cq_1_051

Phy 121

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The problem:

A ball accelerates at 8 cm/s^2 for 3 seconds, starting with velocity 12 cm/s.

• What will be its velocity after the 3 seconds has elapsed?

answer/question/discussion: ->->->->->->->->->->->-> :

Acceleration is equaled to change in velocity divided by change in time. Therefore:

Acceleration= (final velocity-initial velocity)/(final time- initial time)

8 cm/sec^2=(final velocity- 12 cm/sec)/(3 sec)

Final Velocity= 36 cm/sec

• Assuming that acceleration is constant, what will be its average velocity during this interval?

answer/question/discussion: ->->->->->->->->->->->-> :

vAve= `ds/`dt

To find the final and initial positions we have to multiply the velocities by the time interval:

36 cm/sec * 3 sec= 108 cm

12 cm/sec * 3 sec= 36 cm

vAve= (108 cm- 36 cm)/ 3 sec= 24 cm/sec

• How far will it travel during this interval?

answer/question/discussion: ->->->->->->->->->->->-> :

We can find the interval of distance by subtracting the initial position from the final position:

(108 cm- 36 cm)= 72 cm

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20 minutes

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&#Very good responses. Let me know if you have questions. &#