A man wishes to cross a river of width 120 m by a motorboat. His rowing speed in still water is `3 m//s` and his maximum walking speed is `1m//s`. The river flows with velocity of `4 m//s`.
(a) Find the path which he should take to get to the point directly opposite to his starting point in the shortest time.
(b) Also, find the time which he takes to reach his destination.

Explanation

To solve the problem, we need to analyze the situation step by step.

### Given Data:
- Width of the river (d) = 120 m
- Speed of the boat in still water (v_b) = 3 m/s
- Speed of the river (v_r) = 4 m/s
- Walking speed of the man (v_w) = 1 m/s

### (a) Finding the Path to Take
1. **Understanding the Boat's Motion**:
- The boat's speed can be broken down into two components:
- \( v_{b_x} = v_b \cos \theta \) (across the river)
- \( v_{b_y} = v_b \sin \theta \) (downstream due to the river's current)

2. **Setting Up the Equations**:
- The man wants to reach the point directly opposite his starting point. This means that the downstream drift caused by the river must be countered by the upstream component of the boat's velocity.
- The effective downstream velocity due to the river and the boat's motion is:
\[ v_{effective} = v_r - v_{b_y} = 4 - 3 \sin \theta \]

3. **Time to Cross the River**:
- The time taken to cross the river can be expressed as:
\[ t_1 = \frac{d}{v_{b_x}} = \frac{120}{3 \cos \theta} \]

4. **Distance Drifted Downstream**:
- The distance drifted downstream while crossing is:
\[ x = (v_r - v_{b_y}) \cdot t_1 = (4 - 3 \sin \theta) \cdot \frac{120}{3 \cos \theta} \]

5. **Walking to the Destination**:
- After reaching the point downstream, the man will walk back to the opposite point with a speed of 1 m/s. The time taken to walk back is:
\[ t_2 = \frac{x}{v_w} = \frac{(4 - 3 \sin \theta) \cdot \frac{120}{3 \cos \theta}}{1} \]

6. **Total Time**:
- The total time \( T \) taken to reach the destination is:
\[ T = t_1 + t_2 = \frac{120}{3 \cos \theta} + (4 - 3 \sin \theta) \cdot \frac{120}{3 \cos \theta} \]
- Simplifying this gives:
\[ T = \frac{120}{3 \cos \theta} \left(1 + (4 - 3 \sin \theta)\right) = \frac{120(5 - 3 \sin \theta)}{3 \cos \theta} \]

7. **Minimizing Time**:
- To minimize time, take the derivative of \( T \) with respect to \( \theta \) and set it to zero:
\[ \frac{dT}{d\theta} = 0 \]
- This leads to finding \( \sin \theta = \frac{3}{5} \) and consequently \( \cos \theta = \frac{4}{5} \).

### (b) Finding the Time Taken
1. **Substituting Values**:
- Substitute \( \sin \theta \) and \( \cos \theta \) into the total time equation:
\[ T = \frac{120(5 - 3 \cdot \frac{3}{5})}{3 \cdot \frac{4}{5}} = \frac{120(5 - \frac{9}{5})}{\frac{12}{5}} = \frac{120 \cdot \frac{16}{5}}{\frac{12}{5}} = \frac{120 \cdot 16}{12} = 160 \text{ seconds} \]

### Final Answer:
- **Path**: The man should row at an angle \( \theta \) such that \( \sin \theta = \frac{3}{5} \) and \( \cos \theta = \frac{4}{5} \).
- **Time Taken**: 160 seconds or 2 minutes and 40 seconds.