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 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)

Acceleration of the particle is given as
$f=f_{0}\left(1-\frac{t}{T}\right) \ldots (i) $
$\therefore $ Velocity of particle is given as,
$v=\int f d t=\int f_{0}\left(1-\frac{t}{T}\right) d t$
$=f_{0}\left(t-\frac{t^{2}}{2 T}\right)+c \ldots (ii) $
Given that, at $t=0, v=0$
Hence, from Eq. (i), we get
$0=f_{0}(0-0)+c $
$\Rightarrow c=0$
$\therefore $ From Eq. (ii), we have
$v=f_{0}\left(t-\frac{t^{2}}{2 T}\right) \ldots(iii)$
Suppose, acceleration of the particle becomes zero at time $t=t_{1}$.
Hence, from Eq. (i),
$0=f_{0}\left(1-\frac{t_{1}}{T}\right) $
$\Rightarrow 1-\frac{t_{1}}{T}=0$
$ \Rightarrow t_{1}=T$
Hence, velocity of particle at $t=t_{1}=T$, from Eq. (iii),
$v=f_{0}\left(t_{1}-\frac{t_{1}^{2}}{2 T}\right)$
$\Rightarrow v=f_{0}\left(T-\frac{T^{2}}{2 T}\right)$
$=f_{0}\left(T-\frac{T}{2}\right)=\frac{f_{0} T}{2}$