JEE PYQ: Motion in a Straight Line - Question ID 485b36ceb375 (JEE Main 2022)

Core Formula & Concept:

This problem tests the kinematic equations for motion under constant acceleration. When an object starts from rest ($u = 0$) and moves with constant acceleration $a$, the distance $s$ covered in time $t$ is given by:

$s = ut + \frac{1}{2} a t^2$ Since $u = 0$, this simplifies to: $s = \frac{1}{2} a t^2$

The key insight is that under constant acceleration, the distances covered in successive equal time intervals follow a simple arithmetic progression. Specifically, if the distance covered in the first $t$ seconds is $s_1$, then the distance covered in the next $t$ seconds ($s_2$) will be $3s_1$.

Step-by-Step Derivation:

Step 1: Express the distance in the first $t$ seconds.

Given that the toy starts from rest and travels $10\,m$ in $t$ seconds: $s_1 = \frac{1}{2} a t^2 = 10\,m$

Step 2: Find the distance covered in $2t$ seconds.

Using the same equation for time $2t$: $s_{\text{total}} = \frac{1}{2} a (2t)^2 = \frac{1}{2} a \cdot 4 t^2 = 4 \cdot \left( \frac{1}{2} a t^2 \right) = 4 s_1 = 4 \times 10 = 40\,m$

Step 3: Compute the distance covered in the next $t$ seconds.

The distance covered in the next $t$ seconds (i.e., from $t$ to $2t$) is: $s_2 = s_{\text{total}} - s_1 = 40\,m - 10\,m = 30\,m$

Thus, the toy travels $30\,m$ in the next $t$ seconds.

Common Traps & Exam Tip:

Trap 1: Assuming uniform velocity. Many students mistakenly treat this as uniform motion, leading them to think the distance in the next $t$ seconds is the same ($10\,m$). This is incorrect because the toy is accelerating.

Trap 2: Incorrect ratio application. Some students recall that distances in successive equal time intervals under constant acceleration are in the ratio $1:3:5:...$ but misapply it. The correct ratio for the first two intervals is $1:3$, meaning the second interval covers $3$ times the first.

Exam Tip: Always write down the kinematic equation explicitly. For constant acceleration problems, the distance covered in the $n$-th second (or interval) can be derived using: $s_n = u + \frac{a}{2} (2n - 1)$ For $u = 0$, this simplifies to $s_n = \frac{a}{2} (2n - 1)$. For the first interval ($n=1$), $s_1 = \frac{a}{2}$, and for the second ($n=2$), $s_2 = \frac{3a}{2} = 3 s_1$. This confirms the result.