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)

Given : At time $t=0$,
velocity, $v=0$.
Acceleration $f=f_{0}\left(1-\frac{t}{T}\right)$
At $f=0, 0=f_{0}\left(1-\frac{t}{T}\right)$
Since $f_{0}$ is a constant,
$\therefore 1-\frac{t}{T}=0 $
or $ t=T \text {. }$
Also, acceleration $f=\frac{d v}{d t}$
$\therefore \int\limits_{0}^{v_{x}} d v=\int\limits_{t=0}^{t=T} f d t=\int\limits_{0}^{T} f_{0}\left(1-\frac{t}{T}\right) d t $
$\therefore v_{x}=\left[f_{0} t-\frac{f_{0} t^{2}}{2 T}\right]_{0}^{T}$
$=f_{0} T-\frac{f_{0} T^{2}}{2 T}=\frac{1}{2} f_{0} T$