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32. The velocity v of a particle is given by the equation v=61 – 68, where v is in m/sec and t is time in seconds then:(A) at 1 = 0, velocity is maximum(C) minimum velocity is zero2(B) at 1=3velocity is minimum(D) minimum velocity is –2 m/sec |
| Answer» GIVEN v=6t 2 −6t 3 . DIFFERENTIATING v w.r.t. t, we have Dtdv =12t−18t 2 .Putting dv/dt=0, we will get the values of t at which v is maximum or minimum. Therefore,12t−18t 2 =0⇒t=0,2/3sTo the distinguish between points of maxima and minima, we need the second DERIVATIVE of v.dt 2 d 2 v =12−36tNow dt 2 d 2 v ∣ t=0 =12>0.So t=0 is a point of minima.dt 2 d 2 v ∣ t =2/3s=12−36× 32 =−12<0So t=2/3s is a point of maxima.Hence, the minimum value of v is 0 MS −1 (by putting t=0 s in v).The maximum value of v is6× 94 −6× 278 = 38 − 916 = 98 ms −1 (by putting t= 32 s in v).Answered Bytoppr61 ViewsHow satisfied are you with the answer?This will help us to improve betteranswrGet Instant Solutions, 24x7No Signup requiredgirlMore Questions by difficultyEASYMEDIUMHARD | |