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The relation between time `t` and distance `x` of a particle is given by `t=Ax^(2)+Bx`, where `A` and `B` are constants. Then the `(A)` velocity is given by `v=2Ax+B` `(B)` velocity is given by `v=(2Ax+B)^(-1)` `(C )` retardation is given by `2Av^(2)` `(D)` retardation is given by `2Bv^(2)` Select correct statement `:-`A. Only `C`B. Only `D`C. Only `A` & `D`D. Only `B` & `C` |
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Answer» Correct Answer - D Differentiating `t-Ax^(2)+Bx` w.r.t. time `1=(2Ax+B)vrArrv=(2Ax+B)^(-1)` Again differentiating w.r.t. time `(dv)/(dt)=(-1)(2Ax+B)^(-2)(2A(dx)/(dt))=-2Av^(2)` `rArr` Ratardation `=-(dv)/(dt)=2Av^(2)` |
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