Question

A body is at rest at $x = 0$. At $t = 0$, it starts moving in the positive $x$-direction with a constant acceleration. At the same instant another body passes through $x = 0$ moving in the positive $x$-direction with a constant speed. The position of the first body is given by $x_1(t)$ after time $t$ and that of the second body by $x_2(t)$ after the same time interval. Which of the following graphs correctly describes $(x_1 - x_2)$ as a function of time $t$?

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

For $1^{st}$ particle:
It starts moving $(u_1 = 0)$ with constant acceleration
$x_1 = x_1 (t) = u_1 t + \frac{1}{2} at^2 $
$ = \frac{1}{2} at^2 \,\,...(i)$
For $2^{nd}$ particle:
It is moving with constant velocity $(v)$
$x_2 = x_2(t) = vt \,\,...(ii)$
Relative position of particle ‘$1$’ w.r.t. ‘$2$’
Hence $x_1 - x_2 = x_{1,2} = \frac{1}{2} at^2 - vt\,\,...(iii)$
Hence graph should be parabola.
Differentiating equation $(iii)$ w.r.t time, we get the
relative velocity of particle $1$ w.r.t. $2$
or $|v_{1,2}| = \frac{dx_{1,2}}{dt} = at - v$
As particle ‘$1$’ starts moving from rest, hence at $t = 0$
$v_{1, 2}$ should be negative.
It means the slope of the parabola at $t = 0$ should be negative. The parabola should open up. Hence graph ‘$b$' fulfil the requirement