Ram Shyam are walking on two perpendicular tracks with speed `3ms^(-1) and 4 ms^(-1)`, respectively. At a certain moment (say t = 0s), Ram and Shyam are at 20 m and 40 m away from the intersection of tracks, respectively, and moving towards the intersection of the tracks.
The time t when they are at shortest distance from each other subsequently is -

Options

A. 8.8 s
B. 12 s
C. 15 s
D. 44 s

Explanation

To solve the problem, we need to determine the time at which Ram and Shyam are at the shortest distance from each other while walking towards an intersection on perpendicular tracks.

### Step-by-Step Solution:

1. **Identify Initial Positions and Velocities**:
- Ram is initially 20 m away from the intersection and moves towards it with a speed of \(3 \, \text{m/s}\).
- Shyam is initially 40 m away from the intersection and moves towards it with a speed of \(4 \, \text{m/s}\).

2. **Set Up the Coordinate System**:
- Let the intersection point be the origin (0, 0).
- Ram's position at time \(t\) seconds is given by \( (20 - 3t, 0) \).
- Shyam's position at time \(t\) seconds is given by \( (0, 40 - 4t) \).

3. **Calculate the Distance Between Ram and Shyam**:
- The distance \(D\) between Ram and Shyam at time \(t\) is given by the distance formula:
\[ D(t) = \sqrt{(20 - 3t)^2 + (40 - 4t)^2} \]

4. **Simplify the Distance Equation**:
- Expanding the equation:
\[ D(t) = \sqrt{(20 - 3t)^2 + (40 - 4t)^2} = \sqrt{(400 - 120t + 9t^2) + (1600 - 320t + 16t^2)} \]
\[ D(t) = \sqrt{25t^2 - 440t + 2000} \]

5. **Find the Minimum Distance**:
- To find the time \(t\) when the distance \(D(t)\) is minimized, we can minimize the square of the distance \(D^2(t)\):
\[ D^2(t) = 25t^2 - 440t + 2000 \]
- Differentiate \(D^2(t)\) with respect to \(t\) and set the derivative to zero:
\[ \frac{d(D^2)}{dt} = 50t - 440 = 0 \]
\[ 50t = 440 \implies t = \frac{440}{50} = 8.8 \, \text{s} \]

6. **Conclusion**:
- The time \(t\) when Ram and Shyam are at the shortest distance from each other is \(8.8\) seconds.