On a long horizontally moving belt, a child runs to and fro with a speed 9 km h`""^(-1)` (with respect to the belt) between his father and mother located 50 m apart. The belt moves with speed of 4 km h`""^(-1)`. For an observer on a stationary platform, the speed of the child runing in the direction of motion of the belt is

Options

A. `4km h^(-1)`
B. `5km h^(-1)`
C. `9 km h^(-1)`
D. `13 km h^(-1)`

Answer

To solve the problem, we need to determine the speed of the child running on a moving belt as observed from a stationary platform. Here’s how we can break it down step by step:

### Step 1: Understand the given information
- The speed of the child with respect to the belt is \( V_{child/belt} = 9 \) km/h.
- The speed of the belt with respect to the ground is \( V_{belt/ground} = 4 \) km/h.
- The child runs back and forth between his father and mother, who are 50 m apart, but this distance does not affect the speed calculation.

### Step 2: Determine the speed of the child with respect to the ground
To find the speed of the child with respect to a stationary observer (the ground), we need to add the speed of the belt to the speed of the child with respect to the belt. Since the child is running in the same direction as the belt, we can use the following formula:

\[ V_{child/ground} = V_{child/belt} + V_{belt/ground} \]

### Step 3: Substitute the values
Now, we substitute the given values into the equation:

\[ V_{child/ground} = 9 \text{ km/h} + 4 \text{ km/h} \]

### Step 4: Perform the addition
Now, we perform the addition:

\[ V_{child/ground} = 13 \text{ km/h} \]

### Conclusion
Thus, the speed of the child running in the direction of motion of the belt, as observed from a stationary platform, is **13 km/h**.

### Final Answer
The correct option is **4 (13 km/h)**.

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