Question:
A swimmer can swim at \(5\,\text{m s}^{-1}\) in still water. River flows at \(3\,\text{m s}^{-1}\). To cross the river in shortest time, the angle with respect to the perpendicular to the flow is
A swimmer can swim at \(5\,\text{m s}^{-1}\) in still water. River flows at \(3\,\text{m s}^{-1}\). To cross the river in shortest time, the angle with respect to the perpendicular to the flow is
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For \textbf{minimum time crossing}, always swim perpendicular to the river flow.
For \textbf{reaching the point directly opposite}, swim at an upstream angle so that the river drift is compensated.
Updated On: Jul 9, 2026
- \(0^\circ\)
- \[ \sin^{-1}\left(\frac{3}{5}\right) \]
- \[ \tan^{-1}\left(\frac{3}{5}\right) \]
- \(90^\circ\)
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The Correct Option is A
Solution and Explanation
Concept:
The time taken to cross a river depends on the component of the swimmer's velocity perpendicular to the river bank.
\[
t=\frac{\text{Width of river}}{\text{Perpendicular component of velocity}}
\]
To cross the river in minimum time, the perpendicular component of velocity must be maximum.
Step 1: Assume the swimmer makes an angle \(\theta\) with the perpendicular to the flow. The swimmer's speed in still water is \[ v=5\,\text{m s}^{-1}. \] Hence the component perpendicular to the flow is \[ v_\perp=5\cos\theta. \]
Step 2: Write the expression for crossing time. If the width of the river is \(d\), \[ t=\frac{d}{5\cos\theta}. \]
Step 3: Find the condition for minimum time. For minimum crossing time, \[ \cos\theta \] must be maximum. The maximum value of \(\cos\theta\) is \[ 1. \] This occurs when \[ \theta=0^\circ. \]
Step 4: Obtain the answer. Therefore, the swimmer should swim exactly perpendicular to the river flow. \[ \boxed{\theta=0^\circ} \] \[ \boxed{\text{Answer = (A)}} \]
Step 1: Assume the swimmer makes an angle \(\theta\) with the perpendicular to the flow. The swimmer's speed in still water is \[ v=5\,\text{m s}^{-1}. \] Hence the component perpendicular to the flow is \[ v_\perp=5\cos\theta. \]
Step 2: Write the expression for crossing time. If the width of the river is \(d\), \[ t=\frac{d}{5\cos\theta}. \]
Step 3: Find the condition for minimum time. For minimum crossing time, \[ \cos\theta \] must be maximum. The maximum value of \(\cos\theta\) is \[ 1. \] This occurs when \[ \theta=0^\circ. \]
Step 4: Obtain the answer. Therefore, the swimmer should swim exactly perpendicular to the river flow. \[ \boxed{\theta=0^\circ} \] \[ \boxed{\text{Answer = (A)}} \]
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