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The relation between time t and distance...

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^(3)`
`(D)` retardation is given by `2Bv^(2)`
Select correct statement `:-`

A

Only `C`

B

Only `D`

C

Only `A` & `D`

D

Only `B` & `C`

Text Solution

Verified by Experts

The correct Answer is:
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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