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Question

Find the average velocity of the position function s(t)=2t24t on the interval from t=0 sec to t=5 sec.

A
4 m/s
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B
5 m/s
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C
3 m/s
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D
6 m/s
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Solution

The correct option is D 6 m/s
Given s(t)=2t24t
So, at t=0,s(t)=0, it means particle has started its motion from origin, so, we will check whether it has returned to origin or not
s(t)=0 2t24t=0
2t(t2)=0 t=0,2 sec
Thus, the particle will return to origin at t=2 sec, afterwards it will continue its motion.
So, the displacement of particle from t=2 sec to t=5 sec will be
s=2[t2]524[t]52=2(254)4(52)=4212=30 m
Hence, average velocity is ¯v=Total displacementTotal time=305=6 m/s

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