Width of a river is `30 m,` velocity is ` 4 m//s` and rowing velocity is `5 m//s`
(a) Make the velocity diagram for crossing the river in shortest time. Then, find this shortest time, net velocity of boatman and drigt along the river.
(b) Can the boatman reach a point just oppsite on the other shore? If yes then make the velocity diagram, the direction in which the should row his boat and the time taken to cross the river in this case.
(c) How long will it iake hom to row `10 m` up the stream and then back to his starting point?

Explanation

Let's solve the problem step by step.

### Given Data:
- Width of the river (AB) = 30 m
- Velocity of the river (VR) = 4 m/s
- Rowing velocity of the boat (VB) = 5 m/s

### Part (a): Crossing the River in Shortest Time

1. **Velocity Diagram**:
- The boatman rows directly across the river (perpendicular to the riverbank) to minimize the time taken.
- The rowing velocity (VB) is directed straight across (AB), and the river's velocity (VR) is directed downstream (BC).

 *(Illustrative purpose only)*

2. **Time Taken to Cross the River**:
- The time taken to cross the river can be calculated using the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Velocity in the direction of distance}} = \frac{30 \text{ m}}{5 \text{ m/s}} = 6 \text{ seconds} \]

3. **Net Velocity of the Boatman**:
- The net velocity (V_net) can be calculated using the Pythagorean theorem:
\[ V_{\text{net}} = \sqrt{(V_B^2 + V_R^2)} = \sqrt{(5^2 + 4^2)} = \sqrt{25 + 16} = \sqrt{41} \approx 6.4 \text{ m/s} \]

4. **Drift Along the River**:
- The drift (D) can be calculated as:
\[ D = V_R \times \text{Time} = 4 \text{ m/s} \times 6 \text{ s} = 24 \text{ m} \]

### Part (b): Reaching a Point Just Opposite on the Other Shore

1. **Can the Boatman Reach Directly Opposite?**:
- Yes, since the rowing velocity is greater than the river velocity, he can reach the point directly opposite.

2. **Velocity Diagram**:
- The boatman should row at an angle upstream to counteract the river's current. The angle θ can be found using:
\[ \sin \theta = \frac{V_R}{V_B} = \frac{4}{5} \]
- Thus, the angle θ is:
\[ \theta = \sin^{-1}\left(\frac{4}{5}\right) \approx 53^\circ \]

3. **Effective Velocity Across the River**:
- The effective velocity across the river (V_eff) can be calculated as:
\[ V_{\text{eff}} = \sqrt{V_B^2 - V_R^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \text{ m/s} \]

4. **Time Taken to Cross the River**:
- The time taken to cross the river when rowing at this angle is:
\[ \text{Time} = \frac{30 \text{ m}}{3 \text{ m/s}} = 10 \text{ seconds} \]

### Part (c): Rowing 10 m Upstream and Back

1. **Rowing Upstream**:
- The effective velocity upstream (V_up) is:
\[ V_{\text{up}} = V_B - V_R = 5 \text{ m/s} - 4 \text{ m/s} = 1 \text{ m/s} \]
- Time taken to row 10 m upstream:
\[ \text{Time}_{\text{up}} = \frac{10 \text{ m}}{1 \text{ m/s}} = 10 \text{ seconds} \]

2. **Rowing Downstream**:
- The effective velocity downstream (V_down) is:
\[ V_{\text{down}} = V_B + V_R = 5 \text{ m/s} + 4 \text{ m/s} = 9 \text{ m/s} \]
- Time taken to row 10 m downstream:
\[ \text{Time}_{\text{down}} = \frac{10 \text{ m}}{9 \text{ m/s}} \approx 1.11 \text{ seconds} \]

3. **Total Time**:
- The total time taken for the round trip is:
\[ \text{Total Time} = \text{Time}_{\text{up}} + \text{Time}_{\text{down}} = 10 \text{ seconds} + 1.11 \text{ seconds} \approx 11.11 \text{ seconds} \]

### Summary of Results:
- **(a)** Shortest time to cross the river: 6 seconds, net velocity: \( \sqrt{41} \approx 6.4 \text{ m/s} \), drift: 24 m.
- **(b)** Time to reach directly opposite: 10 seconds, angle: \( 53^\circ \).
- **(c)** Total time for 10 m upstream and back: approximately 11.11 seconds.