Understanding Boat Motion in a Stream

This problem involves understanding the concept of relative velocity, specifically how a boat moves in a flowing stream. We need to calculate two main things: the time it takes for the boat to cross the stream and the angle its effective path makes with the direction of the stream.

Given Information for Boat Travel

Let's list down the important information provided in the question regarding the boat and stream:

• Velocity of the stream (\(v_s\)) = 20 kmph
• Velocity of the boat in a direction perpendicular to the stream (\(v_b\)) = 20 kmph (This is the component of the boat's effective velocity relative to the ground, perpendicular to the stream's flow.)
• Width of the stream (d) = 2 km

Calculating Time Taken to Cross the Stream

The time taken for the boat to cross the stream depends only on the component of the boat's velocity that is directed perpendicular to the stream's flow and the width of the stream. The stream's velocity affects how far downstream the boat drifts, but it does not affect the time it takes to cover the perpendicular distance (the width of the stream) to the other bank.

The formula for time (t) is:

\[t = \frac{\text{Width of stream}}{\text{Velocity of boat perpendicular to stream}}\]

Substituting the given values into the formula:

\[t = \frac{d}{v_b} = \frac{2 \text{ km}}{20 \text{ kmph}}\]

\[t = \frac{1}{10} \text{ hour}\]

To convert this time from hours into minutes, we multiply by 60, since there are 60 minutes in an hour:

\[t = \frac{1}{10} \times 60 \text{ minutes}\]

\[t = 6 \text{ minutes}\]

So, the time taken for the boat to cross the stream is 6 minutes.

Determining Angle with the Stream Direction

When the boat travels across the stream, it has two velocity components relative to the ground:

1. A velocity component perpendicular to the stream, which is the boat's own effective velocity component across the stream.
2. A velocity component parallel to the stream, which is due to the stream's current carrying the boat downstream.

Let's denote these components:

• Velocity component perpendicular to the stream (\(V_y\)) = 20 kmph (This is the effective speed at which the boat covers the width of the stream).
• Velocity component parallel to the stream (\(V_x\)) = Velocity of the stream (\(v_s\)) = 20 kmph (This is the speed at which the boat is carried downstream).

The effective path of the boat relative to the ground is the resultant of these two perpendicular velocity components. We can visualize these two velocities as the adjacent and opposite sides of a right-angled triangle. The angle \(\theta\) that the boat's resultant path makes with the direction of the stream can be found using the tangent function, which relates the opposite side (perpendicular velocity) to the adjacent side (parallel velocity):

\[\tan \theta = \frac{\text{Velocity perpendicular to stream}}{\text{Velocity parallel to stream}}\]

\[\tan \theta = \frac{V_y}{V_x}\]

Substituting the known velocity values into the equation:

\[\tan \theta = \frac{20 \text{ kmph}}{20 \text{ kmph}}\]

\[\tan \theta = 1\]

To find the angle \(\theta\), we take the inverse tangent (arctangent) of 1:

\[\theta = \arctan(1)\]

\[\theta = 45^{\circ}\]

Therefore, the angle at which the boat's effective path makes with the direction of the stream is 45 degrees.

Summary of Results

Combining our calculations for the time taken and the angle made with the stream direction, we get the final answer: