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Question

# The acceleration of a particle, starting from rest, varies with time according to the relation a=−sω2sinωt. The displacement of this particle at time t will be:

A
ssinωt
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B
sωcosωt
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C
sωsinωt
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D
12(sω2sinωt)t2
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Solution

## The correct option is A ssinωtGiven, a= -sω2sinωtdvdt=−sω2sinωt=>dv=-=−sω2sinωtdtIntegrating both side, from time t=0,t∫v0dv=∫t0(−sω2sinωt)dtv=sω.cosω(t−0)+c1=sω.cosωt+c1We know that, v=dsdtdsdt= sω.cosωt+c1ds=sω.cosωtdt+c1Integrating te above w.r.t time,∫s0ds=∫t0(sω.cosωωt)dts=s.sinωt+c1+c2=s.sinωt+c Eliminating integrating constant, c,we get,displacement S=s.sinωt

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