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_(o)T`
C. `(1)/(2)f_(o)T^(2)`
D. `f_(o)T`

Answer

A

Explanation

When f=0 means `f_(0)(1-(t)/(T))=0`
`rArr=1-(t)/(T)=0rArr t=T`
Now, `v=int_(o)^(t)fdt=int_(0)^(T)(f_(0)-(f_(0)t)/(T))dt=[f_(0)t-(f_(0)t^(2))/(2T)]_(0)^(T)`
`=f_(0)T-(f_(0)T^(2))/(2T)-0=f_(0)T-(f_(0)T)/(2)=(f_(0)T)/(2)`