = 4 m/s//^2//
 * a = ∆v/∆t**

= SECTION 5 = <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">table is not lined up straight <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing graph of Vi vs Braking Distance. did not answer investigation Qs. || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing ch1 mini challenge plan, checking up Qs and physicswords || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing half of the calculations for eqs of motion || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing total stopping distance activity and create a problem || What do you see? a car stoping short because of a moose in the middle of the road
 * <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;">7/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">0/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">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;">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;">**44** ||


 * objective:** to determine the effect of initial speed on braking distance

<span style="-webkit-border-horizontal-spacing: 0px; -webkit-border-vertical-spacing: 0px; border-collapse: separate;">Vi= Length of flag / time in gate || Go on Excel and make a scatter plot Title: Braking distance vs. Initial Velocity Y axis: Braking distance (m) X axis: Vi (m/s)
 * What do you think?**
 * What factors must you consider to determine if you will be able to stop in the distance between you and the animal to avoid hitting it?
 * Weather
 * How old the car is (tires)
 * Night time
 * How fast you're going
 * How fast your reaction time is
 * Investigation 10/18**
 * ||  || Vi(m/s) || BrakingDistance(m) ||
 * 5m/s || 9.43m ||
 * 1m/s || 11.3m ||
 * 1.1m/s || 11.9m ||
 * 1.3m/s || 8.7m ||
 * 2.6m/s || 12.8m ||
 * 6.5m/s || 25.01m ||
 * 1.07m/s || 8.85m ||
 * 1.07m/s || 8.85m ||


 * Calculate Breaking Distance**
 * Honda Civic Stopping Distances **

1) Calculate the stopping distances for Kibala’s Honda Civic


 * Initial Velocity || Stopping Distance ||
 * 10 mph || 3.42 ||
 * 20 mph || 13.67 ||
 * 30 mph || 30.75 ||
 * 40 mph || 54.67 ||
 * 50 mph || 85.42 ||
 * 60 mph || 123 feet ||
 * 70 mph || 167.28 ||
 * 80 mph || 218.94 ||
 * 90 mph || 276.75 ||
 * 100 mph || 341.94 ||

4) If Initial Velocity is doubled how does stopping distance change? Increased by 4 5) If the Initial Velocity is multiplied 4 times how does the stopping distance change? Increased by 16 6) If the Initial Velocity is halved how does the Stopping Distance Change? Decreased by 4 7) If the initial Velocity is quartered how does the stopping distance change? Decreased by 16 8) What speed would you need to have a stopping distance of a mile? Stopping distance of 393.11

1. As the initial speed increases the braking distance increases. 2. Automobile B is safer because its braking distance is shorter, meaning that in any situation where they have to brake quickly to avoid an accident they'll have a better chance of stopping soon than automobile A. 3. a) 5m b) 80m c) 45 m d) 125m 4. 9m is added on so the total distance is 39m ∆d= 1/2 (Vi +Vf) t Vf= Vi + at d= (Vi)(t) + 1/2 at^2 Vf^2= Vi^2 + 2ad 1. **d** is not used **a**= -4.1 **t**= ? **Vi**= 9 **Vf**= 0 2. **d** is not used **a**= 2.5 **t**= ? **Vi**= 7.0 **Vf**= 12.0 3. **d** is not used **a**= -0.50 **t**= ? **Vi**= 13.5 **Vf**= 0 4. **d** is not used **a**= ? **t**= 1500 **Vi**= -1.2 **Vf**= -6.5 5a. **a**= 4.7 x 10^-3 **t**= **Vi**= **Vf**= b. **Vi**= 1.7 **Vf**= ? 1. **a** is not used **t**= 6.5 (s) **d**= ? **Vi**= 0 (m/s) **Vf**= 73.7 (km/h) = 6.58 m/s 2. **a** is not used **t**= 2.50 **d**= ? **Vi**= 15.0 **Vf**= 0
 * 10/23 Physics to Go**
 * pg. 88-89**
 * 10/24**
 * Active Physics Plus 10/24**
 * Average Acceleration**
 * Vf= Vi + at
 * 0= 9 + (-4.1)(t)
 * -9= (-4.1)(t)
 * **t= 2.2 seconds**
 * Vf= Vi + at
 * 12= 7 + (2.5)(t)
 * 5= (2.5)(t)
 * **t= 2 seconds**
 * Vf= Vi + at
 * 0= 13.5 + (-0.50)(t)
 * -13.5= (-0.50)(t)
 * **t= 27 seconds**
 * Vf= Vi + at
 * -6.5= -1.2 + (a)(1500)
 * -5.3= (a)(1500)
 * **a= 0.004 m/s^2**
 * answer is 1.41 m/s
 * answer is 3.11 m/s
 * Displacement with Constant Uniform Acceleration**
 * 23.7 km/hr x 1000 m/km x 1 hr/ 60 min x 1 min/ 60 sec = 6.58 m/s\
 * d= 1/2 (Vi + Vf) t
 * d= 1/2 (0 + 6.58)(6.5)
 * d= 1/2 (6.58)(6.5)
 * d= 3.29 (6.5)
 * **d= 21.4 m**
 * d= 1/2 (Vi + Vf) t
 * d= 1/2 (15 + 0) (2.5)
 * d= 7.5 (2.5)
 * **d= 18.75 m**

4. **a** is not used **t**= ? **d**= 99 **Vi**= 78 (km/hr) = 21.7 (m/s) **Vf**= 0
 * d= 1/2 (Vi + Vf) t
 * 99= 1/2 (21.7 + 0) t
 * 99= 4.6 (t)
 * t=
 * answer is 9.14 seconds

Average Acceleration 1. t = 2.20s 2. t = 2s 3. t = 27s 4. a = -0.212 m/s^2 5. a) Vf = 1.41 m/s b) Vf = 3.11 m/s Displacement with constant uniform acceleration 1. d = 21.385m 2. d = 18.75m 3. d = 1000m, no 4. t = 9.14s 5. Vf = -4.57 Velocity and displacement with uniform acceleration 1. d = 29.7m 2. d = 59m 3. d = -18.75m 4. d = 31.35m Final velocity after any displacement 2. a) Vf = 21 m/s b) Vf = 15.78 m/s c) Vf = 12.50 m/s 3. a) Vf = 15.91 m/s b) t = 6.92s 4. d = 88.22m 5. a = 2.31 m/s^2 6. d = -7.41m
 * 10/24**
 * Equations of Motion**


 * 10/25**
 * Practice Conversation**
 * km/h to m/s**

100km/?h x 1000m/1km 100/?h x 1000m/1 x ?/60min x


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


 * 1.** No acceleration during (d)reaction


 * 11/2**
 * HOMEWORK**



Section 6:
<span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">+12 EC <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing everything except go zone and stop zone prediction sheets || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing questions 1-5 || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">missing stop and go zone of a real intersection || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">investigation partly present but untitled ||
 * <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;">0/10
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Investigate: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsTalk: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">0/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsPlus: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">10/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">PhysicsToGo: || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">0/20
 * <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">Wiki || <span style="font-family: 'Courier New',Courier,monospace; font-size: 15px;">5/10
 * <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;">**37*.75 = 27.75/75** ||

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 || shrink ||
 * ^  ||^   || Decrease tr || expand ||
 * 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
 * Variable |||| Change || **Predicted** shrink or expansion of GO ZONE ||
 * ty || yellow-light time || Increase ty || shrink ||
 * ^  ||^   || Decrease ty || expand ||
 * tr || reaction time || Increase tr || expand ||
 * ^  ||^   || Decrease tr || shrink ||
 * 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 || shrink ||


 * 11/10 Active Physics Plus Questions:**

Question 1:
 * The go zone is greater than expected.

Question 2:
 * ty = increase
 * tr = no effect
 * w = increase
 * Vi = increase
 * a = no effect

Question 3:
 * Driver w/ worn out brakes will have to brake first.
 * Driver w/ worn out brakes has a larger stopping distance.
 * Driver w/ worn out brakes has a stop zone that is pushed further back from the intersection.

Question 4:
 * The faster driver has a larger total stopping distance.
 * The driver who's slower has a stop zone further back from the intersection.

Question 5:
 * The non-drunk driver has a larger stopping distance.
 * The drunk driver has a stop zone further back from the intersection.

Question 6:
 * The speed affects the stop zone more than the go zone because of the person's reaction time.




 * 11/15 Investigation**
 * //Part 1//**


 * Radius (cm) || Max Speed (cm/s) ||
 * 3 || 1.57 ||
 * 20 || 8.18 ||

//**Part 2**//

//**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
 * Physics Talk**

CHECKING UP QUESTIONS 1. What is the direction of the force that keeps an object moving in a circle? 2. What is the name of the force that keeps an object moving in a circle? 3. Name the force that keeps an automobile moving in a circular path on a road. 4. Explain how the velocity of an object can change even if the speed is not changing. 5. Describe 3 situations in which acceleration can take place. 6. What is the force that keeps Earth moving in a circle around the Sun?
 * The direction of the force is positive
 * The force is called **centripetal force**
 * The force is named **friction**
 * An object can start going backwards
 * Going down a hill, pressing the gas and being pushed by another car
 * The force is called **gravity**


 * Hypothesis:** The car would go faster on a dry road because there it more friction.

//**Part 3**//

You can achieve a larger maximum speed with the heavier vehicle because in the end, it takes a quicker time for 10 revolutions than for the lighter vehicle

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


 * 11/16 Homework**


 * Vocab Words:**

__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


 * Checking Up Questions**

1. The force is pulling on the object in a straight line when moving in a circle. 2. Centripetal force keeps an object moving in a circle. 3. Friction keeps an automobile moving in a circular path on a road. 4. The velocity can change even if the speed is not changing when the car changes changes direction. 5. Acceleration can change when a car changes velocity, speed, or direction. 6. The centripetal force is gravity that keeps Earth moving in a circle around the Sun.

**11/16**

Change of velocity=Acceleration

__**Centripetal Acceleration**__-a change in velocity not by a change in speed but just by a change in direction. Acceleration and Force are **__unidirectional__**. They point in the same direction.


 * 11/16 Active Physics Plus**

1) (2)(π)(6400km)/(24 h)= 1675.52 km/h -- 465.42 m/s 2) (2)(1.5x10^8)(π)/(24) = 39269908.17 km/h -- 10908307.82 m/s 3) (2)(π)(15)/(.016)= 5890.49 cm/s 4a) The car will have to slow down and the radius will be smaller. 4b) The car might lose traction if the road is too slippery. 4c) If the turn is too tight and the road is too slippery the car might spin out because there won't be enough friction. 5) The moon traveling around the earth is gravity. When you stir pasta in a pot, the spoon keeps the pasta moving in a circular motion. Clothes that spin in the dryer are kept moving by the motion of the dryer, it rotates in a circle. 6)
 * [[image:acpjlevine/PTG_#6.png caption="PTG_#6.png"]] ||
 * PTG_#6.png ||

7) Depending on the conditions of the road, the road controls your tires. The road has to have enough friction to make sure that your tires aren't sliding all over the road. 8) a=v^2/r a=(270)^2/1000 a=72.9 m/s^2
 * [[image:acpjlevine/PTG_#8.png caption="PTG_#8.png"]] ||
 * PTG_#8.png ||

9) Yes both explanations are correct. In both examples a sharp left turn was made and the person ended up sliding to the right and hitting their right shoulder on the side of the door. 10) Friction on the road and wind resistance keep the car moving in a circular motion and pulled in towards the center. 11) They are especially dangerous because you are going fast around the wide turns and then the turns get tighter and tighter and you have to adjust your speed, but if you don't it can be very dangerous. 12) If the curve bends to the right you would end up in the land of oncoming traffic but if it bends to the left you would end up in the ditch.