The river 500 m wide is flowing with a current of 4kph. A boat starts from one bank of the river wishes to cross the river at right angle to stream direction. Boatman can row the boat at 8 kph. In which direction he should row the boat. What time he ''ll take to cross the river ?

Explanation

To solve the problem step by step, we will analyze the situation involving the river, the boat, and the velocities involved.

### Step 1: Understand the Problem
We have a river that is 500 meters wide, flowing with a current of 4 km/h. A boat can row at a speed of 8 km/h. The goal is to determine the angle at which the boat should row to cross the river directly (perpendicular to the current) and the time it will take to cross.

### Step 2: Convert Units
First, we need to ensure all units are consistent. The width of the river is given in meters, and the speeds are in kilometers per hour. We can convert the width of the river from meters to kilometers:

\[ 500 \text{ meters} = 0.5 \text{ kilometers} \]

### Step 3: Set Up the Velocity Components
Let:
- \( V_{bg} \) = velocity of the boat with respect to the ground = 8 km/h
- \( V_r \) = velocity of the river = 4 km/h

To cross the river directly, the boat must have a component of its velocity that cancels out the current of the river.

### Step 4: Use Trigonometry
Let \( \theta \) be the angle at which the boat should row relative to the direction of the current. The velocity of the boat can be broken down into two components:
- Perpendicular to the river (across the river): \( V_{bg} \sin(\theta) \)
- Parallel to the river (downstream): \( V_{bg} \cos(\theta) \)

To ensure that the boat crosses directly, the downstream component must equal the river's current:

\[ V_{bg} \cos(\theta) = V_r \]

Substituting the known values:

\[ 8 \cos(\theta) = 4 \]

### Step 5: Solve for \( \theta \)
Rearranging the equation gives:

\[ \cos(\theta) = \frac{4}{8} = \frac{1}{2} \]

From this, we find:

\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Step 6: Calculate the Time to Cross the River
Now, we need to calculate the time taken to cross the river. The effective speed of the boat across the river is given by:

\[ V_{bg} \sin(\theta) = 8 \sin(60^\circ) \]

Using \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):

\[ V_{bg} \sin(60^\circ) = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3} \text{ km/h} \]

Now, we can calculate the time taken to cross the river:

\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{0.5 \text{ km}}{4\sqrt{3} \text{ km/h}} \]

Calculating this gives:

\[ \text{Time} = \frac{0.5}{4\sqrt{3}} \approx \frac{0.5}{6.928} \approx 0.072 \text{ hours} \]

To convert this into seconds:

\[ 0.072 \text{ hours} \times 3600 \text{ seconds/hour} \approx 259.2 \text{ seconds} \]

### Final Answers
- The boat should row at an angle of \( 60^\circ \) upstream relative to the current.
- The time taken to cross the river is approximately \( 259.2 \) seconds.