Physics Q29 – JEE-MAIN 2025

Home Physics Motion in a PlaneQuestion #11

A river is flowing from west to east direction with speed of

9 {km h}^{- 1}

. If a boat capable of moving at a maximum speed of

27 {km h}^{- 1}

in still water, crosses the river in half a minute, while moving with maximum speed at an angle of

150^{\circ}

to direction of river flow, then the width of the river is:

Clue 1 of 3: Decompose Velocity

To find the width of the river, you need to consider only the component of the boat's velocity that is perpendicular to the river flow.

🥷

Ninja Strategy•Correct Vector Decomposition

The key is to correctly identify that the component of the boat's velocity perpendicular to the river flow is $v_{b} \sin (150^{\circ})$ , not $v_{b} \sin (60^{\circ})$ or the river's speed.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify Given Values and Target✦ Active

Given the speed of the river $v_{r} = 9 {km h}^{- 1}$ , the maximum speed of the boat in still water $v_{b} = 27 {km h}^{- 1}$ , the time to cross the river $t = 0.5 minute$ , and the angle of the boat with the river flow $θ = 150^{\circ}$ . We need to find the width of the river, $w$ .

First, convert the time to hours for consistency with speeds: $t = 0.5 min = \frac{0.5}{60} h = \frac{1}{120} h$ .

Step 2: Calculate Velocity Component Perpendicular to River Flow○ Expand

The width of the river depends only on the component of the boat's velocity perpendicular to the river flow. This component is given by $v_{y} = v_{b} \sin (θ)$ .

v_{y} = 27 {km h}^{- 1} \times \sin (150^{\circ})

\sin (150^{\circ}) = \sin (180^{\circ} - 30^{\circ}) = \sin (30^{\circ}) = \frac{1}{2}

v_{y} = 27 {km h}^{- 1} \times \frac{1}{2} = 13.5 {km h}^{- 1}

💡 Teacher's Secret Hint

Remember that the angle is given with respect to the river flow, so $\sin (θ)$ gives the component perpendicular to the flow.

Step 3: Calculate the Width of the River○ Expand

The width of the river $w$ is the product of the perpendicular velocity component and the time taken to cross.

w = v_{y} \times t

w = 13.5 {km h}^{- 1} \times \frac{1}{120} h

w = \frac{13.5}{120} km

w = \frac{135}{1200} km = \frac{9}{80} km

Convert the width from kilometers to meters:

w = \frac{9}{80} \times 1000 m

w = 9 \times \frac{100}{8} m = 9 \times 12.5 m

w = 112.5 m

💡 Teacher's Secret Hint

Always ensure all units are consistent before performing calculations. Converting to meters at the end is a good practice for final answers.

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Physics Question 29 – JEE-MAIN 2025

Clue 1 of 3: Decompose Velocity

🧠 Full Solution Path