Physics Q27 – JEE-MAIN 2026

Home Physics Motion in a PlaneQuestion #303

Two cars

A

and

B

are moving in the same direction along a straight line with speeds

100 km/h

and

80 km/h

, respectively such that car

A

is moving ahead of car

B

. A person in car

B

throws a stone with a speed

v

so that it hits the car

A

with a speed of

5 m/s

. The value of

v

is _______

km/h

Clue 1 of 3: Relative Velocity Concept

When objects move in the same direction, their relative speed is the difference between their individual speeds. When an object is thrown from a moving frame, its velocity relative to the ground is the sum of its velocity relative to the frame and the frame's velocity.

🧠 Full Solution Path

Step 1: Convert speeds to consistent units✦ Active

First, convert the speeds of cars A and B from $km/h$ to $m/s$ for consistency with the stone's impact speed. The conversion factor is $1 km/h = \frac{5}{18} m/s$ .

V_{A} = 100 km/h = 100 \times \frac{5}{18} m/s = \frac{500}{18} m/s = \frac{250}{9} m/s

V_{B} = 80 km/h = 80 \times \frac{5}{18} m/s = \frac{400}{18} m/s = \frac{200}{9} m/s

The speed of the stone relative to car A is given as $V_{S A} = 5 m/s$ .

Step 2: Formulate the relative velocity equation○ Expand

Let $v^{'}$ be the speed of the stone relative to car B (in $m/s$ ). Since the stone is thrown forward to hit car A, its velocity relative to the ground ( $V_{S}$ ) is the sum of car B's velocity and its own velocity relative to car B.

V_{S} = V_{B} + v^{'}

The speed of the stone relative to car A ( $V_{S A}$ ) is the difference between the stone's velocity relative to the ground and car A's velocity relative to the ground.

V_{S A} = V_{S} - V_{A}

Substitute the expression for $V_{S}$ into the equation for $V_{S A}$ :

V_{S A} = (V_{B} + v^{'}) - V_{A}

💡 Teacher's Secret Hint

Ensure you correctly identify the direction of relative velocities. Since car A is ahead and the stone is thrown to hit it, the stone's speed relative to the ground must be greater than car A's speed.

Step 3: Solve for

v^{'}

and convert to

km/h

○ Expand

Substitute the known values into the relative velocity equation and solve for $v^{'}$ .

5 = (\frac{200}{9} + v^{'}) - \frac{250}{9}

5 = v^{'} - \frac{50}{9}

v^{'} = 5 + \frac{50}{9} = \frac{45 + 50}{9} = \frac{95}{9} m/s

Finally, convert $v^{'}$ from $m/s$ back to $km/h$ .

v = \frac{95}{9} m/s \times \frac{18}{5} km/h per m/s = 19 \times 2 km/h = 38 km/h

💡 Teacher's Secret Hint

Double-check your unit conversions, especially when converting back to the requested unit.

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Physics Question 27 – JEE-MAIN 2026

Clue 1 of 3: Relative Velocity Concept

🧠 Full Solution Path