cq_1_031

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Phy 121

Your 'cq_1_03.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** CQ_1_03.1_labelMessages.txt **

The problem:

A ball starts with velocity 0 and accelerates uniformly down a ramp of length 30 cm, covering the distance in 5 seconds.

• What is its average velocity?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

vAve=’ds/’dt=(s_1-s_0)/(t_1-t_0)=(30cm-0cm)/(5s-0s)=30cm/5s=6cm/s

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• If the acceleration of the ball is uniform then its average velocity is equal to the average of its initial and final velocities.

You know its average velocity, and you know the initial velocity is zero.

What therefore must be the final velocity?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

vAve=(v_f+v_0)/2

6cm/s=(v_f+0)/2

6cm/s=v_f/2

6cm/s*2=v_f

12cm/s=v_f

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• By how much did its velocity therefore change?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

‘dv=v_f - v_0=12cm/s - 0cm/s=12cm/s

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• At what average rate did its velocity change with respect to clock time?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

aAve=’dv/’dt=(v_f-v_0)/(t_1-t_0)=(23cm/s-0cm/s)/(5s-0s)=(23cm/s)/(5s)=4.6cm/s^2

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It's not clear where 23 cm/s came from. You determined earlier that vf was 12 cm/s.

I suspect a typo. Your procedure and reasoning are fine.

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• What would a graph of its velocity vs. clock time look like? Give the best description you can.

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

For uniform acceleration this would be a line that increases at a constant rate. The slope of this line would equal the acceleration. The slope would be 4.6cm/s.

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The units of the slope wouldn't be cm/s. Again, looks like a typo, since I'm pretty sure you understand the units involved here.

The slope would be the same as the average rate of change of velocity, but you appear to have an error in that calculation.

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Good work, but check my notes.

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