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20. The tortoise and the hare are in a road race to defend the honor of their breed. The
tortoise crawls the entire 1000. m distance at a speed of 0.2000 m/s while the rabbit
runs the first 200.0 m at 2.000 m/s The rabbit then stops to take a nap for 1.300 hr and
awakens to finish the last 800.0 m with an average speed of 3.000 m/s. Who wins the
race and by how much time?

T=D/V

T=1000 meters/0.2000 meters per second

T=5000 seconds

5000/60

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It would take around an hour and 20 minutes for the tortoise to finish the race, if
going at a constant speed of 0.2000 meters per second

T=200 meters /2 meters per second
T= 100 seconds

It would take the hare one minute and 40 seconds to run the first 200 meters of the
1000 meter race

60*1.3=78

The hare stops to take a nap for 78 minutes making the first 200 meters with the
nap to be 79 minutes and 40 seconds.

T=800 meters/ 3 meters per second
T=266.67
266.67/60
T=4.4445

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It would take the hare around 83 minutes and 40 seconds to finish the 1000 meter
race.

In the 1000 meter race, the tortoise would beat the hare by 3 minutes with the
tortoise running a constant pace of 0.2000 meters per second, and with the hare
running 2 meters per second, then stopping to nap, and then finishing by running 3
meters per second.

Two physics professors challenge each other to a 100. m race across the football field.
The loser will grade the winner's physics labs for one month. Dr. Rice runs the race in
10.40 s. Dr. De La Paz runs the first 25.0 m with an average speed of 10.0 m/s, the next
50.0 m with an average speed of 9.50 m/s, and the last 25.0 m with an average speed of
11.1 m/s. Who gets stuck grading physics labs for the next month?

Dr Rice=10.4 seconds
Dr De La Paz
Time=Distance/Velocity
T=25 meters/10 meters per second
T= 2.5 seconds
T=50 meters/9.5 meters per second

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T=5.26 seconds
T=25 meters/11.1 meters per second
T=2.25 seconds
2.5+5.26+2.25=10.01

Dr. Rice would be grading the chemistry labs for the next month as they ran the 100
meters 0.39 hundreths of a second slower than Dr. De La Paz.

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Presentation: Weird Weather in Siberia

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Name: Date:
________________ _______
______________ _______
_______
_

Hypothesis: Velocit Velocit
Independent
Variable: Dist. 1 Time 1 y 1 Dist. 2 Time 2 y 2 Acceleration
Dependent
Variable: 61 0.56 109 61 0.53 115 11
104
Write Units --> 61 0.87 70 61 0.51 123
6
Trial 61 0.93 65 61 0.87 70 30
angle 1 = 61 0.78 78 61 0.63 97
opp/hyp 127
23/61=0.377 100
angle 1= 22 188
Degrees 130
angle 1 = 22
Degrees
angle 1 = 22
Degrees

avg.

angle 2 61 0.98 62 61 0.49 124
13/61=0.2131 61 1.09 56 61 0.55 111
angle 2= 12 61 1.17 52 61 0.4 127
Degrees 61 1.08 56 61 0.5 121
angle 2= 12
Degrees
angle 2= 12
Degrees

avg.

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angle 3 61 1.95 31 61 2.03 30 -0.5
3/61=0.049 61 1.77 34 61 2.92 21 -4
angle 3= 3 61 1.92 32 61 3.57 17 -4
Degrees 61 1.88 32 61 2.84 22
angle 3= 3 -3.5
Degrees
angle 3= 3
Degrees

avg.

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Acceleration Worksheet. Name:
Date: ______________
_________________________

14.2 Acceleration

Acceleration is the rate of change in the speed of an object. To determine the rate of
acceleration, you use the formula below. The units for acceleration are meters per second per
second or m/s2.

A positive value for acceleration shows speeding up, and negative value for acceleration shows
slowing down. Slowing down is also called deceleration.

The acceleration formula can be rearranged to solve for other variables such as final speed (v2)
and time (t).

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EXAMPLES

1. A skater increases her velocity from 2.0 m/s to 10.0 m/s in 3.0 seconds. What is the skater’s
acceleration?

Looking for Solution
Acceleration of the skater

The acceleration of the skater is 2.7 meters per
second per second.

Given
Beginning speed = 2.0 m/s
Final speed = 10.0 m/s
Change in time = 3 seconds

Relationship

2. A car accelerates at a rate of 3.0 m/s2. If its original speed is 8.0 m/s, how many seconds will it
take the car to reach a final speed of 25.0 m/s?

Looking for Solution
The time to reach the final speed.

`

The time for the car to reach its final speed is 5.7
seconds.

Given
Beginning speed = 8.0 m/s; Final speed = 25.0 m/s
Acceleration = 3.0 m/s2
Relationship

1. While traveling along a highway a driver slows from 24 m/s to 15 m/s in 12 seconds. What is
the automobile’s acceleration? (Remember that a negative value indicates a slowing down or
deceleration.)

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A = (V2 - V1)/T
A = (15 m/s - 24 m/s)/12 Sec.
A = -9 m/s/12 sec.
A = -¾m/s2

2. A parachute on a racing dragster opens and changes the speed of the car from 85 m/s to 45 m/s
in a period of 4.5 seconds. What is the acceleration of the dragster?

A=(V2-V1)/T
A=(45 m/s -85 m/s)/4.5 seconds

A=-40 m/s/4.5 seconds

A=-9 m/s2

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The car decelerates by around 9 meters per second.

3. The table below includes data for a ball rolling down a hill. Fill in the missing data values in
the table and determine the acceleration of the rolling ball.

Time (seconds) Speed (km/h)

0 (start) 0 (start)

23

46

69

8 12

10 15

Acceleration = ___________________________

A=(V2-V1)/T
A=(15 km/hour-0 km/hour)/12 seconds

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A=15 km/hour/12 seconds

A=1.25 km/hour/sec

The ball accelerates at a rate of 1.25 kilometers an hour
each second.

4. A car traveling at a speed of 30.0 m/s encounters an emergency and comes to a complete stop.
How much time will it take for the car to stop if it decelerates at -4.0 m/s2?

T=(V2-V1)/A
T=(0-30)/-4

T=-30/-4

T= 7.5

It would take 7.5 seconds for a car to stop if traveling at a

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speed of 30 m/s and it decelerates at a rate of -4.0 m/s2

5. If a car can go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0
seconds if its starting speed were 50 mi/hr?

A=(V2-V1)/T
A=60-0/8

A=7.5 mi/hr/s

The car accelerates at a rate of 7.5 miles per hour per
second.

V2= V1+ (A * T)
V2= 50+ (7.5 *5)
V2=87.5 miles per hour
If the car started at a speed of 50 miles per hour and

accelerated at a rate of 7.5 miles per hour per second,
after 5 seconds the car would be going 87.5 miles per
hour.

6. A cart rolling down an incline for 5.0 seconds has an acceleration of 4.0 m/s2. If the cart has a

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beginning speed of 2.0 m/s, what is its final speed?

V2= V1 + (A*T)
V2= 2 m/s+ (4 m/s2*5 seconds)
V2= 2 m/s+ 20 m/s
V2= 22 m/s
The cart’s final speed would be 22 meters per second.

7. A helicopter’s speed increases from 25 m/s to 60 m/s in 5 seconds. What is the acceleration of
this helicopter?

A=(V2-V1)/T

A= (60 m/s- 25 m/s)/5
A= 35 m/s/5 seconds
A= 7 m/s2
The acceleration of the helicopter is 7 meters per second

8. As she climbs a hill, a cyclist slows down from 25 mi/hr to 6 mi/hr in 10 seconds. What is her
deceleration?

A=(V2-V1)/T
A= (6 mi/hr - 25 mi/hr)/10 seconds

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A= (-19 mi/hr)/10 seconds
A=- 1.9 mi/hr/sec
As she goes up the hill, she is decelerating at a rate of 1.9
miles per hour per second.

9. A motorcycle traveling at 25 m/s accelerates at a rate of 7.0 m/s2 for 6.0 seconds. What is the
final speed of the motorcycle?

V2= V1 + (A*T)
V2= 25 m/s + (7.0 m/s2* 6 sec.)
V2= 25 m/s+ 42 m/s
V2=67 m/s
The final speed of the motorcycle would be 67 meters per
second

10. A car starting from rest accelerates at a rate of 8.0 m/s/s. What is its final speed at the end of
4.0 seconds?

V2= V1 + (A*T)
V2= 0+ (8 m/s2*4 sec.)
V2= 0+32 m/s
V2=32 m/s
After four seconds, the speed of the car would be 32

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meters per second.

11. After traveling for 6.0 seconds, a runner reaches a speed of 10 m/s. What is the runner’s
acceleration?

A= (V2-V1)/T
A=(10 m/s - 0 m/s)/6 sec
A=10 m/s/6 sec
A= 1.6667 m/s2
The runner accelerates at a rate of 1 ⅔ meters per second.

12. A cyclist accelerates at a rate of 7.0 m/s2. How long will it take the cyclist to reach a speed of
18 m/s?

T= (V2 - V1)/A
T= (18 m/s - 0 m/s)/7 m/s2

T= 18 m/s/7 m/s2

T= 2.57 seconds

It would take a little over 2 and a half seconds for the cyclist
to accelerate to a rate of 18 meters per second.

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13. A skateboarder traveling at 7.0 meters per second rolls to a stop at the top of a ramp in 3.0
seconds. What is the skateboarder’s acceleration?

A= (V2-V1)/T
A= (7-0)/3
A=2.3333333
The acceleration of the skateboarder is about 2 and ⅓
meters per second.

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Projectile Motion Simulator

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Quiz Review:

QUIZ REVIEW: Motion

Name: ________________________ Date:
___________

Formulas:

A= v2 −v1 V2 = V1 + (a * T) T= V2−V1
T2 a

1. After traveling for 14.0 seconds, a bicyclist reaches a speed of 89 m/s. What is the runner’s
acceleration?

A=(V2-V1)/T
A=(89 m/s-0 m/s)/14 sec
A= 89 m/s/14/sec
A= 6.357 m/s2
The bicyclist accelerates at a rate of 6.357 meters per second

2. A car starting from rest accelerates at a rate of 18.0 m/s2. What is its final speed at the end of
5.0 seconds?

V2= V1 +(A*T)
V2= 0 + (18 m/s2 * 5 sec)
V2= 90 m/s
The car’s speed after 5 seconds is 90 meters per second.

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A cyclist accelerates at a rate of 16.0 m/s2. How long will it take the cyclist to reach a speed of
49 m/s?

T=(V2-V1)/A
T=(49 m/s - 0 m/s)/16 m/s2
T= 49 m/s/16 m/s2
T=3.0625
It would take 3.0625 seconds for a cyclist to reach a speed of 49 m/s
if they accelerated at a rate of 16 meters per second.

Make Drawings of Each Problem
3. During an Apollo moon landing, reflecting panels were placed on the moon. This allowed
earth-based astronomers to shoot laser beams at the moon's surface to determine its distance.
The reflected laser beam was observed 6.5 seconds after the laser pulse was sent. The speed of
light is 3.0 × 108 m/s. What was the distance between the astronomers and the moon?
*Is this distance correct? Research the distance to the moon.
No, the correct distance to the moon is 384,400,000 meters or 3.844*108

D=V*T
D= ½(3.0*108[6.5])
D=9.75*108

3. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for French
class for the third time this week. She must get from one side of the school to the other by
hurrying down three different hallways. She runs down the first hallway, a distance of 65.0
m, at a speed of 5.2 m/s. The second hallway is filled with students, and she covers its 32.0
m length at an average speed of 1.46 m/s. The final hallway is empty, and Suzette sprints its
60.0 m length at a speed of 7.3 m/s.
a. Does Suzette make it to class on time or does she get detention for being late again?

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T=D/V
T=65 m/5.2 m/s
T=12.5 seconds
Suzette ran down the first hallway in 12.5 seconds

T=D/V
T=32m/1.46m/s
T=22 seconds
Suzette ran down the second hallway in 22 seconds
T=D/V
T=60m/7.3 m/s
T=8 seconds
It took Suzette 8 seconds to run down the final hallway. This means
that she would make it to class in 42.5 seconds meaning that she
wouldn’t be late.

4. The tortoise and the hare are in a road race to defend the honor of their breed. The tortoise
crawls the entire 1000. m distance at a speed of 0.35 m/s while the rabbit runs the first 200.0
m at 1.85 m/s The rabbit then stops to take a nap for 1.200 hr and awakens to finish the last
800.0 m with an average speed of 4.2 m/s. Who wins the race and by how much time?

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Tortoise Math:
T=D/V
T=1000 meters/0.35 meters per second
T=2857.14285714 seconds
2857.1428714/60=47.61
It took the tortoise a little over 47 minutes and 37 seconds for
the tortoise to finish the race.

Hare Math: 229/310
T=D/V
T=200/1.85
T=108.10810810
It took the hare 1 minute and 48 seconds to run the first 200
meters of the race, this would mean that the hare would finish
the race in a total of exactly 9 minutes.
1.2 Hours= 1 hour 12 minutes
Total time= 1 hour 13 minutes and 48 seconds
T=800/4.2

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T=190.47619047
It took the hare 3 minutes and 10 seconds to run the last 800
meters of the race. If the hare had been running at this pace the
entire time, it would’ve finished the race in a total of 3 minutes
and 58 seconds.
Total time= 1 hour 16 minutes and 58 seconds
The tortoise would’ve finished the race before the hare by a
little over 30 minutes.

5. What is the Acceleration of the Cart on the Ramp? Determine the Angle of the Ramp (A).

Angle Chart: https://drive.google.com/open?id=0B4RmhXJlHvo1YXZhcDNMSDNSMXc

Which Angle had the greatest Acceleration? Write a Conclusion based on your findings.
Create a Graph if you have time.

Height of Dist Time Velocity Velocity
Ramp 2 Acceleration
.1 1 1 Dist. 2 Time 2
(Opposite)

250 250 m 5 sec. 50 m/s A=V2-V1/T
m 10 sec. 25 m/s

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50 m A=50-25/5
A=5m/s2
Angle=
Opp/Hypot A=V2-V1/T
A=125-50/2
∠a=50/500
∠a=0.1 A=75/2
∠a=6° A=37.5m/s2

Angle=
Opp/Hypot

∠b=100/250

∠b=0.4

∠b=24° 250 50 m/s 250 m 2 sec. 125 m/s
m 5 sec.
100 m

Use a2 + b2 = c2 to determine the length of side b

b2=-a2+c2

b2=-502+2502

b2=2500+62500

b2=65000

Graph:

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X-axis = Angle
Y-axis = acceleration

Conclusion:
Purpose of experiment, hypothesis, brief description of experiment, data evidence to support
hypothesis, conclusion

EXTRA CREDIT:
Light from another star in the galaxy reaches the earth in 46 minutes. The speed of light is 3.0
× 108 m/s. In kilometers, how far is the earth from the star?

Answer must be in scientific notation

D= V*T
D= (3.0*108 m/s)(2.76*103 sec)
D=8.28*1011 m
D=8.28*108 km

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QUIZ: Motion

Name: Martin Roberg Date: 3/19
Class: Galileo

Formulas:

A= v2 −v1 V2 = V1 + (a * T) T= V2−V1
T2 a

1. After traveling for 18.0 seconds, a fighter jet reaches a speed of 125 m/s. What is the
fighter jet’s acceleration?

A=V2-V1/T

A=125 m/s -0 m/s /18 sec

A=6.944444444 m/s2

The Fighter Jet accelerates at a constant rate of 6.944444444
meters per second

2. A Ford Mustang GT started at 15 m/s and accelerates at a rate of 18.0 m/s2. What is its final
speed at the end of 7.0 seconds?

V2=V1 + (A*T)
V2=15 m/s + (18 m/s2*7 sec)

V2=15 m/s+126 m/s

V2=141 m/s

The final speed at the end of 7 seconds for a Ford Mustang GT that
started at 15 m/s and accelerated at the rate of 18 m/s2 is 141 m/s

3. A cyclist accelerates at a rate of 12.0 m/s2. How long will it take the cyclist to reach a speed
of 115 m/s?

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T=V2-V1/A
T=115 m/s -0 m/s /12 m/s2

T=115 m/s/12 m/s

T=9.583333333 seconds

If a cyclist was accelerating at a rate of 12 m/s2, it would take
9.583333333 seconds for the cyclist to hit a speed of 115 m/s

4. During an Apollo moon landing, reflecting panels were placed on the moon. This allowed
earth-based astronomers to shoot laser beams at the moon's surface to determine its distance.
The reflected laser beam was observed 7 seconds after the laser pulse was sent. The speed of
light is 3.0 × 108 m/s.

D=V*T

D=3.0*108 m/s *7 sec

D=2,100,000,000

D=2.1*109 meters

How accurate was this measurement if the actual distance to the Moon is 384,000,000 meters?
Explain:

384000000= 3.84*108

This measurement wasn’t accurate. It was off by 1,716,000,000
meters or by 1.716*109

*Is this distance correct? Research the distance to the moon.

5. It is now 12:59 p.m.., but when the bell rings at 1:00 pm. Suzette will be late for French class
for the third time this week. She must get from one side of the school to the other by
hurrying down three different hallways. She runs down the first hallway, a distance of 45.0
m, at a speed of 3.2 m/s. The second hallway is filled with students, and she covers its 65.0
m length at an average speed of 2.16 m/s. The final hallway is empty, and Suzette sprints
its 75.0 m length at a speed of 9.5 m/s.

a. Does Suzette make it to class on time or does she get detention for being late again?

T=D/V T=D/V T=D/V

T=45 m/ 3.2 m/s T= 65 m/2.16 m/s T=75 m/9.5 m/s

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T=14.0625 seconds T=30.092 seconds T=7.894 seconds

14.0625+30.092+7.894=52.0485

Suzette took a little over 52 seconds to run down the hallway. This

means that she would get to class with a little less than 7 seconds left

meaning she would get to class on time.

6. The tortoise and the hare are in a road race to defend the honor of their breed. The tortoise

crawls the entire 1000. m distance at a speed of 0.65 m/s while the rabbit runs the first
200.0 m at 1.45 m/s. The rabbit then stops to take a nap for 0.79 hr and awakens to finish
the last 800.0 m with an average speed of 12.5 m/s. Who wins the race and by how much
time?

Tortoise Math:
T=D/V
T=1000 m/0.65 m/s
T=1538.461 sec
1538/60=25.63333333
It took the tortoise a little over 25 and a half minutes to finish the
1000 m race.

Hare Math

T=D/V 1 hour= 3600 seconds T=D/V

T=200 m/1.45 m/s 0.79*3600=2844 seconds T=800 m/12.5 m/s

T=137.931 seconds T=64 seconds

137.931+2844+64

Total time= 3045.931 seconds

3045.931/60

50.7655

It took the hare around 50 minutes and 45 seconds to finish the 1000

m race. This would mean that it took the hare about double the time

to finish the race compared to the time of the tortoise.

7. What is the Acceleration of the Cart on the Ramp? Determine the Angle of the Ramp (A).

Angle Chart: https://drive.google.com/open?id=0B4RmhXJlHvo1YXZhcDNMSDNSMXc

Which Angle had the greatest Acceleration? Write a Conclusion based on your findings.

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Create a Graph if you have time.

Height of Time Velocity Time 2 Velocity
Ramp 1 1 Dist. 2 2 Acceleration

(Opposite) Angle Dist. 1

A=V2-V1/T

V=D/T A=66.667 m/s

V=D/T -50 m/s/9 sec
V=600

V=600 m/9 sec 16.667 m/s/9

m/12 sec V=66.667 sec

125 m 6 600 m 12 sec. V=50 m/s 600 m 9 sec. m/s A=1.851 m/s2

A=V2-V1/T

V=D/T A=300 m/s- 75

V=D/T V=600 m/s/2 sec

V=600 m/2 sec A=225 m/s/2

m/8 sec V=300 sec

250 m 12 600 m 8 sec. V=75 m/s 600 m 2 sec. m/s A=112.5 m/s

Angle=Opposite/Hypotenuse

Angle=125 m/ 1200 m

Angle=0.1041666667

Angle=6o

Angle=250 m/1200 m

Angle=0.2083333333

Angle=12o

Use a2 + b2 = c2 to determine the length of side b

a2+b2=c2

b2=-a2+c2

b2=-1252+12002

b2=-15625+1440000

b2=1424375

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b=√1424375
b=1193.47

a2+b2=c2
b2=-a2+c2
b2=2502+12002
b2=1377500
b=√1377500
b=1173.669

Graph:
X-axis = Angle
Y-axis = acceleration

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Conclusion:

Purpose of experiment, hypothesis, brief description of experiment, data evidence to support
hypothesis, conclusion

The purpose of this experiment was to find which angle would have a higher acceleration, 6
degrees or 12 degrees. Some of the constants in this experiment were the same cart and the
same hypotenuse. This was done so that the data can be more exact as if two different carts or
two different distances would change the outcome of the experiment. The control of this
experiment was the height of the ramp or the opposite. This changed the angle of which the cart
was going down the ramp. This was done to have different accelerations between the two trials.
My hypothesis was that the greater the angle of the ramp, the greater the acceleration would be.
The data was recorded by timing the cart in two different 600m intervals, this being done
because the hypotenuse was 1200 m. After recording the time, there was enough information to
find the velocity of the cart, which is needed in the acceleration formula. To find the velocity of
the cart for angle 6 for the first 600 m, I divided the distance by the time. The equation for this
was V= 600m/12 sec. This showed that the first velocity of the cart for the 6-degree angle was
50 m/s. With the 12-degree angle, the equation was V=600 m/8 sec. This showed that the first
velocity of the cart for the 12-degree angle was 75 m/s, 25 greater than the 6-degree angle. To
find the acceleration, both the first and second velocities are needed so the same formula was
carried over to the second half of the 1200 m ramp. This left me with the equations V=600 m/9
sec for the 6-degree angle and V=600 m/2 sec for the second angle. From this, I found that the
final velocity for the 6-degree angle was 66.667 m/s while the final velocity for the 12-degree
angle was 300 m/s, more than 5 times that of the 6-degree angle. To find the acceleration for
the two different ramps I used the formula A=V2-V1/T which gave me an acceleration of 1.851
m/s2 for the 6-degree angle and 112.5 m/s2 for the 12-degree angle. In conclusion, this showed
definitively that my hypothesis of the greater the angle is, the quicker the cart would accelerate
was in fact true. This was clearly shown by the data with the acceleration of the cart while on
the 12-degree ramp being far larger than the acceleration of the 6-degree ramp. The data clearly
shows that for the entire time that the cart was going down the two different ramps, with a
larger angle the cart was always going faster on the 12-degree ramp.

EXTRA CREDIT:
Light from another star in the galaxy reaches the earth in 32 minutes. The speed of
light is 3.0 × 108 meters/second. In kilometers, how far is the earth from the star?
(Hint: Convert to seconds)(Hint 2: Think about meters to kilometers)
Answer must be in scientific notation
60 seconds=1 minute
1920 seconds=32 minutes

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1000 m= 1 km
32*60=1920

D=V*T
D=3.0*108 m/s *1920 seconds
D=576000000000 meters
D=5.76*1011 meters
576000000000/1000
D=576000000 km
D=5.76*108 km

The earth is 5.76*108 km from the star.

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Quarter 4

GPE Project

Potential Energy Project 2019

Chapter 4: Energy (pg. 100-115)
Due: Friday April 12, 2019

Video Playlist - Use these videos as a resource

1. Define and make note cards or QUIZLET for the following words:
https://quizlet.com/_6d9sqk

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Energy Joules/Kilojoules Chemical Potential Law of Conservation

The ability to do 1 kilojoule = 1000 J, Energy of Energy
work or cause symbol kJ.
change the energy stored the law that states
a unit of energy; in the chemical that energy cannot
used to express bonds of a be created or
both energy and substance destroyed but can
work; symbol j be changed from
one form to another

Kinetic Energy Newton Elastic Potential Gravity

the energy an A unit of measure Energy A force that pulls
object has due to that equals the objects toward
its motion force required to the potential energy each other
accelerate 1 of an object that is
kilogram of mass at stretched or
1 meter per second compressed
per second

Potential Energy Gravitational Potential Total Mechanical Weight

stored energy Energy Energy: A measure of the
force of gravity on
Potential energy the sum of kinetic an object
that depends on and potential
the height of an energy: E= U + K
object

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Resource: http://www.physicsclassroom.com/class/energy/Lesson-1/Potential-Energy 246/310

Gravitational Potential Energy

2. Read Textbook: pg. 100-115
Directions: Make 10-15 Text Dependent questions
1.Question: What is required to change itself or it’s environment?
Answer: Energy
2.Question: Can any action be done without the use of energy?
Answer: No
3.Question: How many different forms of energy are there?
Answer: There is four different types of energy
4.Question: What are the different forms of energy?
Answer:Electrical, Chemical, Radiant, Thermal
5.Question:Making toast is an example of what kind of energy?
Answer: Electrical energy
6.Question: What is an example of radiant energy?
Answer: Energy from the sun travelling to warm up plants
7. Question: What is the formula for Kinetic Energy?
Answer: KE=½mv2
8. Question: What is the unit used for energy?
Answer: Joules
9.Question: Does energy always have to involve motion?
Answer: No
10.Question: What is stored energy due to the position of an object called?
Answer: Potential Energy
11.Question: What is Chemical Potential Energy?
Answer: Energy stored in chemical bonds
12. Question: What is the formula for Gravitational Potential Energy?
Answer: GPE=(mass)(gravity)(height)

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3. Determine the Gravitational Potential Energy (GPE) different masses (g) at 3 different
heights.

Object Mass (grams) Mass (kg)

You 59,000 g 59 kg

Calculator 116 g 0.116 kg

100 mL beaker 44.5 g 0.0445 kg

Graduated cylinder 50 g 0.05 kg

African elephant 5,443,000 g 5443 kg

Tesla or similar car 2,250,000 g 2250 kg

Cruise ship 8.799692e10 g 87996920 kg

baseball 108.5 g 0.1085 kg

*2.2 lbs = 1 kg
130 lbs= 59 kg

3. Data Table:

Your data table will need: Object, mass, gravity, height, GPE

Directions: Research your favorite roller coaster and found out the height of the tallest hill.
*Picture of Roller Coaster
Kingda Ka (World's Tallest Roller Coaster)

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Object Mass (kg) Gravity (m/s2) Height (m) GPE (Joules)
You 59 kg 9.8 (10) 139
9.8 (10) 139 59*9.8*139=
Calculator 0.116 kg 9.8 (10) 139 80369.8 Joules
9.8 (10) 139 59*10*139=
100 mL beaker 0.0445 kg 9.8 (10) 139 82010 Joules
9.8 (10) 139
Graduated 0.05 kg 9.8 (10) 139 0.116*9.8*139=
cylinder 158.0152 Joules
9.8 (10) 139 0.116*10*139=
African elephant 5443 kg 161.24 Joules

Tesla or similar 2250 kg 0.0445*9.8*139=
car 60.6179 Joules
0.0445*10*139=
Cruise ship 87996920 kg 61.855 Joules

baseball 0.1085 kg 0.05*9.8*139=
68.11 Joules
0.05*10*139=
69.5 Joules

5443*9.8*139=
7414454.6 Joules
5443*10*139=
7565770 Joules

2250*9.8*139=
3064950 Joules
2250*10*139=
3127500 Joules

87996920*9.8*139=
1.198694044E11
Joules
87996920*10*139=
1.223157188E11
Joules

0.1085*9.8*139=
147.7987 Joules
0.1085*10*139=
150.815 Joules

Graph: Object and GPE

4. Determine the GPE of one of the masses on the following planets:
E10= Earth gravity 10 m/s2
E9.8= Earth gravity 9.8 m/s2

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Table: Gravitational Potential Energy on other Planets (GPE)

Planet GPE on Earth % difference Gravity on that GPE on other
(J) from Earth planet Planet (J)
(Research)

Jupiter GPE=mgh 147.9% 24.79 m/s2 GPE=mgh
GPE=50 kg *
GPE=50 kg* 10 greater 24.79 m/s2 *
139m
m/s2 *139 m GPE=172290.5
Joules
GPE=69500

Joules

Saturn GPE=mgh 0.044% 10.44 m/s2 GPE=mgh
GPE= 50 kg*
GPE=50 kg* 10 greater 10.44 m/s2* 139
m
m/s2 *139 m GPE=72558
Joules
GPE=69500
GPE=mgh
Joules GPE= 50 kg*
274 m/s2*139 m
The Sun GPE=mgh 185.915% 274 m/s2 GPE=1904300
1.62 m/s2 Joules
The Moon GPE=50 kg* 10 greater 0.002512 m/s2
GPE=mgh
Large Star m/s2 *139 m GPE= 50 kg*
(VY Canis Majoris) 1.62 m/s2*139 m
GPE=69500 GPE= 11259
Joules
Joules
GPE=mgh
GPE=mgh 144.234% GPE= 50
kg*0.002512
GPE=50 kg* 10 less than m/s2*139 m
GPE=17.4584
m/s2 *139 m Joules

GPE=69500 GPE=mgh
GPE= 50
Joules kg*11.7
m/s2*139 m
GPE=mgh 199.9% less GPE=81315
Joules
GPE=50 kg* 10 than
GPE=mgh
m/s2 *139 m GPE=50 kg*6.1

GPE=69500

Joules

Star Wars Planet #1 GPE=mgh 17% greater 11.7 m/s2

GPE=50 kg* 10

m/s2 *139 m

GPE=69500

Joules

Star Wars Planet #2 GPE=mgh 39% less 6.1 m/s2

GPE=50 kg* 10

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m/s2 *139 m m/s2*139 m
GPE=69500 GPE=42395
Joules Joules

Star Wars Planet #3 GPE=mgh 82% greater 18.2 m/s2 GPE=mgh
GPE=50 kg*18.2
GPE=50 kg* 10 m/s2*139 m
GPE=126490
m/s2 *139 m Joules

GPE=69500

Joules

*Use the height of your favorite Roller Coaster. You will use this to figure out the Velocity
at the bottom of the hill on the Star Wars Planets.

A. Velocity of Roller Coaster on Earth
Example:
GPE = KE
mgh = .5mV2
(15g(9.8 m/s2)(57m) = 0.5(15g)v2
8550 = 7.5v2
√1140 = √v2
33.8 m/s = v
The roller coaster would travel 33.8 m/s at the bottom of the tallest hill.
B. Jupiter
GPE = KE
mgh = .5mv2
172290.5=.5(50 kg)v2
172290.5/25=25v2/25
v2=6891.62
v=√6891.62
v=83.01
The roller coaster would be traveling at a rate of 83 m/s at the bottom of the tallest hill.

C. Star Wars Planet #2:
GPE = KE
Mgh = .5mv2
42395=.5(50 kg)v2
42395.5/25=25v2/25
v2=1695.82
v=√1695.82
v=41.18
The roller coaster would be traveling at a rate of 41.2 m/s at the bottom of the tallest hill.
D. Saturn

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