Question

A particle moves along x-axis and its acceleration at any time $t$ is $a=2sin\left(\pi t\right)$, where $t$ is in seconds and $a$ is in $m/s^2$. The initial velocity of particle (at time $t=0$) is $u=0$.
Then the magnitude of displacement (in meters) by the particle from time $t=0$ to $t=t$ will be

• A. $\frac{2}{{\pi }^2}\sin \pi t-\frac{2t}{\pi }$
• B. $-\frac{2}{{\pi }^2}\sin \pi t+\frac{2t}{\pi }$
• C. $\frac{2t}{\pi }$
• D. None of these

Answer 1

The correct option is B $-\frac{2}{{\pi }^2}\sin \pi t+\frac{2t}{\pi }$

$a=\sin \pi t$ $\therefore \int dv=\int 2\sin \pi tdt$
$v=-\frac{2}{\pi }\cos \pi t+C$
At $t=0,v=0(\because u=0)$
$\therefore C=\frac{2}{\pi }$
$\Rightarrow v=\frac{2}{\pi }(1-\cos \pi t)$
Note: Velocity is always non-negative as $\cos \theta \le 1$. Hence, particle always moves along positive x - direction.
Since velocity, $v=\frac{dS}{dt}$
$\therefore$ Distance from time $t=0$ to $t$, $S=\underset{0}{\overset{t}{\int }}\frac{2}{\pi }(1-\cos \pi t)dt=\frac{2}{\pi }{[t-\frac{1}{\pi }\sin \pi t]}_0^t=\frac{2}{\pi }t-\frac{2}{{\pi }^2}\sin \pi t$
Since the direction of velocity of particle is not changing, we can say
Displacement of the particle = Distance covered by the particle
$\Rightarrow$ Displacement from time $t=0$ to $t\text{s}=\frac{2t}{\pi }-\frac{2}{{\pi }^2}\sin \pi t$