The deceleration experienced by a moving motor boat, 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 velocity" v "at a time t after the cut off is
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a
v01+2ktv02
b
v02k
c
v01+2kt
d
v02kt
answer is A.
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Detailed Solution
given dvdt=−kv3integrting ∫v0v v−3dv=−k∫0t dtv−2−2v0v=−k(t)0t 1v2v0v=2ktapplying the limits,1v2−1v02=2kt1v2=1v02+2kt1v2=1+2ktv02v02v2=v021+2ktv02v=v01+2ktv02=velocity at a time t after the cut off