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

Answer

d

Explanation

Acceleration
`f=(dv)/(dt)=f_(0)(1+(t)/(T))`
or `dv=f_(0)(1-(t)/(T))*dt`….(i)
Integrating Eq .(i) on bothe sides , we get
`v=f_(0)t-(f_(0))/(T)*(t^(2))/(2)+C` ….(ii) After applying boundary conditions
v=0att=0
we get ,C=0
`rArrv=f_(0)t-(f_(0))/(T)*(t^(2))/(2)`.... (iii)
As `f=f_(0)(1-(t)/(T))`
When `f_(0)=0`, t=T Substituting t=T in Eq .(iii) ,then velocity
`V_(X)=f_(0)T-(f_(0))/(T)-(f_(0))/(T).(T^(2))/(2)=(1)/(2)f_(0)T`