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SECTION SEVEN
Section7
Points
WDYSee/Think:
/10
Investigate:
/20
PhysicsTalk:
/20
PhysicsPlus:
/20
PhysicsToGo:
/20
Wiki
/10
TOTAL POINTS
75
Learning Outcomes:
Recognize the need for a centripetal force when rounding a curve
a=10^2/24
a= 4.17m/s^2

SECTION SEVEN
Learning Outcomes:
Recognize the need for a centripetal force when rounding a curve
Predict the effect of an inadequate centripetal force
Relate speed to centripetal force
What Do You See/What Do You Think
November 14th
{Screen_shot_2011-11-15_at_9.07.13_PM.png} Screen_shot_2011-11-15_at_9.07.13_PM.png
Screen_shot_2011-11-15_at_9.07.13_PM.png
I see a car driving on top of a winding cliff. The car is going fast so its turns are sharp resulting in the car almost falling off.Why is the sign indicating to slow down?It says to slow down because driving on a sharp turn to quickly could result in losing control of the vehicle and falling off the cliff.How is the amount you should slow down determined?The amount is determined by the size of your car and how sharp the turn is.
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{section_7_notes2.jpg} section_7_notes2.jpg
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InvestigationNovember 15th
Part 1: Can a car go faster around a wide turn or sharp turn?
Hypothesis: Answer the question posed in the title of Part 1 and give a reason for this hypothesis
Procedure:
1. Choose two radii to place your circular mass on, one towards the center of your Lazy Susan and one towards the edge. The radius towards the center will simulate the car on a sharp turn while the radius towards the edge will simulate the car on a wide turn.
Sharp Turn: 4.5 cm
Wide Turn: 24.5 cm
2. Spin the Lazy Susan with the mass at the inner radius for 10 revolutions then spin the Lazy Susan with the mass at the outer radius for 10 revolutions, recording the time each takes to make 10 revolutions.
OKAY!
3. Calculate the time it took the Lazy Susan to make one revolution for each radii.
Sharp Turn: 1.781 s
Wide Turn: 3.049 s
4. Use the formula C=2πr to solve for the circumference of the path that the weight traveled for each radii. Next, Divide the Circumference for each path by the time for the weight to make one revolution for the inner and outer radius. This will tell you the speed of the weight while it traveled around the Lazy Susan.
{Screen_shot_2011-11-15_at_8.33.57_AM.png} Screen_shot_2011-11-15_at_8.33.57_AM.png
Screen_shot_2011-11-15_at_8.33.57_AM.png
Sharp Turn:
v = (2)(3.14)(4.5)/1.781
v = 28.26 cm/1.781s
v = 15.88 cm/s
Wide Turn:
v = (2)(3.14)(14.5)/3.049
v = 91.06 cm/3.049s
v = 29.87 cm/s
5. Create a table to organize your data for Part 1.
Radius
Max Speed
Sharp
4.5 cm
15.88 cm/s
Wide
14.5 cm
29.87 cm/s
6. Can you achieve a larger maximum speed on a wide turn (large radius) or a sharp turn (small radius)? How does your data show this?
You can achieve a larger maximum speed on a larger radius, or wide turn. Our data shows this because the wide turn has a larger maximum speed by 13.98 cm/s.
Part 2: Can a Car Go Faster Turning on an Icy Surface or Dry Surface?
Hypothesis: Answer the Question posed in the Title of Part 2 and give a reason for this hypothesis.
Procedure:
1. Spin the Lazy Susan with the mass on the Wood Surface for 10 revolutions then spin the Lazy Susan with the mass at the same radius on the sandpaper for 10 revolutions, recording the time each takes to make 10 revolutions.
Wood Surface: 3.13 s
Sandpaper Surface: 1.58 s
Radius: 14.5 cm
2. Calculate the time it took the Lazy Susan to make one revolution for each surface
Wood Surface:
v = (2)(3.14)(14.5)/3.13
v = 91.06 cm/3.13s
v = 29.1 cm/s
Sandpaper Surface:
v = (2)(3.14)(14.5)/1.58
v = 91.06 cm/1.58s
v = 57.63 cm/s
3. Use the formula C=2πr to solve for the circumference of the path that the weight traveled for each radii. Next, Divide the Circumference for each path by the time for the weight to make one revolution for the inner and outer radius. This will tell you the speed of the weight while it traveled around the Lazy Susan.
{Screen_shot_2011-11-16_at_8.51.42_AM.png} Screen_shot_2011-11-16_at_8.51.42_AM.png
Screen_shot_2011-11-16_at_8.51.42_AM.png
4. Create a table similar to below to organize your data for Part 2
Radius
Max. Speed
Wood
14.5 s
29.1 cm/s
Sandpaper
14.5 s
57.63 cm/s
5. Can you achieve a larger maximum speed on sandpaper (normal asphalt) or on the wood surface (slippery road conditions)? How does your data show this?
You can go faster on normal asphalt than on slippery road conditions. The difference between the two maximum speeds is 28.53.
Part 3: Can a more massive vehicle go faster turning or a less massive vehicle go faster turning?
Hypothesis: Answer the Question posed in the Title of Part 3 and give a reason for this hypothesis.
Procedure:
1. Spin the Lazy Susan with the lighter mass on the Wood Surface for 10 revolutions, then spin the Lazy Susan with the heavier mass at the same radius on the sandpaper for 10 revolutions, recording the time each takes to make 10 revolutions.
Lighter Mass: 1.64 s
Heavier Mass: 1.84 s
Radius: 4.5 cm
2. Calculate the time it took the Lazy Susan to make one revolution for each mass.
Lighter Mass:
v = (2)(3.14)(4.5)/1.64
v = 28.26 cm/1.64s
v = 17.23 cm/s
Heavier Mass:
v = (2)(3.14)(4.5)/1.84
v = 28.26 cm/1.84s
v = 15.35 cm/s
3. Create a table similar to below to organize your data for Part 2
Radius
Max. Speed
Lighter
4.5 cm
17.23 cm/s
Heavier
4.5 cm
15.35 cm/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?
The mass is irrelevant.
Homework:November 15th
Physics Words
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
Checking Up Questions
1. What is the direction of the force that keeps an object moving in a circle?
The force is always toward the center of the circle.
2. What is the name of a force that keeps an object moving in a circle?
The force is call centripetal force.
3. Name the force that keeps an automobile moving in a circular path on a road.
The force that keeps an automobile moving is friction between the tires and the road.
4. Explain how the velocity of an object can change even if the speed is not changing.
A car can move in a circle at a constant speed, but the velocity changes because it is changing direction.
5. Describe three situations in which acceleration can take place.
Acceleration can take place when an object speeds up, slows down, or changes direction.
6. What is the force that keeps the Earth moving in a circle around the Sun?
The force that keeps Earth moving around the Sun is gravity.
Centripetal Acceleration: a change in velocity caused not by a change in speed. but just by a change in direction.
Physics To GoNovember 16th
1. C=2πr
C=2π(6400)
C=40212.39/ 24
V=1675.52km/h
1675.52 km x 1000m = 1675516.082m/3600s
V=465.42m/s
2. C=2π(1.5x10^8) / 24h
= 39269908.17km/h-
3. C=2π(15)
C=94.25 / 60
V= 1.57cm/sec
4. a. The automobile can go faster if the curve is tighter.
b. The automobile should go slower if the road surface becomes slippery.
c. The automobile should go slower around the curve because the road is slippery and there won't be as much friction to hold the car now.
5. An example is a race car racing on the track. The track is curved to allow the drivers to make sure they don't slide. Gravity is pulling them to always stay in the curved path,
7. The driver turns the steering wheel so the wheels will turn on a curved road. However, it is the road that actually moves the entire automobile and forces it to continue following the wheels. The road has a force (gravity, centripetal force) that is making the car follow the road.
8. a=v^2/r
a= 270^2 / 1000
a= 72.9m/s^2
9. The fist explanation is not correct because if you turn to the left, the force will make your body go to the left too. The second explanation is correct because your body wanted to go straight but the force/friction made your body go to the left as well. Also, the passenger door had a centripetal force on it because that was what made it continue in the circular path.
10. Centripetal force acts on the race car to go in a circular path when it is on the track.
11. When you are traveling on a wider turn (gentle turn), you can gain more maximum speed. However, on sharp turns, you need to go slower. Drivers need to realize this and they need to know to slow down on these types of curves to prevent sliding off the road and getting into a dangerous accident.
12. Curve bends right: You would end up into the lane of oncoming traffic because your car would slide on the road and go straight into the other lane. Curve bends left: You would end up into the ditch because, as said before, your car would continue to slide and accelerate off the road.
Active Physics Plus: November 16th
Centripetal Force (1-4)
{Screen_shot_2011-11-17_at_8.26.27_AM.png} Screen_shot_2011-11-17_at_8.26.27_AM.png
Centripetal Acceleration:(5-7)
{Screen_shot_2011-11-17_at_8.28.25_AM.png} Screen_shot_2011-11-17_at_8.28.25_AM.png
Force: measured in Newtons (N)
F=mv^2/r
(kg)(m^2/s^2) x (1/m)= N
1).
f=13,720 N
m= 2000 kg
r= 10m
v= ?
13720=2000v^2/10
68.6=v^2
V= 8.28 m/s
2).
f=6860 N
m=2000 kg
r= 10m
v= ?
6860=2000v^2/10
34.3=v^2
V=5.86m/s
3).
f= 20,580 N
m=3000 kg
r= ?
v= 10 m/s
20580=(3000)(10)^2/ r
20580=300000/r
r=14.58m
4).
v=5m/s
m=2200 kg
f= 6000 N
r=?
6000=2200(5^2)/r
r=9.17m
5).
a=?
v=10m/s
r=12
a=v^2/r
a=10^2/12
a= 8.33m/s^2
6).
a= ?
v= 20m/s
r= 12
a=v^2/r
a=20^2/12
a= 33.3m/s^2
7).
r= 24
v=10m/s
a=?
a=10^2/24
a= 4.17m/s^2

Chapter 1: Section 6
Learning Outcomes:
Investigate the factors that affect the STOP and GO Zones at intersections with traffic lights
Investigate the factors that result in an Overlap Zone or a Dilemma Zone at intersections with traffic lights
Use a computer simulation to mathematically model the situations that can occur at an intersection with traffic lights
What do you see/What do you think 11/3
{Screen_shot_2011-10-26_at_11.57.56_AM.png} Screen_shot_2011-10-26_at_11.57.56_AM.png
What do you see?I see the red car stopping short at a red light and stops in the middle on an intersection. The dog is flying out of the car. The green car is speeding up and going through the red light.
What do you think?The yellow light should have been longer because the red car couldn't stop in time. It had to stop in the middle of the intersection. The green car is doing an illegal act because it is going through the red light.
Stop Zone/Go ZoneNovember 14th
{Go_Zone.jpg} Go_Zone.jpg
{Stop_Zone.jpg}
Investigation:Friday, November 4th {Screen_shot_2011-11-04_at_8.17.50_AM.jpg} Screen_shot_2011-11-04_at_8.17.50_AM.jpg
3a.)Will automobile B be able to make it through the intersection?Yes, because automobile B cannot just come to a sudden stop with cars behind him3b.)Is automobile B in the go zone?yes B is, because B has no time to stop, or does not have enough distance left until the middle of the intersection.3c.)Would any automobile closer to the intersection that automobile A be in the Go Zone?Yes, because the closer you are with a yellow light, the more reason you have to continue through.3d.)Is automobile C in the go zone?No, and if C decided to continue, then they would
{Screen_shot_2011-11-04_at_8.25.45_AM.jpg} Screen_shot_2011-11-04_at_8.25.45_AM.jpg
4a.)Is autopmobile E in the stop zone?Yes, because it is too far away from the yellow light to stop.4b.)Is automobile F in the stop zone? Explain your answer.No automobile F is in the Go zone, because they have enough room to stop.
PG 95-96 PART B {Screen_shot_2011-11-08_at_8.18.51_AM.png} Screen_shot_2011-11-08_at_8.18.51_AM.png
A-YesB-YesC-NoD-No
{Screen_shot_2011-11-08_at_8.22.04_AM.png} Screen_shot_2011-11-08_at_8.22.04_AM.pngE-YesF-YesH-NoG-No
{Screen_shot_2011-11-08_at_8.24.13_AM.png} Screen_shot_2011-11-08_at_8.24.13_AM.pngJ-YesL-NoM-YesK-Yes
Go Zone/Stop Zone Notes:November 17th
GoZone
StopZone
Vi (speed limit)
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 Prediction/Actual Results:
{Screen_shot_2011-11-07_at_5.27.18_PM.png} Screen_shot_2011-11-07_at_5.27.18_PM.png
Stop Zone Prediction/Actual Results:
{Screen_shot_2011-11-07_at_5.30.37_PM.png} Screen_shot_2011-11-07_at_5.30.37_PM.png
Part B: Yellow Light Dilemma pg. 95November 18th1a. Automobiles A and D should stop, and B and C should go.2a. Automobiles E and F should stop, H has the choice, and G should go.3a. Automobiles J and L should stop, M has the choice, and K should go.4a. Intersection I has a distint STOP and GO Zone. Intersection II has an overlap that could be part of either the STOP or GO Zone. Intersection III has a Dilemma Zone.4b. Your choices are to either go or stop. It would be safe to do either because the zones are unclear.4c. Your choices are to either go or stop. Neither is safe because that area is neither a STOP Zone nor a GO Zone.4d. Intersection II has an Overlap Zone and Intersection III has a Dilemma Zonw.
Go Zone/Stop Zone EquationsNovember 18thGo Zone:
v = d/t
v = d/ty
v = (w + GZ) / ty
GZ = (v)(ty) - w
(The furthest distance at which you can go safely...where the Go Zone ends)
Stop Zone:
SZ = TSD
SZ = dr + dB
SZ = Vitr - (Vi)^2 / 2a
The closest position to the intersection in which you can brake safely...where the Stop Zone begins)
Checking Up Questions
November 18th
1. In this section, the spreadsheet is referred to as a model. What makes it a model?
A model is a mathematical equation to better understand a problem.
2. In your own words, describe what is meant by the GO Zone.
The GO Zone includes all positions where the automobile can safely proceed through the intersection when the light turns yellow.
3. In your own words, describe what is meant by the STOP Zone.
The STOP Zone includes all positions where the car can safely stop before reaching the intersection.
4. Describe what is meant by the Overlap Zone.
In the Overlap Zone, when the light turns yellow, the driver of the vehicle has the choice of either stopping or going through the intersection safely before the light turns red.
5. Describe what is meant by the Dilemma Zone.
The Dilemma Zone is a space between the STOP and GO Zones. In this region, a driver traveling the speed limit cannot safely stop before the intersection or pass through the intersction completely before the light turns red. Only occurs in a poorly designed intersection.
Active Physics Plus:
November 19th
1. Given a car approaching an intersection of width 25 m and yellow light time 3 seconds on a road of speed limit 45 mph (20 m/s) with a maximum deceleration of -5 m/s2 and a driver with reaction time .7 seconds. Diagram the intersection, calculate the Stop Zone and Go Zone, and decide if the intersection is safe or dangerous and explain why.
w = 25 m
ty = 3 s
Vi = 20 m/s
a = -5 m/s
tr = .7 s
--
GZ = (Vi)(ty) - w
GZ = (20)(3) - 25
GZ = 35 m
--
SZ = (Vi)(tr) - (Vi^2/(2a)
SZ = (20)(.7) - (20^2)/(2)(-5)
SZ = 54 m
{APP_chap_1_sect_6_#1.jpg} APP_chap_1_sect_6_#1.jpg
2. Design 3 separate realistic and safe intersections. A realistic intersection has an overlap zone that is 10-30 m long. Re-compute the Stop Zones and Go Zones using the stop zone and go zone formulas then diagram the 3 safe intersections and photo them along with the calculations into your wiki. Check your intersections using the excel sheet to make sure its safe before you calculate the stop zone and go zone using the formulas.
{2a.jpg} 2a.jpg
{2a.png} 2a.png
{2b.jpg} 2b.jpg
{2b.png} 2b.png
{2c.png} 2c.png
3. Design 1 safe intersections for a cognizant driver of reaction time .5 sec and maximum deceleration of -6 m/s2. Choose the other 3 variables(speed limit, yellow light time and width of intersection) to make a safe and realistic intersection. A realistic intersection has an overlap zone that is 10-30 m long. You can use the excel sheet and do not have to use the manual formulas to calculate Stop Zone and Go Zone. Diagram them and list the 1 input variables under the photo of the diagram in your wiki. {3_APP_chap_1_sect_6.jpg} 3_APP_chap_1_sect_6.jpg
{Screen_shot_2011-11-09_at_11.47.39_AM.png} Screen_shot_2011-11-09_at_11.47.39_AM.png
4. Design 1 safe intersections for a fatigued driver with reaction time 2 seconds and maximum deceleration of -4 m/s2. Choose the other 3 variables(speed limit, yellow light time and width of intersection) to make a safe and realistic intersection. A realistic intersection has an overlap zone that is 10-30 m long. You can use the excel sheet and do not have to use the manual formulas to calculate Stop Zone and Go Zone. Diagram them and list the 5 input variables under the photo of the diagram in your wiki.
{4_APP_chap_1_sect_6.jpg} 4_APP_chap_1_sect_6.jpg
{Screen_shot_2011-11-11_at_8.21.48_AM.png} Screen_shot_2011-11-11_at_8.21.48_AM.png
HW 10/9
1. The Go Zone would shrink because it would take longer to cross the intersection if it were wider.
2. The driver could change the velocity to increase the Go Zone.
3. The driver with bad brakes has to stop first, has a longer TSD, and has a Stop Zone pushed back farther away from the intersection.
4. The faster driver has a longer TSD and has a Stop Zone pushed back farther from the intersection.
5. The drunk driver has a longer TSD and has a Stop Zone pushed back farther from the intersection.
6. Speed affects the Stop Zone more because the variable "v" is used twice in the SZ equation. The Stop Zone is extended if you are going faster.
Go Zone/Stop Zone of a Real Intersection
November 13th
Intersection 5: Kinderkamack and Prospect Ave. in Woodcliff Lake
{Screen_shot_2011-11-13_at_12.58.55_PM.png} Screen_shot_2011-11-13_at_12.58.55_PM.png
Variables:
Yellow light time: 4 sec
Reaction time: 0.4 sec
Speed limit: 40 mph = 17.9 m/s
Negative acceleration: -5 m/s^2
Width of intersection: about 18 m
Calculate and Diagram:
GZ = Vity - w
GZ = (17.9)(4) - 18
GZ = 53.6 m
SZ = Vitr - (Vi^2)/(2a)
SZ = (17.9)(0.4) - (17.9^2) / (2)(-5)
SZ = 39.2 m
{Screen_shot_2011-11-13_at_2.13.24_PM.png} Screen_shot_2011-11-13_at_2.13.24_PM.png
Double the speed limit:GZ = Vity - wGZ = (35.8)(4) - 18GZ = 125.2 mSZ = Vitr - (Vi^2)/(2a)SZ = (35.8)(0.4) - (35.8^2)/(2)(-5)SZ = 142.5 m
{Screen_shot_2011-11-13_at_2.31.43_PM.png} Screen_shot_2011-11-13_at_2.31.43_PM.png
Doubling the speed limit makes a less safe intersection and causes a Dilemma Zone to form. It is extremely dangerous to you and others if you double the speed limit because you would be going twice as fast and it would take you much longer to stop, which is why the ticket is such a large fine. The Stop Zone changes more when speed is doubled because it is pushed back farther from the intersection, and the faster you go the harder it is to stop closer to the intersection.

Chapter 1: Section 5
Learning Outcomes
Tuesday, October 18th
-Plan and carry out an experiment
-Determine breaking distance
-Examine accelerated motion.
What do you see / What do you think?
Tuesday, October 18th
{Screen_shot_2011-10-18_at_8.26.06_AM.png} Screen_shot_2011-10-18_at_8.26.06_AM.png
-It looks like the vehicle could not stop in time, in result of going to fast.
-The deer clearly came out of no where.
-It appears that the car slammed on the breaks, causing it to skid and hit the mouse.
-The factors that would determine the stop would be how much weight is in the car, how fast you are going, and how big the car actually is.
Investigation
Tuesday, October 18th
Objective: to determine the effect of initial speed on breaking (stopping) distance.
Vi(m/s)
BreakingDistance(m)
.59m/s
2.8m
.94
5.15m
1.27
11.32
1.06
6.5
.8
4
.97
5
1.23
11
Make Graph (scatter plot)
Smooth curve through the graph.
YAXIS: Breaking Distance vs. XAXIS: Vi(m/s)
Vi=Length of flag/time in gate
{Screen_shot_2011-10-19_at_9.21.04_AM.png} Screen_shot_2011-10-19_at_9.21.04_AM.png
Sec5 Investigation Questions:
1.) If initial velocity is doubled by what factor does breaking distance increase.
-If the initial velocity is doubled, our information shows that the breaking distance will quadruple.
2) If initial velocity is tripled by what factor does braking distance increase.
-If the initial velocity is tripled, assuming that the proportion of Vi:BD stays the same, it would be 3^2, so it would be multiplied by 9.
3) If initial velocity is quadrupled (x4) by what factor does braking distance increase.
-If the Initial velocity is quadruples the breaking distance would be multiplied by 16.
4) Do #8 in book on page77
-b) No the breaking distance is going to be different.
-c) The breaking distances were squared of our speed.
5) If the sports car changed its speed to 30 mph what do you expect its braking distance to be? (Hint: if you half the speed by what factor will the braking distance change?)
-Decrease speed is 1/2 --> square of a 1/2. When you square 30 mph (speed), you get 900 miles for your breaking distance.
6) If the touring sedan changed its speed to 30 mph what do you expect its braking distance to be?
-The breaking distance is 900 miles as well because you square the speed (30mph).
7) If each car (sports car and touring sedan) decreased its speed to 15 mph what would their braking distances be? (Hint: if you quarter speed by what factor will the braking distance change?)
15^2 = 225 miles
Homework:
Wednesday, October 19th
Physics Words: Negative acceleration: a change in the velocity with respect to time of an object by decreasing speed in the positive direction or increasing speed in the negative direction
Checking Up Questions:
1. If a vehicle is traveling at constant velocity and the comes to a sudden stop, has it undergone negative acceleration or positive acceleration? Explain your answer.
The vehicle has undergone negative acceleration because the car is decreasing speed in a positive direction.
2. Explain how you know that increasing the velocity of an automobile increases the braking distance.
The faster the automobile goes, the more distance it needs to make a complete stop.
3. Why is the term negative acceleration used instead of deceleration?
In order to be clear about meanings in physics, the terms positive acceleration and negative acceleration are used instead of acceleration and deceleration. Positive represents one direction and negative represents the opposite direction. (An object could have a negative acceleration by decreasing its speed in the positive direction or increasing its speed in the negative direction.)
Notes:
Thursday, October 20th
equation 1Vi (2) = ratio of initial velocitiesVi (1)
equation 2Braking Distance (2) = ratio of braking distancesBraking Distance (1)
equation 3
(Vi (2))2 = braking distance (2)
(Vi (1))2 _braking distance (1)
Honda Civic Stopping Distance:__
Thursday, October 20th
Initial Velocity
Stopping Distance
10 mph
3.42 feet
20 mph
13.67 feet
30 mph
30.75 feet
40 mph
54.67 feet
50 mph
85.42 feet
60 mph
123 feet
70 mph
167.42 feet
80 mph
218.67 feet
90 mph
276.75 feet
100 mph
341.67 feet
4) If Initial Velocity is doubled how does stopping distance change?
the stopping distance is multiplied by 4 (use equation 3)
5) If the Initial Velocity is multiplied 4 times how does the stopping distance change?
16 times
6) If the Initial Velocity is halved how does the Stopping Distance Change?
the stopping distance is divided by 4 (stopping distance / 4 )
OR
multiply by .25
7) If the initial Velocity is quartered how does the stopping distance change?
cut it by 1/16 (use equation 2)
(10mph)2= times 1/16
(40mph)2
8) What speed would you need to have a stopping distance of a mile?
393.1 mph
Homework Physics to Go pg. 88-89 #1-8Monday, October 24th1. A student measured the braking distance of her automobile and recorded the data in the table. Plot the data on a graph and describe the relationship that exists between initial speed and braking distance.
As the initial speed increases, the braking distance increases.
{Screen_shot_2011-10-20_at_11.00.17_AM.png} Screen_shot_2011-10-20_at_11.00.17_AM.png
2. Below is a graph of the braking distances in relation to initial speed for two automobiles. Compare qualitatively (without using numbers) the braking distances when each automobile is going at a slow speed and then again at a higher speed. Which automobile is safer? Why? How did you determine what “safer” means in this question?
{Screen_shot_2011-10-20_at_10.55.20_AM.png} Screen_shot_2011-10-20_at_10.55.20_AM.png
^Automobile A is safer because it has a slower velocity and a lesser braking distance. The "safer" automobile is the one that would prevent you from getting into an accident if you needed to brake suddenly.
3. An automobile is able to stop in 20 m when traveling at 30 mi/h. How much distance will it require to stop when traveling at the following:
a) 15 mi/h? (half of 30 mi/h)(Vi[2]) / (Vi[1]) = (BD[2]) / (BD[1])(30/15)^2 = (20/x)x = 5m
b) 60 mi/h? (twice 30 mi/h)(30/60)^2 = (20/x)x = 80m
c) 45 mi/h? (three times 15 mi/h)(30/45)^2 = (20/x)x = 45m
d) 75 mi/h? (five times 15 mi/h)(30/75)^2 = (20/x)x = 125 m
4. An automobile traveling at 10 m/s requires a braking distance of 30 m. If the driver requires 0.9 s reaction time, what additional distance will the automobile travel before stopping? What is the total stopping distance, including both the reaction distance and the braking distance?
5. Consult the information for the sports car at the end of this chapter. This shows the stopping distance. How far would you expect this automobile to travel until coming to rest when brakes are applied at a speed of 30 mi/h?
6. Use the information for the sedan at the end of this chapter. Find the braking distances for 50 mi/h and 25 mi/h. Draw a graph using the different braking distances. Plot the speeds on the horizontal axis and the braking distances on the vertical axis.
7. Does the braking information for the sedan include the driver's reaction time? If it does not, then how much distance is added to the total braking distance, supporting that the driver has a 1/2 s reaction time? Who should let the consumer know about the 1/2 s reaction time--the information sheet or a drvier training manual?
8. Apply what you learned in this section to write a statement explaining the factors that affect stopping distance. The total stopping distance includes the distance you travel during your reaction time, plus the braking distance. What do you now know about stopping that will make you a safer driver? The factors that affect stopping distance are weather, distractions, reaction time, and weight of car. I now know that it takes more than just reacting to make the car come to a stop, so you have to be more aware of your surroundings and follow the speed limit.
{Photo_on_2011-10-24_at_08.03_#2.jpg} Photo_on_2011-10-24_at_08.03_#2.jpg
{Photo_on_2011-10-24_at_08.03.jpg} Photo_on_2011-10-24_at_08.03.jpg
Active Physics Plus:Monday, October 24th
Equations of Motion:d = .5(Vi+Vf)tVf = Vi + atd = Vit + (.5)(a)(t^2)Vf^2 = Vi^2 + 2ad
Average acceleration
1. As the shuttle bus comes to a sudden station to avoid hitting a dog, it accelerates uniformly at -4.1 m/s^2 as it slows from 9.0 m/s to 0 m/s. Find the time interval of acceleration for the bus.
a = -4.1 m/s^2t = ?Vi = 9 m/sVf = 0 m/sdon't need dVf = Vi + at0 = 9 + (-4.1)(t)-9 = -4.1tt = 2.2 s
2. At car traveling at 7.0 m/s accelerates uniformly at 2.5 m/s^2 to reach a speed of 12.0 m/s. How long does it take for this acceleration to occur?
Vi = 7 m/sVf = 12 m/sa = 2.5 m/s^2t = ?Vf = Vi + at12 = 7 + 2.5tt = 2 s
3. With an average acceleration of -0.50 m/s^2, how long will it take a cyclist to bring a bicycle with an initial speed of 13.5 m/s to a complete stop?
a = -0.5 m/s^2Vi = 13.5 m/sVf = 0 m/st = ?Vf = Vi + at0 = 13.5 + -0.5tt = 27 s
4. Turner's treadmill runs with a velocity of -1.2 m/s and speeds up at regular intervals during a half-hour workout. After 25 min, the treadmill has a velocity of -6.5 m/s. What is the average acceleration of the treadmill during this period?
Vi = -1.2 m/st = 25 min = 1500 sVf = -6.5 m/sa = ?Vf = Vi + at-6.5 = -1.2 + 1500aa = -0.004 m/s^2
5. Suppose a treadmill has an average acceleration of 4.7 * 10^-3 m/s^2
a. How much does its speed change after 5.0 min?a = 4.7 * 10^-3 m/s^2Vi = 0 m/sVf = ?t = 5 min = 300 sVf = Vi +atVf = 0 + (300)(4.7 * 10^-3)Vf = 1.41 m/s
b. If the treadmill's initial speed is 1.7 m/s, what will its final speed be?
Vi = 1.7 m/sVf = ?a = 4.7 * 10^-3 m/s^2t = 300 sVf = Vi + atVf = 1.7 + (4.7 * 10^-3)(300)Vf = 3.11 m/s
Displacement with constant uniform acceleration
1. A car accelerates uniformly from rest to a speed of 23.7 km/h in 6.5 s. Find the distance the car travels during this time.
t = 6.5sd = ?Vi = 0 m/sVf = 23.7 km/h = (23.6 km / 1 hr) * (1000 m / 1 km) * (1 hr / 60 min) * (1 min / 60 sec) = 6.58 m/sd = .5(Vi+Vf)td = (.5)(0 + 6.58)(6.5)d = 21.385 m
2. When Maggie applies the brakes of her car, the car slows uniformly from 15.0 m/s to 0 m/s in 2.50 s. How many meters before a stop sign must she apply her brakes in order to stop at the sign?
Vi = 15 m/sVf = 0 m/st = 2.5 sd = ?d = (.5)(Vi+Vf)td = (.5)(15 + 0)(2.5)d = 18.75 m
3. A jet plane lands with a speed of 100 m/s an can accelerate uniformly at a maximum rate of -5.0 m/s^2 as it comes to rest. Can this plane land at an airport where the runaway is 0.80 km long?
Vi = 100 m/sa = -5 m/s^2Vf = 0 m/sd = ?Vf^2 = Vi^2 + 2ad0 = 100^2 + 2(-5)(d)d = 1000 mNo it cannot land.
4. A driver in a car traveling at a speed of 78 km/h sees at cat 101 m away on the road/ How long with it take for the car to acceleration uniformly to a stop in exactly 99 m?
t = ?d = 99 mVi = 78 km/h = 21.67 m/sVf = 0 km/hd = (.5)(Vi+Vf)t99 = (.5)(21.67 + 0)(t)t = 9.14 s
5. A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3.2 km in 3.5 min. How fast is the car moving after this time?
Vi = 6.4 m/sVf = ?d = 3.2 km = 3200 mt = 3.5 min = 210 sd = (.5)(Vi+Vf)t3200 = (.5)(6.4 + Vf)(210)Vf = 24.08 m/s
Velocity and displacement with uniform acceleration
1. A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s^2 for 3.6 s. Find the final speed and the displacement of the car during this time.
a = 0.92 m/st = 3.6 sd = ?Vi = 23.7 km/h = 6.58 m/sVf = ? = 9.9 m/s(23.7 km / 1 hr) * (1000 m / 1 km) * (1 hr / 60 min) * (1 min / 60 sec) = 6.58 m/sVf = Vi + atVf = 6.58 + (0.92)(3.6)Vf = 9.9 m/sFind displacement (d)d = (.5)(Vi + Vf)(t)d = (.5)( 6.58 + 9.9)(3.6)d = 29.7 m
2. An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.0 m/s^2. Find the final speed and the displacement after 5.0 s.
Vi = 4.3 m/sa = 3 m/s^2t = 5 sVf = ? = 19.3d = ?Vf = Vi + atVf = 4.3 + (3)(5)Vf = 19.3 m/sd = (.5)(Vi + Vf)(t)d = (.5)(4.3 + 19.3)(5)d = 59 m
3. A car starts from rest and travels for 5.0 s with a uniform acceleration of -1.5 m/s^2. What is the final velocity of the car? How far does the car travel in this time interval?
t = 5 sa = -1.5 m/s^2Vi = 0Vf = ? = -7.5 m/sd = ?Vf = Vi + atVf = 0 + (-1.5)(5)Vf = -7.5 m/sd = (.5)(Vi + Vf)(t)d = (.5)(0 + -7.5)(5)d = -18.75 m
4. A driver of a car traveling at 15.0 m/s applies the brakes, causing a uniform acceleration of -2.0 m/s^2. How long does it take the car to accelerates to a final speed of 10.0 m/s? How far has the car moved during the braking period?
a = -2 m/s^2Vi = 15 m/sVf = 10 m/st = ? = 2.5 m/sd = ?Vf = Vi + at10 = 15 + (-2)(t)t = 2.5 sd = (.5)(Vi + Vf)(t)d = (.5)(15 + 10)(2.5) d = 31.25 m
Final velocity after any displacement
2. A car traveling at 7.0 m/s accelerates uniformly at the at the rate of 0.8 m/s^2 for a distance of 245 m.
a = 0.8 m/s^2d = 245 mVi = 7 m/sVf = ?
a. What is tits velocity at the end of the acceleration?Vf^2 = Vi^2 + 2adVf^2 = (7)^2 + (2)(0.8)(245)Vf^2 = 441 m^2/s^2Vf = 21 m/s
b. What is its velocity after it accelerates for 125 m?Vf^2 = Vi^2 + 2adVf^2 = (7)^2 + (2)(0.8)(125)Vf = 15.78 m/s
c. What is its velocity after it accelerates for 67 m?Vf^2 = Vi^2 + 2adVf^2 = (7)^2 + (2)(0.8)(67)Vf = 12.5 m/s
3. A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s^2.
Vi = 0 m/sVf = ? = 15.9 m/sa = 2.3 m/s^2
a. What is the speed of the car after its has traveled 55 m?
Vf^2 = Vi^2 + 2adVf^2 = 0 + 2(2.3)(55)Vf = 15.9 m/sb. How long does it take the car to travel 55 m?d = (.5)(Vi + Vf)(t)55 = (.5)(0 + 15.9)(t)t = 6.9 s
4. A certain car is capable of acceleration at a uniform rate of 0.85 m/s^2. What is the magnitude of the car's displacement as it accelerates uniformly from a speed of 83 km/h to one of 94 km/h?
a = 0.85 m/s^2Vi = 83 km/h = 23.1 m/sVf = 94 km/h = 26.1 m/sd = ?Vf^2 = Vi^2 + 2ad26.1^2 = 23.1^2 + 2(0.85)(d)d = 86.9 m
5. An aircraft has a liftoff speed of 120 km/h. What minimum uniform acceleration does this require if the aircraft to be airborne after a takeoff run of 240 m?
Vi = 0 m/sVf = 120 km/h = 33.33 m/sa = ?d = 240 mVf^2 = Vi^2 + 2ad33.33^2 = 0 + 2(a)(240)a = 2.27 m/s^2
6. A motorboat acceleration uniformly from a velocity of 6.5 m/s to the west to a velocity of 1.5 m/s to the west. If its acceleration was 2.7 m/s^2 to the east, how far did it travel during the acceleration?
Vi = 6.5 m/sVf = 1.5 m/sa = 2.7 m/s^2d = ?Vf^2 = Vi^2 + 2ad1.5^2 = 6.5^2 + (2)(2.7)(d)d = -7.41 m
Mini Challenge Plan
Thursday, October 27th
Nicky: making keynote, storyline, equation 2 (with Kate)Mel: graph 1, equation 1, responsibilities Kate: graph 2, equation 2 (with Nicky)
Allison: graph 3, equation 3
Storyline: Allison was driving home from the football game when the following happened:
1. (NICKY) Allison was driving us home from the football game and we were approaching an intersection at 20 m/s. The light turned yellow when she was 35 meters away. Does she have enough time to stop if she has an acceleration of -2 m/s?
A= -2 m/sD= 50 mT= ?Vi= 20 m/sVf= 0 m/sEquation: vf=vi+at
2.(ALLISON) Allison is driving at 70 m/s and is approaching a curve that has a speed limit of 11 m/s. She has 15 meters to brake. What should her deceleration be?
A= ?D= 15 mT= XVi= 70 m/sVf= 11 m/sEquation: vf^2= Vi^2+2ad
3. (MEL) Allison has a reaction time of .5 seconds, which includes moving her foot from gas to brake. They are going through a school zone at 11 m/s and the car has a maximum deceleration of -5 m/s^2. What is Allison’s reaction distance, braking distance, and total stopping distance?
A= -5 m/s^2T= .5 sD= ?Equation: v= d/t
Graphs:Distance vs Time (Allison)Velocity vs Time (Kate and Nicky)Braking Distance vs Initial Velocity (Mel)
Equations:Braking DistanceDecelerationv=d/t
Total Stopping Distance ActivityFriday, October 28th
1) Draw a diagram of a car moving down a road and braking for an object.
{tots_stop_distamce.jpg} tots_stop_distamce.jpg
2) Assuming we are traveling forward with a positive velocity, then we would have a negative acceleration when we are braking.
Calculating Reaction Distance
1) Finding reaction distance using diagram abovev=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=1sd=10m
3) v=20m/s, t=1sd=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.
Prep for Chapter Challenge: Friday, October 28th
1) tr=.5s, vi=11, a=-4, dr=?, db=?, TSD = ?11 = d/.5 (v=reaction distance/reaction time)reaction distance = 5.5 m0 = 121 + 2(-4)dbraking distance = 15.13 mtotal distance = 20.63 m
2) vi=27, dr=?, db =?, TDP =?, a=-4, t=.527=d/.5reaction distance = 13.5 m0 = 729 + 2(-4)dbraking distance = 91.13 mtotal distance = 104.63 m
3) vi=27, t=1, dr=?,db=?, TSD =?, a=-4reaction distance = 27 m0=729 + 2(-4)dbraking distance = 91.13 mtotal distance = 118.13 m
4) t=.5, vi=27, dr=?, db=?, TSD=?, a=-227=d/.5reaction distance = 13.5 m0 = 729 + 2(-2)dbraking distance = -182.25 mtotal distance = 195.75 m
Create a problemWednesday, November 2nd
-Solve for dr, db TSD.-Embed givens in problem.-Don't embed givens that don't need to be mentioned.
My Problem:
A teenager speeds up to a yellow light. What would the reaction time, braking distance , and total stopping distance be for the teenager be if They are traveling at 20 m/s (44 mph) and have a reaction time of .4s, and a deceleration of -5 m/s^2
tr = .4 sVi = 20 m/sa= -5 m/s^2V= d/t20 m/s (.4 s) = dreaction/ .4s (.4s)dreaction = 8 mVf^2 = Vi^2 +2ad-(20 m/s)^2 / -10 = dbrakingdbraking = 40 mTSD = 48 m.
Alex's Problem:
Your driving in your brand new customized car that Bill Trembath bought you . You're going down your street and have a reaction time of .5 seconds and traveling at 25 m/s with a deceleration of -6 m/s^2. What is your reaction distance , braking distance, and total stopping distance?
tr = .5 secondsVi = 25 m/sa= -6 m/s^2V= d/t25 m/s (.5 s) = d/ (.5 seconds) (.5 s)dreaction = 12.5 m
Vf^2= Vi^2 + 2ad-(25 m/s)^2 / -12 = dbraking52.08 m= dbrakingTSD = 64.58 m

Chapter 1: Section 4
Learning Outcomes
Wednesday, October 5th
-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.
What do you see/ What do you think?
Wednesday, October 5th
I see a red cat that is speeding up within a matter of a small amount of time. The Man and the dog look like they were crossing the street, and then they ran because the red car started to speed up.
What are some differences and similarities of the motion of these 2 vehicles as each goes from a stop to f 30 mi/h?
The car reaches the top speed faster than the bus. the car has greater friction with the road.
Investigation
Wednesday, October 5th
-Tangent line: line that only touches a curve at one point.
-Slope of the tangent: the instantaneous velocity at a specific point in time.
1. If you were to place the cart at the top of the ramp and release it to freely move down the ramp would it move through the first half of the distance in the same amount of time as the second half of the distance? Why or why not?
No it would not because the car builds up momentum and speed as it moves down the hill.
2. Below are four different distance vs time graphs. Copy and paste the graphs with their correct descriptions of motion into your wiki.
{Screen_shot_2011-10-05_at_9.13.09_AM.png} Screen_shot_2011-10-05_at_9.13.09_AM.png
Screen_shot_2011-10-05_at_9.13.09_AM.png
{Screen_shot_2011-10-05_at_9.13.18_AM.png} Screen_shot_2011-10-05_at_9.13.18_AM.png
Screen_shot_2011-10-05_at_9.13.18_AM.png
--------------> Moves slowly and then accelerates.
{Screen_shot_2011-10-05_at_9.13.13_AM.png} Screen_shot_2011-10-05_at_9.13.13_AM.png
Screen_shot_2011-10-05_at_9.13.13_AM.png
{Screen_shot_2011-10-05_at_9.13.24_AM.png} Screen_shot_2011-10-05_at_9.13.24_AM.png
Screen_shot_2011-10-05_at_9.13.24_AM.png
-------------->Moves fast and then slows down.
How is the velocity changing with respect to time?
As time increases, velocity increases.
Tangent Line- The tangent line is at the same position and time in the distance and velocity graphs.
Slope=
Tangent Line 1
{1a.png} 1a.png
1a.png
Tangent Line 2
{2a.png} 2a.png
2a.png
Velocity
{1b_&_2a.png} 1b_&_2a.png
1b_&_2a.png
Velocity Time Graph-
{1b_&_2a.png} 1b_&_2a.png
1b_&_2a.png
(1 meter per second)
.664 is the
Calculate Acceleration:
Physics Talk:
Wednesday, October 5th
Physics Words:
Vector: a quantity that has both magnitude and direction
Negative acceleration: a decrease in velocity with respect to time. The object can slow down or speed up.
Positive Acceleration: an increase in velocity with respect to time. The object can speed up or slow down.
1. Acceleration = change in velocity/change in time
a= ∆v
∆t
2. Meters per seconds squared; (m/s)/s
3. A vector quantity is a quantity that has both magnitude and direction. Scalar quantity is described by magnitude alone.
4. Constant velocity graph:
{Screen_shot_2011-10-05_at_3.25.22_PM.png} Screen_shot_2011-10-05_at_3.25.22_PM.png
Screen_shot_2011-10-05_at_3.25.22_PM.png
Constant acceleration graph:
{Screen_shot_2011-10-05_at_3.25.40_PM.png} Screen_shot_2011-10-05_at_3.25.40_PM.png
Screen_shot_2011-10-05_at_3.25.40_PM.png
5. The slope of a graph represents the rate of acceleration or deceleration.
(1 meter per second)
.664
Calculate Acceleration-
rise/run
∆v/∆t
.6/1.1= .55 meters/second squared
3. Predict-
{Photo_on_2011-10-06_at_08.17.jpg} Photo_on_2011-10-06_at_08.17.jpg
Tangent Line- Velocity increases as time goes on.
Homework:
Thursday, September 6th
Physics Words:
Acceleration: the change in velocity with respect to a change in time
Vector: a quantity has both magnitude and direction
Negative acceleration: a decrease in velocity with respect to time. The object can slow down (20 m/s to 10 m/s or speed up (-20 m/s to -30 m/s)
Positive acceleration: an increase in velocity with respect to time. The object can speed up (20 m/s to 30 m/s) or slow down (-20 m/s to -10 m/s)
Tangent line: a straight line that touches a curve in only one point
Checking Up Questions:
1. Give the defining equation for acceleration in words, and by using symbols.
Acceleration is the change in velocity with respect to time. a=∆v/∆t
2. What is an SI unit for measuring acceleration? Use words and unit symbols to decribe the unit.
The SI unit is (m/s)/s or m/s^2, (km/h)/s, or ft/s^2.
3. What is the difference between a vector and a scalar quantity?
A quantity that involves both direction and size (magnitude is called a vector quantity (ex. velocity...indicates a change in position over a period of time and the direction). A quantity that has size, but not direction, is called a scalar quantity (ex. speed...only indicates the change in position over a period of time in a straight line).
4. Sketch a distance-time graph for
a) constant velocity:
Activity: Car vs Truck vs Bus
Friday, October 7th
{Photo_on_2011-10-10_at_16.59.jpg} Photo_on_2011-10-10_at_16.59.jpg
DoNow: Physics Talk Review
Friday, October 7th
1. a= 30/5= 6m/s^2
2. Vectors: Velocity, Force (push and pull)
Scalar: Speed, Calories, Anything you can count. A vector involves direction and speed. A scalar involves size but no direction.
Physics To Go #1-16 pg. 68
Due Thursday, October 13th
1.) Can a situation exist in which an object has zero acceleration and nonzero velocity? Explain.
No, an object can never have 0 acceleration with a nonzero velocity whenever you are moving, that is your velottiy. You divide that by the time it took you to travel that speed and you would get acceleration. You would never get 0 unless you had 0m/s for your velocity.
2.)Can a situation exist in which an object has zero velocity and nonzero acceleration, even for an instant? Explain.
No, an object can never have 0 velocity because for it have 0 velocity it would need to also have 0 acceleration because anything divided by 0 is zero.
3.)If two automobiles have the same acceleration, do they have the same velocity? Why or why not?
They don't need to have the same velocity because they could have the same acceleration but could not have achieved it in the same velocity. For example, they could both have 5 m/s for acceleration but one velocity can be 25m and the other one could be 15m.
4.)If two automobiles have the same velocity, do they have the same acceleration? Why or why not?
They don't need to have the same acceleration to have the same velocity because they could have achieved that velocity in a different amount of time. 20m could have been achieved in 5s or 2s leaving it with different accelerations.
5.)Can an accelerating automobile be overtaken by an automobile moving with constant velocity?
No because eventually the accelerating automobile will catch up.
6.) Is it correct to refer to speed-limit signs instead of velocity-limit signs? Why or why not? What units are assumed for speed limit signs in the U.S.?
We refer to speed-limit signs because velocity only refers to speed in a certain direction where speed limit signs are in speeds in every direction. It is better for drivers to see. The units associated with speed-limit signs are miles per hour.
7.) Suppose an automobile were accelerating at 2mi/hr every 5s and could keep accelerating for 2 min at that rate.
a). a=∆v/∆t
a=2mi/h
v= ?
t= 120s/5s= 24s
2=∆v/24
∆v= 48mi/s
b). d= vt
d= (48
8.) At an international auto race, a race car leaves the pit after a refueling stop and accelerates uniformly to a speed of 75 m/s in 9s to rejoin the race.
A. What is the racecars acceleration at this time?
a=∆v/∆t
a=75/9
a=8.3 mi/s
B. What was the race cars average speed during the acceleration?
C. How far does the race car go during the time it is accelerating?
D. A second race car leaves after its pit stop and accelerates to 75 mi/s in 8s. Compared to the first face car, what is this race cars acceleration, average speed during the acceleration, and distance traveled?
9.) a). a=∆v/∆t
a= (.6-4.5)/(1.3-0)
a= -3 m/s
b). d=1/2at^2
d=1/2(-3)(1.3)^2
c). d=1/2(3)(1.1)62
d= 1.8
d= Trial 2
10.) a). Around 12m/s
b). a=∆v/∆t
a= (9)/(7.5)
a= 1.2m/s
c). It would have picked up speed and the final velocity would have been greater.
11.) a). B
b). D
c). E
d). A
e). F
f). C
12.) a). ef,
b). ac
c). ce
d). eg
e). 700m
f). 0m. Back at the starting point
13.) a). a=∆v/∆t
a= (250)/(30)
a= 8.3m/h/s (After each seconds he is gaining 8.3 mi/h.)
b). ∆t= 45 sec
Vf= ?
Vi= 0mi/hr
a= 8.33 mi/hr/sec
8.33=Vf-0 / 45
Vf= 374.85= 375mi/h
c). Vf= 500 mi/h
Vi= 250 mi/h
∆t= ?
a= 8.33 mi/hr/sec
8.33= (500)-(250) / ∆t
8.33= 250/ ∆t
∆t= 30.01 seconds x2 (because of going from 0 to 500 mi/h)
60 seconds
d). d= 1/2at^2
d=1/2(8.33)(60)^2
d=15000 m
14.) a). Vf= ?
a= 9.8 m/s^2
Vi= 0 m/s
∆t= 4.516 s
9.8= Vf-0 / 4.516
Vf= 44.26 m/s
b). 100m=1/2(9.8m/s^2)∆t^2
t= 4.51 seconds
c). a=∆v/∆t
9.8m/s^2= ∆v/ 10s
∆v= 98 m/s
d). d=1/2(9.8)(10)^2
d= 490m
e). The answers would have to be around 6 times less on the Moon than on Earth.
(16)
a). 1.0 seconds
d= 1/2(4)(1)^2
d= 2 meters
b). 2.0 seconds
d=1/2(4)(2)^2
d= 8 meters
c). 3.0 seconds
d=1/2(4)(3)^2
d= 18 meters
d). 4.0 seconds
d=1/2(4)(4)^2
d= 32 meters
Speeding Up / Slowing Down
Wednesday, October 12th
Speeding Up
{Photo_on_2011-10-12_at_08.02.jpg} Photo_on_2011-10-12_at_08.02.jpg
Slowing Down
{Photo_on_2011-10-12_at_08.01.jpg} Photo_on_2011-10-12_at_08.01.jpg
Questions/Answers for speeding up and slowing down:
{Photo_on_2011-10-12_at_08.01_#2.jpg} Photo_on_2011-10-12_at_08.01_#2.jpg
Active Physics Plus:
Wednesday, October 12th
d= 1/2at^2 + Vit
1. Vf = 20 m/s
Vi= 7 m/s
a= 3 m/s^2
∆t= ?
a= Vf-Vi
∆t
3 m/s^2= 20 m/s - 7 m/s ––> ∆t (3m/s^2) = (20 m/s - 7 m/s) ––> ∆t = 4.33 sec
1 3 3 m/s^2 3m/s^2
2. a= 1.5 m/s^2
∆t= 10 seconds
Vi= 7 m/s
Vf= ?
1.5 = Vf - 7 m/s ––> 1.5(10) = Vf - 7 m/s ––> Vf = 22 m/s
10s
3a. a= ∆v/∆t
Vf=20 m/s
Vi= 0 m/s
∆t= 5 sec
a=?
a= 20 m/s - 0 m/s ––> a= 4 m/s^2
5 sec
3b. d= 1/2at^2
d= ?
a= 4 m/s^2
t= 5 sec
d= 1/12 (4 m/s^2)(5 sec) ––> d= 50 m
4. a= 4 m/s^2
∆t= 10 sec
d=?
d= 1/2at^2
d= (1/2)(4 m/s^2)(10 sec) ––> d= 200 m