Question

A particle moving along $x$ -axis has acceleration $f$, at time $t$, given by $f = f _{0}\left(1-\frac{ t }{ T }\right)$, 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 $\left( v _{ x }\right)$ is:

• A. $\frac{1}{2}f_0 T^2$
• B. $f_0 T^2$
• C. $\frac{1}{2}f_0 T$
• D. $f_0 T$

Answer 1

Answer: (C)

$\frac{1}{2}f_0 T$

Answer 2

Acceleration $\frac{ dv }{ dt }= f = f _{0}\left(1-\frac{ t }{ T }\right)$
$\Rightarrow \displaystyle\int_{0}^{ v } d v = f _{0} \displaystyle\int_{0}^{ T }\left(1-\frac{ t }{ T }\right) dt$
$\Rightarrow v = f _{0}\left( t -\frac{ t ^{2}}{2 T }\right)_{0}^{ T }= f _{0}\left( T -\frac{ T ^{2}}{2 T }\right)=\frac{1}{2} \,f _{0}\, T$