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.
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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: vAve='ds*'dt, 30cm*5sec=150 cm/sec
vAve is not `ds * `dt, and cm * sec gives you cm * sec, not cm / sec.
vAve is one of the two most basic definitions you need to understand.
vAve is defined as an average rate of change, which involves dividing change in one quantity by change in another.
It follows that vAve = `ds / `dt, which will also result in the correct units.
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: vf=vAve*2, 150cm/sec*2=300cm/sec
By how much did its velocity therefore change?
answer/question/discussion: the vf =twice the ave velocity
At what average rate did its velocity change with respect to clock time?
answer/question/discussion: aAve=change in velocity/change in time, 300cm/sec/5sec=60cm/sec^2
What would a graph of its velocity vs. clock time look like? Give the best description you can.
answer/question/discussion: since acceleration is uniform, the graph would be linear. at t0 and v0, the x, y coordinates would both be 0. at t5 and vf the x, y coordinates would be 5 sec, 300cm/sec. Theconnecting line between would be linear with a slope of 60cm/sec^2
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30 min
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Overall this is good. All your reasoning was correct except your calculation of vAve. Of course that changes all your subsequent results, but the changes will be easy to make.
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