Question

A river of width 200 m is flowing from west to east with a speed of 18 km/h. A boat, moving with speed of 36 km/h in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are _________ and _________} respectively.

• A. 20 s and 100 m
• B. 40 s and 100 m
• C. 40 s and 0 m
• D. 40 s and 200 m

Hint:

Minimum time to cross is always $W/v_{boat}$ regardless of river flow. However, this path results in a drift equal to $v_{river} \times t$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Minimum time to cross a river occurs when the boat is steered perpendicular to the river flow. One-round trip involves crossing and returning to the same bank. During this time, the river flow causes a drift along the bank.

Step 2: Key Formula or Approach:
1. $v_{river} = 18 \text{ km/h} = 18 \times \frac{5}{18} = 5 \text{ m/s}$.
2. $v_{boat} = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10 \text{ m/s}$.
3. Time to cross $t = \frac{W}{v_{boat}}$.

Step 3: Detailed Explanation:
For minimum time, steer boat at 90$^\circ$ to the bank.
Time to reach the other bank:
\[ t_1 = \frac{200 \text{ m}}{10 \text{ m/s}} = 20 \text{ s} \]
Time for the return trip (back to the original bank):
\[ t_2 = \frac{200 \text{ m}}{10 \text{ m/s}} = 20 \text{ s} \]
Total minimum journey time $T = t_1 + t_2 = 40$ s.
Displacement (drift) along the river bank:
The river continues to flow in the same direction (West to East) during both legs of the trip.
Drift during $t_1 = v_{river} \cdot t_1 = 5 \times 20 = 100$ m.
Drift during $t_2 = v_{river} \cdot t_2 = 5 \times 20 = 100$ m.
Total displacement along the bank $= 100 + 100 = 200$ m.

Step 4: Final Answer:
The journey takes 40 s and the displacement is 200 m.