A car is going east wards with a velocity of 8m/s. To the passengers in the car, a train appears to be moving north wards with a velocity of 15 m/s. What is the actual velocity of the train:

Options

A. 7 m/s
B. 17 m/s
C. 23 m/s
D. None of these

Answer

To find the actual velocity of the train, we can use the concept of relative velocity. The car is moving east with a velocity of 8 m/s, and to the passengers in the car, the train appears to be moving north with a velocity of 15 m/s. We can visualize this situation using vector components.

### Step-by-Step Solution:

1. **Define the Velocity Vectors**:
- Let the velocity of the car \( \vec{V}_{car} = 8 \hat{i} \) m/s (east direction).
- Let the velocity of the train \( \vec{V}_{train} \) be represented as \( V_x \hat{i} + V_y \hat{j} \) m/s, where \( V_x \) is the east-west component and \( V_y \) is the north-south component.

2. **Relative Velocity**:
- The relative velocity of the train with respect to the car is given by:
\[ \vec{V}_{relative} = \vec{V}_{train} - \vec{V}_{car} \]
- Since the passengers see the train moving north at 15 m/s, we can write:
\[ \vec{V}_{relative} = 0 \hat{i} + 15 \hat{j} \text{ m/s} \]

3. **Setting Up the Equation**:
- From the relative velocity equation, we have:
\[ V_x \hat{i} + V_y \hat{j} - 8 \hat{i} = 0 \hat{i} + 15 \hat{j} \]
- This gives us two equations:
- For the \( \hat{i} \) component: \( V_x - 8 = 0 \) → \( V_x = 8 \) m/s
- For the \( \hat{j} \) component: \( V_y = 15 \) m/s

4. **Calculating the Magnitude of the Train's Velocity**:
- Now we can find the magnitude of the train's velocity using the Pythagorean theorem:
\[ V_{train} = \sqrt{V_x^2 + V_y^2} = \sqrt{(8)^2 + (15)^2} \]
\[ V_{train} = \sqrt{64 + 225} = \sqrt{289} = 17 \text{ m/s} \]

5. **Direction of the Train's Velocity**:
- The direction (angle \( \theta \)) of the train's velocity can be calculated using:
\[ \tan(\theta) = \frac{V_y}{V_x} = \frac{15}{8} \]
- This gives us the angle of the train's velocity with respect to the east direction.

### Final Answer:
The actual velocity of the train is **17 m/s** at an angle \( \theta \) where \( \tan(\theta) = \frac{15}{8} \).

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