JEE PYQ: Motion in a Straight Line - Question ID 561a65aff286 (JEE Main 2022)

Core Formula & Concept:

When an object moves under constant acceleration (here, gravity $g = 10\ \text{m/s}^2$), its velocity and position at any time $t$ are governed by the kinematic equations:

• Velocity: $v(t) = v_0 + a t$, where $a = g$ (downward).
• Displacement: $s(t) = v_0 t + \tfrac{1}{2} a t^2$.

The bullet is fired vertically downward with initial velocity $v_0 = 100\ \text{m/s}$. It accelerates downward until it hits the ground at $t = 10\ \text{s}$. At that instant, it undergoes a perfectly inelastic collision, bringing its velocity to zero. For the remaining $10\ \text{s}$ (from $t = 10\ \text{s}$ to $t = 20\ \text{s}$), the bullet remains at rest.

Step-by-Step Derivation:

1. Velocity during free fall (0 ≤ t ≤ 10 s)

Since the bullet is moving downward, we take downward as positive. The acceleration is $+g = +10\ \text{m/s}^2$. Thus $v(t) = v_0 + g\,t = 100 + 10\,t.$ At $t = 0$, $v = 100\ \text{m/s}$; at $t = 10\ \text{s}$, $v = 100 + 10\cdot10 = 200\ \text{m/s}$.

2. Impact and rest (10 s ≤ t ≤ 20 s)

At $t = 10\ \text{s}$, the bullet hits the ground and stops instantaneously. For all $t > 10\ \text{s}$, $v(t) = 0$.

3. Sketch of the v–t graph

• From $t = 0$ to $t = 10\ \text{s}$, the velocity increases linearly from $100\ \text{m/s}$ to $200\ \text{m/s}$.
• At $t = 10\ \text{s}$, the velocity drops abruptly to zero.
• From $t = 10\ \text{s}$ to $t = 20\ \text{s}$, the velocity remains zero.

This matches the shape shown in option A: a straight line rising from $100$ to $200$ over the first $10\ \text{s}$, then a vertical drop to zero, followed by a horizontal line at zero.

Common Traps & Exam Tip:

1. Sign convention: Many students take upward as positive and write $v(t) = 100 - 10\,t$. That would give $v(10) = 0$, which is incorrect because the bullet is still moving downward at $t = 10\ \text{s}$. 2. Collision instant: Forgetting that the velocity drops to zero at $t = 10\ \text{s}$ and continuing the line beyond $200\ \text{m/s}$. 3. Graph shape: Confusing the linear rise with a parabolic or exponential curve.

Tip: Always fix a sign convention (here, downward positive) and stick to it throughout the problem.