A particle moving along x-axis has acceleration `f`, at time `t`, given by `f = f_0 (1 - (t)/(T))`, where `f_0` and `T` are constant.
The particle at `t = 0` has zero velocity. In the time interval between `t = 0` and the instant when `f = 0`, the particle's velocity `(v_x)` is :

Options

A. `(1)/(2)f_(0)T^(2)`
B. `(1)/(2)f_(0)T`
C. `f_(0)T^(2)`
D. `f_(0)T`

Answer

b

Explanation

Given acceleration, `f=f_(0)(1-(1)/(T))`
`(dv)/(dt)=f_(0)(1-(t)/(T))`
`v=intf_(0)(1-(t)/(T))dt`
`v=f_(0)(t-(t^(2))/(2T))+C`
Given t=0, v=0
`0=f_(0)(0-0)+C rArr C=0`
`v=f_(0)(t-(t^(2))/(2T))`
Time at which acceleration is zero is given by
`f_(0)(t-(t^(2))/(T))=0`
`rArr t=T`
The particle velocity in the time interval t=0 to t=T
`v_(x)=int_(t=0)^(t=T) f_(0)(1-(t)/(T))dt`
`=[f_(0)(t-(t^(2))/(2T))]_(0)^(T)`
`=f_(0)(T-(T)/(2))-f_(0)(0)`
`=f_(0)((T)/(2))-0`
`=(1)/(2)f_(0)(T)`