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Q.

The distance moved by the particle in time 't' is given by S=t3−4t2+4t+7. At the  instant when its acceleration is '0' the velocity is

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a

65  unit/sec

b

−43  unit/sec

c

48  unit/sec

d

−48  unit/sec

answer is B.

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Detailed Solution

S=t3−4t2+4t+7 v=3t2−8t+4 a=6t–8=0 t=86=43 V=3(169)−8(43)+4 =16−323+4=16−32+123=−43
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