JEE PYQ: Motion in a Straight Line - Question ID f97d835af5d4 (JEE Main 2026)

Core Formula & Concept:

In problems involving relative motion, the key idea is to analyze the velocities of objects from a common reference frame. Here, we have two cars moving in the same direction, and a stone is thrown from car $B$ to hit car $A$. The speed with which the stone hits car $A$ is given in the ground frame, but the throw is made from car $B$, which itself is moving.

The fundamental formula used is the relative velocity relation: $\vec{v}_{object, frame1} = \vec{v}_{object, frame2} + \vec{v}_{frame2, frame1}$ where:

• $\vec{v}_{object, frame1}$ is the velocity of the object (stone) relative to frame 1 (ground).
• $\vec{v}_{object, frame2}$ is the velocity of the object relative to frame 2 (car $B$).
• $\vec{v}_{frame2, frame1}$ is the velocity of frame 2 (car $B$) relative to frame 1 (ground).

Since all motion is along a straight line in the same direction, we can work with scalar speeds, taking the direction of motion as positive.

Step-by-Step Derivation:

Step 1: Convert all speeds to the same unit (m/s)

Given speeds:

• Speed of car $A$, $v_A = 100 \text{ km/h}$
• Speed of car $B$, $v_B = 80 \text{ km/h}$
• Speed of stone relative to car $A$ at impact = $5 \text{ m/s}$

Convert km/h to m/s: $1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}$ So, $v_A = 100 \times \frac{5}{18} = \frac{500}{18} \approx 27.78 \text{ m/s}$ $v_B = 80 \times \frac{5}{18} = \frac{400}{18} \approx 22.22 \text{ m/s}$

Step 2: Understand the impact condition

The stone hits car $A$ with a speed of $5 \text{ m/s}$ relative to the ground. This means the velocity of the stone relative to the ground at impact is: $v_{stone, ground} = v_A \pm \text{relative speed}$ But since the stone is hitting car $A$, and car $A$ is moving faster than car $B$, the stone must be thrown forward from car $B$ to catch up with car $A$. The stone's speed relative to the ground must be such that it matches or exceeds car $A$'s speed to make contact.

However, the question states that the stone hits car $A$ with a speed of $5 \text{ m/s}$. This is ambiguous — does it mean:

• the stone's speed relative to car $A$ is $5 \text{ m/s}$, or
• the stone's speed relative to the ground is $5 \text{ m/s}$?

Interpretation: The phrase "hits the car $A$ with a speed of $5 \text{ m/s}$" is standardly interpreted as the relative speed of the stone with respect to car $A$ at the moment of impact. That is: $|v_{stone, ground} - v_A| = 5 \text{ m/s}$ Since car $A$ is moving faster than car $B$, and the stone is thrown from $B$ toward $A$, the stone must be moving faster than $B$ but slower than $A$ (or faster, depending on throw direction). But since $A$ is ahead and moving faster, the stone must be thrown forward to catch up, so: $v_{stone, ground} = v_A - 5 \text{ (if stone is slower than A)}$ or $v_{stone, ground} = v_A + 5 \text{ (if stone is faster than A)}$ But since $A$ is moving at $27.78 \text{ m/s}$ and $B$ at $22.22 \text{ m/s}$, and the stone is thrown from $B$, it's unlikely the stone exceeds $A$'s speed unless thrown very fast. So the reasonable interpretation is: $v_{stone, ground} = v_A - 5 = 27.78 - 5 = 22.78 \text{ m/s}$ This means the stone is moving at $22.78 \text{ m/s}$ relative to the ground when it hits $A$, and since $A$ is moving at $27.78 \text{ m/s}$, the relative speed is $5 \text{ m/s}$.

Step 3: Use relative velocity to find $v$

The stone is thrown from car $B$, which is moving at $v_B = 22.22 \text{ m/s}$ relative to the ground. Let $v$ be the speed of the stone relative to car $B$ (this is the $v$ we need to find, in km/h later).

The velocity of the stone relative to the ground is: $v_{stone, ground} = v_{stone, B} + v_B$ Assuming the stone is thrown in the direction of motion (forward), then: $v_{stone, ground} = v + v_B$ But we found $v_{stone, ground} = 22.78 \text{ m/s}$, and $v_B = 22.22 \text{ m/s}$, so: $v + 22.22 = 22.78 \Rightarrow v = 22.78 - 22.22 = 0.56 \text{ m/s}$ But this is too small — and doesn't match any option. This suggests our interpretation of the impact speed is incorrect.

Step 4: Reinterpret the impact speed

Let’s consider the alternative interpretation: the stone hits car $A$ with a speed of $5 \text{ m/s}$ relative to car $A$. That is: $v_{stone, A} = v_{stone, ground} - v_A = \pm 5 \text{ m/s}$ Since the stone is thrown from behind (car $B$ is behind car $A$), and car $A$ is moving faster, the stone must be thrown forward to catch up. So the stone's ground speed must be less than $A$'s speed, so: $v_{stone, ground} - v_A = -5 \Rightarrow v_{stone, ground} = v_A - 5 = 27.78 - 5 = 22.78 \text{ m/s}$ This matches our earlier result.

But again, $v = v_{stone, ground} - v_B = 22.78 - 22.22 = 0.56 \text{ m/s}$ — too small.

Step 5: Consider the possibility of backward throw

What if the stone is thrown backward from car $B$? Then: $v_{stone, ground} = v_B - v$ But since $A$ is ahead and moving faster, throwing backward would make the stone slower, increasing the gap — so it can't hit $A$. So this is not possible.

Step 6: Re-examine the question wording

The question says: "throws a stone with a speed $v$ so that it hits the car $A$ with a speed of $5 \text{ m/s}$."

This likely means: the stone's speed relative to car $A$ at impact is $5 \text{ m/s}$. But as we saw, this leads to a very small $v$, which is not among the options.

Step 7: Alternative interpretation — $5 \text{ m/s}$ is the speed of the stone relative to the ground

Suppose the stone hits car $A$ with a speed of $5 \text{ m/s}$ relative to the ground. Then: $v_{stone, ground} = 5 \text{ m/s}$ But car $A$ is moving at $27.78 \text{ m/s}$, so the relative speed of the stone with respect to $A$ is: $v_{stone, A} = v_{stone, ground} - v_A = 5 - 27.78 = -22.78 \text{ m/s}$ The magnitude is $22.78 \text{ m/s}$, which is not $5 \text{ m/s}$. So this interpretation is invalid.

Step 8: Correct interpretation — $5 \text{ m/s}$ is the speed of the stone relative to car $A$ at impact

Let’s stick with: $v_{stone, A} = v_{stone, ground} - v_A = -5 \text{ m/s} \quad \text{(since stone is slower)}$ So: $v_{stone, ground} = v_A - 5 = 27.78 - 5 = 22.78 \text{ m/s}$ Now, the stone is thrown from car $B$, which is moving at $v_B = 22.22 \text{ m/s}$. The stone's speed relative to $B$ is $v$ (in m/s), so: $v_{stone, ground} = v_B + v \quad \text{(if thrown forward)}$ Thus: $v = v_{stone, ground} - v_B = 22.78 - 22.22 = 0.56 \text{ m/s}$ Convert to km/h: $v = 0.56 \times \frac{18}{5} = 2.016 \text{ km/h}$ This is not among the options.

Step 9: Re-evaluate the direction of relative speed

Perhaps the stone hits car $A$ with a speed of $5 \text{ m/s}$ in the opposite direction — i.e., the stone is moving backward relative to $A$. Then: $v_{stone, A} = +5 \text{ m/s} \Rightarrow v_{stone, ground} = v_A + 5 = 27.78 + 5 = 32.78 \text{ m/s}$ Now, the stone is thrown from $B$ at $22.22 \text{ m/s}$. To reach $32.78 \text{ m/s}$, it must be thrown forward with speed: $v = v_{stone, ground} - v_B = 32.78 - 22.22 = 10.56 \text{ m/s}$ Convert to km/h: $v = 10.56 \times \frac{18}{5} = 38.016 \text{ km/h} \approx 38 \text{ km/h}$ This matches option C.

Conclusion:

The correct interpretation is that the stone hits car $A$ with a speed of $5 \text{ m/s}$ relative to car $A$, and in the same direction as $A$'s motion (i.e., the stone is moving faster than $A$). This means: $v_{stone, A} = v_{stone, ground} - v_A = +5 \text{ m/s}$ So: $v_{stone, ground} = v_A + 5 = 27.78 + 5 = 32.78 \text{ m/s}$ Then, the speed of the stone relative to car $B$ is: $v = v_{stone, ground} - v_B = 32.78 - 22.22 = 10.56 \text{ m/s}$ Convert to km/h: $v = 10.56 \times \frac{18}{5} = 38 \text{ km/h}$

Final Answer: The value of $v$ is 38 km/h, which corresponds to option C. Common Traps & Exam Tip:

Trap 1: Misinterpreting the reference frame of the impact speed. Students often confuse whether the $5 \text{ m/s}$ is relative to the ground or to car $A$. The question says "hits the car $A$ with a speed of $5 \text{ m/s}$", which implies relative to $A$.

Trap 2: Incorrect sign in relative velocity. When calculating $v_{stone, A} = v_{stone, ground} - v_A$, the sign depends on direction. If the stone is faster than $A$, $v_{stone, A}$ is positive; if slower, negative. Students often reverse this.

Trap 3: Unit inconsistency. Failing to convert all speeds to the same unit (m/s or km/h) leads to incorrect results. Always convert to consistent units before calculations.

Exam Tip: In relative motion problems, always define the reference frames clearly. Use the formula: $v_{object, frame1} = v_{object, frame2} + v_{frame2, frame1}$ and assign signs based on direction. Draw a diagram if needed.