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

Answer

C

Explanation

Here, `f= f_(0) (1 - (t)/(T)) or, (dv)/(dt) = f_(0) (1- (t)/(T)) or, dv= f_(0) (1- (t)/(T)) dt`
`:. v= int dv= int [f_(0) (1-(t)/(T))] dt or, v= f_(0) (t-(t^(2))/(2T))+C` where C is the constant of integration.
At t= 0, v= 0
`:. 0 = f_(0) (0-(0)/(2T))+ C rArr C= 0`
`:. v= f_(0) (t- (t^(2))/(2T))`
If f= 0, then
`0 = f_(0) (1- (t)/(T)) rArr t= T`
Hence, particle.s velocity in the time interval t=0 and t= T is given by
`v_(x) = underset(t=0)overset(t=T)int dv= underset(t=0)overset(T)int [f_(0) (1-(t)/(T))]dt= (1)/(2) f_(0)T`