JEE PYQ: Motion in a Straight Line - Question ID 13a0147e02c0 (JEE Main 2017)

In problems involving relative motion with constant acceleration, the key idea is to express the positions of both moving objects as functions of time and then find the instant when their positions coincide.

For an object starting from rest and moving with constant acceleration $a$, its position $s(t)$ at time $t$ is given by: $s(t) = \frac{1}{2} a t^2$

When two objects start from different initial positions and move with different accelerations, the relative position between them changes over time. The car catches up with the bus when the distance covered by the car equals the initial separation plus the distance covered by the bus.

Let:

• $s_0 = 200 \, \text{m}$: initial separation (car is 200 m behind the bus).
• $a_b = 2 \, \text{m/s}^2$: acceleration of the bus.
• $a_c = 4 \, \text{m/s}^2$: acceleration of the car.

At time $t$, the position of the bus (starting from rest) is: $s_b(t) = \frac{1}{2} a_b t^2 = \frac{1}{2} \cdot 2 \cdot t^2 = t^2$

The position of the car (also starting from rest, but 200 m behind) is: $s_c(t) = s_0 + \frac{1}{2} a_c t^2 = 200 + \frac{1}{2} \cdot 4 \cdot t^2 = 200 + 2 t^2$

The car catches up with the bus when $s_c(t) = s_b(t)$: $200 + 2 t^2 = t^2$

Rearrange: $2 t^2 - t^2 = -200 \quad \Rightarrow \quad t^2 = 200$ Wait — this leads to $t = \sqrt{200} = 10 \sqrt{2} \, \text{s}$, which matches option C. However, let's double-check the setup.

Actually, the car is behind the bus by 200 m. So when the car catches up, its position equals the bus’s position: $s_c(t) = s_b(t)$ But $s_c(t)$ starts 200 m behind, so: $s_c(t) = \text{initial position of car} + \text{distance covered by car} = -200 + \frac{1}{2} a_c t^2$ Wait — no. Let’s define a coordinate system where the bus starts at $x = 0$, and the car starts at $x = -200 \, \text{m}$. Then: $x_b(t) = \frac{1}{2} a_b t^2 = t^2$ $x_c(t) = -200 + \frac{1}{2} a_c t^2 = -200 + 2 t^2$ The car catches the bus when $x_c(t) = x_b(t)$: $-200 + 2 t^2 = t^2$ $2 t^2 - t^2 = 200$ $t^2 = 200$ $t = \sqrt{200} = 10 \sqrt{2} \, \text{s}$

Thus, the correct time is $10 \sqrt{2} \, \text{s}$, which corresponds to option C.

Students often make two critical mistakes:

1. Incorrect sign for initial separation: Forgetting that the car starts 200 m behind the bus and incorrectly setting $s_c(t) = 200 + \frac{1}{2} a_c t^2$. This leads to $t^2 = -200$, which is impossible.
2. Misapplying relative acceleration: Trying to use $a_{\text{relative}} = a_c - a_b = 2 \, \text{m/s}^2$ and then solving $s = \frac{1}{2} a_{\text{relative}} t^2$, which ignores the initial 200 m gap. This gives $t = \sqrt{200}$, but the setup is flawed because relative acceleration alone doesn’t account for initial separation.

Exam Tip: Always define a clear coordinate system. Let one object start at $x = 0$, and the other at its initial position (positive or negative). Then write position equations for both and set them equal at the catch-up time.