A particle moves in X-direction with acceleration f from rest. This acceleration changes according to `f = f _(0) (1- (t)/(T)).` and T are constant. t = 0 and for if any negative time interval if f = 0, then velocity of particle is ......

Options

A. `1/2 f _(0) T`
B. `1/2 f _(0) t ^(2)`
C. `f _(0) T`
D. `f _(0) T ^(2)`

Answer

A

Explanation

`f = f _(0) [1- (t)/(T)]`
`f = 0 because f _(0) [1 + (t)/(T)] =0`
`implies t = T`
Now `f = (dV)/(dt)`
`therefore (dV)/(dt) = f _(0) [1 - (t)/(T)]`
`int _(0) ^(V _(x)) dV = int _(t =0) ^(t =T) f _(0) [1- (t)/(T)]dt`
`v _(x) = f _(0) [t - (t ^(2))/(2T) ] _(0) ^(T) `
`V_(x) = f_(0) [T- (T ^(2))/(2T)]`
`V_(x) = f _(0) [T- (T)/(2)]`
`V_(x) = 1/2 f_(0)T`