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 $t$?
Answer: (B) see figure
First body (from rest, acceleration $a$): $x_1 = \dfrac{1}{2}at^2$. Second body (constant speed $v$): $x_2 = vt$.
$$x_1 - x_2 = \frac{1}{2}at^2 - vt$$
This is an upward-opening parabola through the origin. For small $t$ the $-vt$ term dominates, so $x_1 - x_2$ first becomes negative (the second body is ahead). It reaches a minimum at $t = v/a$, returns to zero at $t = 2v/a$ (the first body catches up), and then grows positive.
Only graph (2) dips below the axis and then rises.
Solution by Sreeraj P, M.Sc Physics