JEE PYQ: Motion in a Straight Line - Question ID 0b98e9f6f263 (JEE Main 2015)

When two objects move along the same straight line under constant acceleration (here, gravity), their relative position is simply the difference between their individual positions at any time. The key formulas are:

• Position of a stone thrown upward from height $h$ with initial speed $u$: $y(t) = h + u\,t - \tfrac12\,g\,t^2$
• Relative position of stone 2 with respect to stone 1: $\Delta y(t) = y_2(t) - y_1(t)$

Since both stones start from the same height ($h = 240$ m) and experience the same acceleration ($g = 10$ m/s²), the relative motion is uniformly accelerated with initial relative speed $u_2 - u_1$ and zero relative acceleration.

1. Write the individual position equations.
   Stone 1 (initial speed $u_1 = 10$ m/s): $y_1(t) = 240 + 10\,t - 5\,t^2$ Stone 2 (initial speed $u_2 = 40$ m/s): $y_2(t) = 240 + 40\,t - 5\,t^2$
2. Form the relative position.
   $\Delta y(t) = y_2(t) - y_1(t) = (40\,t - 5\,t^2) - (10\,t - 5\,t^2) = 30\,t$
   The quadratic terms cancel, leaving a linear function of time.
3. Determine the domain of validity.
   Both stones eventually hit the ground ($y = 0$). Solve for each stone’s impact time:
   • Stone 1: $240 + 10\,t - 5\,t^2 = 0$ $t^2 - 2\,t - 48 = 0 \quad\Longrightarrow\quad t = 8\ \text{s}$
   • Stone 2: $240 + 40\,t - 5\,t^2 = 0$ $t^2 - 8\,t - 48 = 0 \quad\Longrightarrow\quad t = 12\ \text{s}$
   For $t > 8$ s, stone 1 is already on the ground and its position is fixed at $y_1 = 0$. Thus for $t \in [8,12]$ s: $\Delta y(t) = y_2(t) - 0 = 240 + 40\,t - 5\,t^2$ This segment is a downward-opening parabola.
4. Sketch the graph.
   
   • $0 \le t \le 8$ s: straight line through the origin with slope $30$ m/s.
   • $8 < t \le 12$ s: parabolic arc starting at $\Delta y(8) = 240$ m and ending at $\Delta y(12) = 0$.
   • For $t > 12$ s both stones are on the ground, so $\Delta y = 0$.
   The only option that shows a straight line rising to a peak at $t = 8$ s and then a smooth parabolic drop to zero at $t = 12$ s is Option A.

Students often forget that the relative acceleration is zero only while both stones are in flight. As soon as the first stone hits the ground, the relative motion switches from linear to parabolic. Always check the domain of validity of your equations and match the graph’s shape to the physical phases of motion.