JEE PYQ: Motion in a Straight Line - Question ID 4ec63bc8cf0d (JEE Main 2023)

Core Formula & Concept:

This problem involves uniformly decelerated motion in a straight line. The key physics concepts and formulas are:

• Kinematic equation for velocity, displacement, and acceleration: $v^2 = u^2 + 2 a s$ where $v$ = final velocity, $u$ = initial velocity, $a$ = acceleration (negative for deceleration), $s$ = displacement.
• Retardation (deceleration): The brakes produce a constant negative acceleration (retardation) $a$, which we can determine from the first scenario.
• Consistency of retardation: The same braking force implies the same magnitude of retardation in both scenarios.

Step-by-Step Derivation:

Step 1: Determine the retardation from the first scenario.

Given: Initial speed $u = 20 \text{ m/s}$, Final speed $v = 0 \text{ m/s}$ (comes to rest), Braking distance $s = 500 \text{ m}$.

Using the kinematic equation: $v^2 = u^2 + 2 a s$ Substitute $v = 0$: $0 = (20)^2 + 2 a (500)$ $0 = 400 + 1000 a$ $1000 a = -400$ $a = -\frac{400}{1000} = -0.4 \text{ m/s}^2$ The negative sign indicates deceleration (retardation). The magnitude of retardation is $0.4 \text{ m/s}^2$.

Step 2: Apply the same retardation in the second scenario.

Now, brakes are applied at half the distance, i.e., $s' = \frac{500}{2} = 250 \text{ m}$. Initial speed remains $u = 20 \text{ m/s}$, Retardation $a = -0.4 \text{ m/s}^2$. We want to find the final speed $v'$ when the train reaches the station (after traveling 250 m under braking).

Using the same kinematic equation: $v'^2 = u^2 + 2 a s'$ Substitute values: $v'^2 = (20)^2 + 2(-0.4)(250)$ $v'^2 = 400 - 200$ $v'^2 = 200$ Thus, $v' = \sqrt{200} \text{ m/s}$

Step 3: Identify $x$.

The question states that the train crosses the station with speed $\sqrt{x} \text{ m/s}$. From above, $v' = \sqrt{200}$, so: $x = 200$

Common Traps & Exam Tip:

• Sign confusion in acceleration: Many students forget that retardation means negative acceleration. Always assign the correct sign based on the direction of motion.
• Misinterpreting "half the distance": Some students mistakenly think the braking distance is halved from the station, not from the original braking point. Clarify the reference point.
• Assuming speed halves when distance halves: Speed does not halve linearly with distance under constant deceleration. Always use the kinematic equations.
• Unit consistency: Ensure all units are consistent (m/s, m, m/s²). Double-check calculations to avoid arithmetic errors.

Exam Tip: When dealing with braking or stopping problems, always start by finding the acceleration (retardation) using the first scenario. Then apply it consistently to other scenarios.