Asst4 Rand1

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course phy 201

4/1 440pm

At clock time t = 7 sec, a ball rolling straight down a hill is moving at 6 m/s and is 47 m from the top of the hill.  It accelerates uniformly at a rate of .6 m/s/s until clock time t = 17 sec.  What is its velocity at this point and what is its average velocity during this time?  How far is it from the starting point at t = 17 sec?Velocity after 17s: .6m/s/s*10s=6m/s+6m/s=12m/s

vAve during acceleration: 6m/s/2=3m/s+6m/s=9m/s

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The average velocity is 9 m/s, but your statement asserts equality of some quantities which are not equal. This can lead to a lot of confusion.

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Distance from start: 9m/s*10sec=90m+47m=137m"

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My preceding comment applies here as well.

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