Q. 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 $t = 0$ has zero velocity. In the time interval between $t = 0$ and the instant when $f = 0,$ the particle’s velocity $(v_{x})$ is
A) $f_{0}T$
B) $\frac{1}{2}f_{0}T^{2}$
C) $f_{0}T^{2}$
D) $\frac{1}{2}f_{0}T$
Explanation
(d) In this problem acceleration is variable
$f = \frac{dv}{dt} = f_{0}\left( 1 - \frac{t}{T} \right)$ …(i)
At $t = 0,\ f = f_{0}$ at $t = T,\ f = 0$ i.e.,
We have to calculate the velocity of the particle in the time from
$t = 0$ to $t = T\ \sec$
From (i) equation, $\frac{dv}{dt} = f_{0}\left( 1 - \frac{t}{T} \right)$;
Integrating both sides, $\int_{}^{}{dv = \int_{0}^{T}f_{0}\left( 1 - \frac{t}{T} \right)dt;}$
$v = f_{0}{\left\lbrack t - \frac{t^{2}}{2T} \right\rbrack_{0}^{T}}^{\ }v = f_{0}\left\lbrack T - \frac{T^{2}}{2T} \right\rbrack;$
$v = f_{0}\left\lbrack T - \frac{T}{2} \right\rbrack;v = f_{0}\frac{T}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.