Question

A river is flowing from west to east direction with speed of \(9\) km/hr. If a boat capable of moving at a maximum speed of \(27\) km/hr 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:

• A. 300 m
• B. 112.5 m
• C. 75 m
• D. \( 112.5 \times \sqrt{3} \) m

Hint:

When solving river crossing problems, decompose the boat's velocity into components: one parallel to the river flow and one perpendicular to it. The perpendicular component gives the speed for crossing the river.

Answer 1

The Correct Option is B

The speed of the boat relative to the river is \( 27 \, \text{km/hr} \), and the boat crosses the river at an angle of \( 150^\circ \) to the direction of the river flow. 
Using the formula for the effective speed component of the boat in the direction perpendicular to the flow of the river: \[ V_{L} = 27 \, \text{km/hr} \times \cos 60^\circ = \frac{27}{2} = 13.5 \, \text{km/hr} \] The time taken to cross the river is \( 30 \, \text{seconds} \) or \( \frac{1}{2} \) minute. Using the formula for distance: \[ S = V_t \times t = 13.5 \, \text{km/hr} \times \frac{30}{60} \, \text{hr} = 13.5 \times \frac{1}{2} = 112.5 \, \text{m} \] Thus, the width of the river is \( 112.5 \, \text{m} \).

Answer 2

This problem involves relative motion, specifically a boat crossing a flowing river. We need to find the width of the river given the speeds of the river and the boat, the direction of the boat's motion relative to the river flow, and the time taken to cross.

Concept Used:

The solution is based on the principle of vector addition for relative velocities. The velocity of the boat with respect to the ground (\( \vec{v}_b \)) is the vector sum of its velocity with respect to the river (\( \vec{v}_{br} \)) and the velocity of the river with respect to the ground (\( \vec{v}_r \)).

\[ \vec{v}_b = \vec{v}_{br} + \vec{v}_r \]

The width of the river is determined by the component of the boat's velocity that is perpendicular to the river flow. If we set up a coordinate system where the river flows along the x-axis, the width (\(W\)) is given by:

\[ W = v_{by} \times t \]

where \( v_{by} \) is the y-component of the boat's velocity with respect to the ground, and \( t \) is the time taken to cross.

Step-by-Step Solution:

Step 1: Define a coordinate system and list the given parameters with consistent units.

Let's set the direction of the river flow (West to East) as the positive x-axis. The direction perpendicular to the flow, across the river, will be the positive y-axis.

• Velocity of the river, \( \vec{v}_r = 9 \hat{i} \, \text{km/hr} \).
• Speed of the boat in still water (relative to the river), \( |\vec{v}_{br}| = 27 \, \text{km/hr} \).
• Angle of the boat's motion with respect to the river flow, \( \theta = 150^\circ \).
• Time taken to cross, \( t = 0.5 \, \text{minutes} \).

We need to convert the time from minutes to hours to maintain unit consistency:

\[ t = 0.5 \, \text{min} \times \frac{1 \, \text{hr}}{60 \, \text{min}} = \frac{0.5}{60} \, \text{hr} = \frac{1}{120} \, \text{hr} \]

Step 2: Resolve the boat's velocity relative to the river into its components.

The velocity of the boat relative to the river, \( \vec{v}_{br} \), has a magnitude of 27 km/hr and is directed at an angle of \( 150^\circ \) to the positive x-axis.

\[ \vec{v}_{br} = |\vec{v}_{br}| (\cos\theta \hat{i} + \sin\theta \hat{j}) \] \[ \vec{v}_{br} = 27 (\cos 150^\circ \hat{i} + \sin 150^\circ \hat{j}) \]

We know that \( \cos 150^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2} \) and \( \sin 150^\circ = \sin 30^\circ = \frac{1}{2} \).

\[ \vec{v}_{br} = 27 \left(-\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}\right) = -13.5\sqrt{3} \hat{i} + 13.5 \hat{j} \, \text{km/hr} \]

Step 3: Determine the component of the boat's velocity responsible for crossing the river.

The velocity of the boat with respect to the ground is \( \vec{v}_b = \vec{v}_{br} + \vec{v}_r \).

\[ \vec{v}_b = (-13.5\sqrt{3} \hat{i} + 13.5 \hat{j}) + (9 \hat{i}) = (9 - 13.5\sqrt{3}) \hat{i} + 13.5 \hat{j} \, \text{km/hr} \]

The component of this velocity perpendicular to the river flow (the y-component) is what determines the time to cross the river. From the vector equation above, this component is:

\[ v_{by} = 13.5 \, \text{km/hr} \]

Final Computation & Result:

Step 4: Calculate the width of the river.

The width of the river \(W\) is the product of the perpendicular component of the velocity and the time taken.

\[ W = v_{by} \times t \] \[ W = 13.5 \, \frac{\text{km}}{\text{hr}} \times \frac{1}{120} \, \text{hr} \] \[ W = \frac{13.5}{120} \, \text{km} = \frac{27/2}{120} \, \text{km} = \frac{27}{240} \, \text{km} \]

Simplifying the fraction:

\[ W = \frac{27 \div 3}{240 \div 3} = \frac{9}{80} \, \text{km} \]

We can also express this in meters:

\[ W = \frac{9}{80} \, \text{km} \times 1000 \, \frac{\text{m}}{\text{km}} = \frac{9000}{80} \, \text{m} = \frac{900}{8} \, \text{m} = 112.5 \, \text{m} \]

The width of the river is \( \frac{9}{80} \) km or 112.5 m.