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Question

The retardation experienced by a moving motor boat after its engine is cut off, is given by dvdt=8v2. If the magnitude of velocity at cut off is v0=3 m/s, the magnitude of the velocity 1 sec after the cut off is (in m/s)


A
325
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B
2425
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C
18
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D
328
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Solution

The correct option is A 325
It is given that, the retardation is  dvdt=8v2

dvv2=8dt

Integrating on both sides we get,
vv0dvv2=t08dt
here the limits are initial velocity which is the velocity at cut off v0=3 m/s and the final velocity be v (v0v), the limit for time be 0t

1vvv0=8t 

1v+1v0=8t
1v1v0=8t
1v=1v0+8t
v=v01+8v0t
v=31+8×3×1=325 m/s

Hence option A is the correct answer

Physics

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