The deceleration experienced by a moving motorboat, after its engine is cut-off is given by dv/dt=−kv3, where k is constant. If v0 is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time t after the cut-off is

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v0/2

v0e−kt

v02v02kt+1

answer is D.

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Detailed Solution

Here dvdt=−kv3 or dvv3=−kdt or ∫v0v dvv3=∫0t −kdt or −12v2v0v=−kt or −12v2+12v02=−kt or v2=v021+2v02kt or v=v02v02kt+1

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