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. (a) ` 1/2 f_0 T^2` .
B. (b) ` f_0T^2`
C. © 1/2 f_0T`
D. (d) ` g_0T`

Answer

C

Explanation

Given, at time ` t= 0`, velocity ` v=0`
` Acceleration, ` f=f_0 91 - t/T)`
When ` f=0, 0=f_0 91 - t/T)`
since `f_0` is constant , therefore , ` 1- t/T =0 ` or t = T`
Also, acceleation, ` f = (dv)/(dt)`
or ` int_0^v_x dv = int_0^T f dt = int_0^T f_0 (1- t/T)dt`
or ` v_x = 9f_0 t - (f_0t^2)/(2 T))_0^T = f_0 T - (f_0T^2)/(2T) = (f_0 T)/2`.