Twp boats are moving along perpendicular paths on a still take at right. One boat moves with a speed of `3 ms^(-1)` and the other boat moves with a speed of `4 ms^(-1)` in the directions such that they collide after some time. At t=0, the boats are 300 m apart. Two boats will collide after time ______.

Options

A. 60 s
B. 30 s
C. 120 s
D. Data is insufficiant

Explanation

To solve the problem, we need to determine the time it takes for two boats moving along perpendicular paths to collide, given their speeds and the initial distance between them.

### Step-by-step Solution:

1. **Identify the Variables**:
- Speed of Boat A, \( v_A = 3 \, \text{m/s} \)
- Speed of Boat B, \( v_B = 4 \, \text{m/s} \)
- Initial distance between the boats, \( d = 300 \, \text{m} \)

2. **Set Up the Problem**:
- Let the time taken for the boats to collide be \( t \).
- The distance covered by Boat A in time \( t \) is \( d_A = v_A \cdot t = 3t \).
- The distance covered by Boat B in time \( t \) is \( d_B = v_B \cdot t = 4t \).

3. **Use the Pythagorean Theorem**:
- Since the boats move along perpendicular paths, we can use the Pythagorean theorem to relate the distances:
\[ d^2 = d_A^2 + d_B^2 \]
Substituting the values we have:
\[ 300^2 = (3t)^2 + (4t)^2 \]

4. **Calculate the Squares**:
- Calculate \( 300^2 \):
\[ 300^2 = 90000 \]
- Calculate \( (3t)^2 \) and \( (4t)^2 \):
\[ (3t)^2 = 9t^2 \]
\[ (4t)^2 = 16t^2 \]

5. **Combine the Equations**:
- Substitute back into the equation:
\[ 90000 = 9t^2 + 16t^2 \]
\[ 90000 = 25t^2 \]

6. **Solve for \( t^2 \)**:
- Rearranging gives:
\[ t^2 = \frac{90000}{25} \]
\[ t^2 = 3600 \]

7. **Find \( t \)**:
- Taking the square root:
\[ t = \sqrt{3600} = 60 \, \text{s} \]

Thus, the two boats will collide after **60 seconds**.