JEE PYQ: Motion in a Straight Line - Question ID f681268c445e (JEE Main 2024)

Core Formula & Concept:

In relative motion along a straight line, the velocity of an object B with respect to another object A is given by: $\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A$ where $\vec{v}_B$ and $\vec{v}_A$ are the velocities of B and A measured with respect to the ground (or any common inertial frame).

Similarly, the velocity of the ground (G) with respect to B is: $\vec{v}_{G/B} = \vec{v}_G - \vec{v}_B$ Since the ground is at rest, $\vec{v}_G = 0$, so: $\vec{v}_{G/B} = -\vec{v}_B$

We choose a consistent direction (e.g., north as positive) and convert all speeds to SI units ($\mathrm{ms}^{-1}$).

Step-by-Step Derivation:

1. Convert speeds to SI units:
Train A speed: $72 \mathrm{~km/h} = 72 \times \frac{1000}{3600} = 20 \mathrm{~ms}^{-1}$ (north, positive).
Train B speed: $108 \mathrm{~km/h} = 108 \times \frac{1000}{3600} = 30 \mathrm{~ms}^{-1}$ (south, negative direction).
So, $\vec{v}_A = +20 \mathrm{~ms}^{-1}$, $\vec{v}_B = -30 \mathrm{~ms}^{-1}$.

2. Velocity of B with respect to A:
$\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A = (-30) - (+20) = -50 \mathrm{~ms}^{-1}$ The negative sign indicates that B appears to move southward relative to A.

3. Velocity of ground with respect to B:
$\vec{v}_{G/B} = \vec{v}_G - \vec{v}_B = 0 - (-30) = +30 \mathrm{~ms}^{-1}$ The positive sign indicates that the ground appears to move northward relative to B.

4. Match with given options:
$\vec{v}_{B/A} = -50 \mathrm{~ms}^{-1}$, $\vec{v}_{G/B} = +30 \mathrm{~ms}^{-1}$ corresponds to option B.

Common Traps & Exam Tip:

Students often confuse the order of subtraction in relative velocity ($\vec{v}_{B/A}$ vs $\vec{v}_{A/B}$) or forget to assign correct signs based on direction. Always define a positive direction (e.g., north) and stick to it throughout the problem. Converting units to SI at the start avoids calculation errors.