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The deceleration experienced by a moving motor boat after its engine is cut off is given by dv/dt =kv^3, where is a constant. Ifv, is the magnitude of the velocity at cut off, the magnitude of velocity at time after the cut off is : |
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Answer» `(v_0)/(2)` `[1/2v^2]_(v_0)^(v)=kt` or `(1)/(2v^2)-(1)/(2v_(0)^(2))=kt` or `v_(0)^(2)-v^(2)=2kt v^(2)v_(0)^(2)` `v^(2)=(v_(0)^(2))/(2ktv_(0)^(2)+1)` or `v=(v_0)/(sqrt(2ktv_(0)^(2)+1)` |
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