A train is moving north with speed 20 m s. If it turns west with same speed

Options

A. The resultant velocity makes an angle of 45° in the south of west direction.
B. The change in velocity will be `20sqrt(2) ms^(-1)`
C. The resultant velocity makes an angle of 60° in the south of west direction.
D. The change in velocity is `10sqrt(2) ms^(-1)`.

Explanation

To solve the problem, we need to analyze the motion of the train as it changes direction. Here's a step-by-step breakdown of the solution:

### Step 1: Identify Initial and Final Velocities
- The train is initially moving north with a speed of \( v_1 = 20 \, \text{m/s} \).
- After turning, the train moves west with a speed of \( v_2 = 20 \, \text{m/s} \).

### Step 2: Represent Velocities as Vectors
- We can represent the initial velocity \( v_1 \) in vector form:
\[ \vec{v_1} = 20 \, \hat{j} \, \text{m/s} \quad \text{(north direction)} \]
- The final velocity \( v_2 \) can be represented as:
\[ \vec{v_2} = -20 \, \hat{i} \, \text{m/s} \quad \text{(west direction)} \]

### Step 3: Calculate Change in Velocity
- The change in velocity \( \Delta \vec{v} \) is given by:
\[ \Delta \vec{v} = \vec{v_2} - \vec{v_1} \]
- Substituting the values:
\[ \Delta \vec{v} = (-20 \, \hat{i}) - (20 \, \hat{j}) = -20 \, \hat{i} - 20 \, \hat{j} \]

### Step 4: Calculate the Magnitude of Change in Velocity
- The magnitude of the change in velocity is calculated using the Pythagorean theorem:
\[ |\Delta \vec{v}| = \sqrt{(-20)^2 + (-20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, \text{m/s} \]

### Step 5: Determine the Direction of the Resultant Velocity
- The resultant velocity vector \( \vec{R} \) can be found using the same components:
\[ \vec{R} = \Delta \vec{v} = -20 \, \hat{i} - 20 \, \hat{j} \]
- To find the angle \( \theta \) that this vector makes with the south-west direction, we can use the tangent function:
\[ \tan(\theta) = \frac{|\text{opposite}|}{|\text{adjacent}|} = \frac{20}{20} = 1 \]
- Therefore, \( \theta = \tan^{-1}(1) = 45^\circ \).

### Conclusion
- The resultant velocity makes an angle of \( 45^\circ \) in the south-west direction.
- The magnitude of the change in velocity is \( 20\sqrt{2} \, \text{m/s} \).

### Final Answers
- The resultant velocity makes an angle of \( 45^\circ \) in the south-west direction.
- The change in velocity is \( 20\sqrt{2} \, \text{m/s} \).