The velocity of a boat in still water is 10 m/s. If water flows in the river with a velocity of 6 m/s what is the difference in times taken to cross the river in the shortest path and the shortest time. The width of the river is 80 m.

Options

A. 1s
B. 10s
C. `(sqrt3)/(2)s`
D. 2 s

Answer

To solve the problem step by step, we need to determine the time taken to cross the river in two scenarios: the shortest path and the shortest time.

### Given Data:
- Velocity of the boat in still water, \( V_b = 10 \, \text{m/s} \)
- Velocity of the river, \( V_r = 6 \, \text{m/s} \)
- Width of the river, \( d = 80 \, \text{m} \)

### Step 1: Calculate the angle \( \theta \)
To find the angle \( \theta \) that the boat needs to make with respect to the riverbank to cross the river in the shortest path, we can use the relationship between the components of the boat's velocity.

The horizontal component of the boat's velocity must equal the velocity of the river:
\[ V_r = V_b \cos(\theta) \]
Substituting the known values:
\[ 6 = 10 \cos(\theta) \]
\[ \cos(\theta) = \frac{6}{10} = \frac{3}{5} \]

### Step 2: Calculate \( \sin(\theta) \)
Using the Pythagorean identity:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Substituting \( \cos(\theta) \):
\[ \sin^2(\theta) + \left(\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2(\theta) + \frac{9}{25} = 1 \]
\[ \sin^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \sin(\theta) = \frac{4}{5} \]

### Step 3: Calculate the time taken to cross the river in the shortest path \( T_s \)
The time taken to cross the river along the shortest path can be calculated using:
\[ T_s = \frac{d}{V_b \sin(\theta)} \]
Substituting the values:
\[ T_s = \frac{80}{10 \cdot \frac{4}{5}} = \frac{80}{8} = 10 \, \text{s} \]

### Step 4: Calculate the time taken to cross the river in the shortest time \( T_t \)
For the shortest time, the boat travels directly across the river:
\[ T_t = \frac{d}{V_b} \]
Substituting the values:
\[ T_t = \frac{80}{10} = 8 \, \text{s} \]

### Step 5: Calculate the difference in time
The difference in time taken to cross the river in the shortest path and the shortest time is:
\[ \Delta T = T_s - T_t = 10 \, \text{s} - 8 \, \text{s} = 2 \, \text{s} \]

### Final Answer:
The difference in times taken to cross the river in the shortest path and the shortest time is \( \Delta T = 2 \, \text{s} \).

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