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=3m/s, the magnitude of the velocity 1sec 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 A325 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=3m/s and the final velocity be v(v0→v), the limit for time be 0→t