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 ${\mathrm{x}}_1\left(\mathrm{t}\right)$ after time t and that of the second body by ${\mathrm{x}}_2\left(\mathrm{t}\right) \mathrm{after} \mathrm{the} \mathrm{same} \mathrm{time} \mathrm{interval}. \mathrm{Which} \mathrm{of} \mathrm{the} \mathrm{following} \mathrm{graphs} \mathrm{correctly} \mathrm{describe} ({\mathrm{x}}_1-{\mathrm{x}}_2) \mathrm{as} \mathrm{a} \mathrm{function} \mathrm{of} \mathrm{time} \mathrm{t}?$

For 1st particle:

It starts moving (${\mathrm{u}}_1 = 0$) with constant acceleration

${\mathrm{x}}_1 = {\mathrm{x}}_1\left(\mathrm{t}\right) = {\mathrm{u}}_1\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 = \frac{1}{2}{\mathrm{at}}^2 .......\left(\mathrm{i}\right)$

For 2nd particle:

It is moving with constant velocity (v)

${\mathrm{x}}_2 = {\mathrm{x}}_2\left(\mathrm{t}\right) = \mathrm{vt} ...\left(\mathrm{ii}\right)$

Relative position of particle ‘1’ w.r.t ‘2’

Hence ${\mathrm{x}}_1-{\mathrm{x}}_2 = {\mathrm{x}}_{1.2} = \frac{1}{2}{\mathrm{at}}^2-\mathrm{vt} ......\left(\mathrm{iii}\right)$

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 $\left|{\mathrm{v}}_{1.2}\right| = \frac{{\mathrm{dx}}_{1.2}}{\mathrm{dt}} = \mathrm{at}-\mathrm{v}$

As particle ‘1’ starts moving from rest, hence at t = 0 ${\mathrm{v}}_{1.2} \mathrm{should} \mathrm{be} \mathrm{negative}.$

It means the slope of the parabola at t = 0 should be negative. The parabola should open up. Hence graph ‘b’ fulfil the requirement.