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

Answer

A

Explanation

Acceleration `(dv)/(dt)=f=_(0)(1-(t)/(T))`
`implies int_(0)^(v)dv=f_(0)int_(0)^(t)(1-(t)/(T))dt`
`=v=f_(0)(t-(t^(2))/(2T))_(0)^(T)=f_(0)(T-(T^(2))/(2T))=(1)/(2)f_(0)T`