13_u3 Review Key.doc 4vy3p

Name Date

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Uniformly Accelerated Particle Model: Review Sheet position (m)

position vs time 50 40 30 20 10 0

0

1

2

3

4

5

6

time (s)

7

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10

1. Use the graph above to answer the following questions: a. Give a written description to describe the motion of this object. At time zero the object is at zero position and is moving quickly in the positive direction. It then slows to a stop in 10 seconds. b. Draw the motion map for the object. Include velocity and acceleration vectors. vel:

v

v

v v

v

0m

acc:

a

a

a

+

a

a

velocity (m/s)

c. Determine the instantaneous velocity of the object at t = 2 s and explain how you did it. I took the slope of the line connecting the position at 0s and the position at 4s. The slope of that line is the instantaneous velocity at the mid-time of 2s. x 32m  0m slope    8.0 ms t 4.0s  0s d. Assume the initial velocity was 10 m/s, determine the acceleration of the object. v 8.0 ms  10.0 ms a   1.0 sm2 t 2.0s  0s

©Modeling Instruction 2013

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15 10 5 (0 2 4 6 8 10 mU3 Accelerated Motion - Review v3.1 -5 / s

e. Sketch a corresponding velocity time graph for the graph above.

©Modeling Instruction 2013

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time (s)

U3 Accelerated Motion - Review v3.1

a. Describe the motion of the object. At time zero the object is moving in the negative direction at 15 m/s. It then slows to a stop in 8 seconds.

velocity (m/s)

2. Use the graph to answer the following questions.

15 0

0

5

10 t (s)

b. Determine the acceleration of the object from the graph. -15 v 0 ms  (15 ms ) m slope    1.9 s 2 1.875 with extra digits t 8.0s  0s

c. Shade the portion of the graph that represents the displacement of the object from 2 to 6 seconds. d. Calculate the velocity of the object at 2 seconds and at 6 seconds. (No you are not allowed to just take the values off the graph.) t = 2.0s&6.0s vi = -15.0 a = 1.875 vf = ?

v 2s : v f  v i  at  15 ms  1.875 sm2 (2.0s )  11.25 ms vi

v 6s : v f  v i  at  15 ms  1.875 sm2 (6.0s )  3.75 ms

e. Calculate the object's displacement from 2 to 6 seconds. t = 4.0s m vi = -11.25 s m a = 1.875 s m vf = -3.75 s x = ? 2

Method 1: Using the area of the red section of the graph:  11.25 ms  3.75 ms h  h2 Atrap  1 w (4.0s )  30.0m 2 2 2 2 Method 2:  x  v i t  21 a(t )  11.25 ms (4.0s )  21 (1.875 sm2 )(4.0s )  30.0m (3.75 ms )2  ( 11.25 ms )2  30.0m Method 3:  v  v  2x  x  2(1.875 sm2 ) 2 f

2 i

3. A car, initially at rest, accelerates at a constant rate of 4.0 m/s2 for 6 s. How fast will the car be traveling at t = 6 s? t = 6.0s v f  v i  at  0  4.0 sm2 (6.0s )  24.0 ms vi = 0 m a = 4.0 s vf = ? 2

4. A tailback initially running at a velocity of 5.0 m/s becomes very tired and slows down at a uniform rate of 0.25 m/s2. How fast will he be running after going an additional 10 meters? x = 10.0m vi = 5.0 v f2  v i2  2ax  v f2  (5.0 ms )2  2( 0.25 sm2 )10.0m a = −0.25 v f  4.5 ms vf = ? vi ©Modeling Instruction 2013

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U3 Accelerated Motion - Review v3.1

5. Draw a motion map along the ramp for the motion of the ball when released from rest. a a vo = 0 cm/s v

x = 0 cm

a v a

a

v

a v

v

x = 50 cm

x = 25 cm

6. Draw a motion map along the ramp for the motion of the ball when released from rest. v

a

a

v

a v

x = 50 cm

a v

a v

a

vo = 0 cm/s a v x = 0 cm

v

x = 25 cm

position (m)

7. Using the graph compare the following quantities for objects A and B. Is A > B, A < B, or A = B.

B A

a. Displacement from 0 to 3 s ___A = B____ How do you know? Begin at 3 time (s) same position, end at same position. b. Displacement from 0 to 1.5 s __ A > B ___ How do you know? A is moving faster the whole 1.5 s. c. Displacement from 1.5 to 3 s __ B > A ___ How do you know? B is moving faster the whole 1.5 s. d. Average velocity from 0 to 3 s ___A =B____ How do you know? Both have the same change in position in the same time. e. Average velocity from 0 to 1.5 s __ A > B ___ How do you know? A has a greater displacement in the 1.5s so it has a greater average velocity. f. Average velocity from 1.5 to 3 s __ B > A ___ How do you know? B has a greater displacement in the 1.5s so it has a greater average velocity. g. Instantaneous velocity at 3 s __ B > A ___ How do you know? B has a steeper slope at 3s. h. Instantaneous velocity at 0 s __ A > B ___ How do you know? A has a steeper slope at 0s. i. If the motion of B is uniformly accelerated, at what time will both graphs have exactly the same slope? Explain. ©Modeling Instruction 2013

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U3 Accelerated Motion - Review v3.1

Both graphs will have the same slope at 1.5s. 1.5s is the mid-time for both and the average velocity is the same as the instantaneous velocity at the mid-time.

©Modeling Instruction 2013

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U3 Accelerated Motion - Review v3.1

time

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position time

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v

vel:start

v

v v a

acc:

time

time

v

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a

©Modeling Instruction 2013

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a start

acc: a

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0mend

a a

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acceleration

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acceleration

acceleration

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vel:

velocity

velocity

velocity

position

position

8. For each of the position vs time graphs shown below, draw the corresponding v vs t, a vs t , and motion map.

vel:

v

v

v

v

v

v

v

0m

a

a

a

+

+ acc:

end

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a

a

a

a

U3 Accelerated Motion - Review v3.1

a