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
$f=f_0\left(1-\frac{t}{T}\right)\dots (i)$
$\therefore$ Velocity of particle is given as,
$v=\int fdt=\int f_0\left(1-\frac{t}{T}\right)dt$
$=f_0\left(t-\frac{t^2}{2T}\right)+c\dots (ii)$
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
$v=f_0\left(t-\frac{t^2}{2T}\right)\dots (iii)$
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),
$v=f_0\left(t_1-\frac{t_1^2}{2T}\right)$
$\Rightarrow v=f_0\left(T-\frac{T^2}{2T}\right)$
$=f_0\left(T-\frac{T}{2}\right)=\frac{f_{0}T}{2}$