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

$x_1=x_1(t)=u_{1}t+\frac{1}{2}a{t}^2$

$=\frac{1}{2}a{t}^2...(i)$

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 $x_1-x_2=x_{1,2}=\frac{1}{2}a{t}^2-vt...(iii)$

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 $|v_{1,2}|=\frac{d{x}_{1,2}}{dt}=at-v$

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