A train is moving due East and a car is moving due North, both with the same speed `30 km h^(-1)`. What is the observed speed and diredction of motion of car to the passsenger in the train ?

Options

A. East - North
B. South - East
C. West - North
D. North - South

Answer

To solve the problem of finding the observed speed and direction of the car from the perspective of a passenger in the train, we can follow these steps:

### Step 1: Define the velocities
- The velocity of the train (V_train) is moving due East at 30 km/h. We can represent this as a vector:
\[ V_{\text{train}} = 30 \hat{i} \text{ km/h} \]
- The velocity of the car (V_car) is moving due North at 30 km/h. We can represent this as a vector:
\[ V_{\text{car}} = 30 \hat{j} \text{ km/h} \]

### Step 2: Calculate the relative velocity of the car with respect to the train
- The relative velocity of the car with respect to the train (V_car/train) is given by:
\[ V_{\text{car/train}} = V_{\text{car}} - V_{\text{train}} \]
- Substituting the values:
\[ V_{\text{car/train}} = 30 \hat{j} - 30 \hat{i} = -30 \hat{i} + 30 \hat{j} \]

### Step 3: Calculate the magnitude of the relative velocity
- The magnitude of the relative velocity can be calculated using the Pythagorean theorem:
\[ |V_{\text{car/train}}| = \sqrt{(-30)^2 + (30)^2} \]
- Simplifying:
\[ |V_{\text{car/train}}| = \sqrt{900 + 900} = \sqrt{1800} = 30\sqrt{2} \approx 42.426 \text{ km/h} \]

### Step 4: Calculate the direction of the relative velocity
- To find the direction, we can use the tangent function:
\[ \tan(\theta) = \frac{V_y}{-V_x} = \frac{30}{-30} = -1 \]
- This implies:
\[ \theta = 135^\circ \]
- The angle is measured from the negative x-axis (West) towards the positive y-axis (North), indicating that the direction is towards the North-West.

### Conclusion
- The observed speed of the car from the passenger's perspective in the train is approximately **42.426 km/h**, and the direction is **135 degrees**, which is towards the North-West.