A train is running at a constant speed of 90 km/h on a straight track. A person standing at the top of a boggey moves in the direction of motion of the train such that he covers 1 meters on the train each second. The speed of the person with respect to ground to -

Options

A. 25 m/s
B. 91 km/s
C. 26 km/h
D. 26 m/s

Answer

To solve the problem, we need to find the speed of the person with respect to the ground. We will follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- Speed of the train (\( V_t \)) = 90 km/h
- Speed of the person on the train (\( V_m \)) = 1 m/s

2. **Convert the speed of the train from km/h to m/s:**
\[ V_t = 90 \text{ km/h} = \frac{90 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{90000}{3600} = 25 \text{ m/s} \]

3. **Determine the speed of the person with respect to the ground:**
The speed of the person with respect to the ground (\( V_{mg} \)) can be found by adding the speed of the person with respect to the train and the speed of the train:
\[ V_{mg} = V_m + V_t \]
Substituting the values we have:
\[ V_{mg} = 1 \text{ m/s} + 25 \text{ m/s} = 26 \text{ m/s} \]

4. **Final Answer:**
The speed of the person with respect to the ground is \( 26 \text{ m/s} \).