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 constants. 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 `f = f_(0)(1-(t)/(T))`
`(d upsilon)/(dt)=f_(0).(1-(t)/(T)) " " (because f=(d upsilon)/(dt))`
`rArr d upsilon = f_(0).(1-(t)/(T))dt`
`rArr int d upsilon = int f_(0)(1-(t)/(T))dt`
`upsilon = f_(0)t-f_(0)(t^(2))/(2T)+c` ...(i)
where, c is constant
when `t = 0, upsilon = 0` thus from Eq. (i) c = 0
`upsilon = f_(0)t-(f_(0))/(T).(t^(2))/(2)` ....(ii)
As `f=f_(0)(1-(t)/(T))`
when, f = 0
`therefore f_(0)(1-(t)/(T))=0 , f_(0) ne 0`
`therefore t = T`
Putting t = T in Eq. (ii)
`upsilon = f_(0)T-(f_(0))/(T).(T^(2))/(2)=(f_(0)R)/(2)`