A particle is moving at `5ms^(-1)` towards east In one second its velocity changes to `5ms^(-1)` towards north. Assuming the acceleration to be uniform, the change in velocity will be directed at

Options

A. `135^(@)` to east
B. `135^(@)` to north
C. `45^(@)` to east
D. `90^(@)` to east

Explanation

To solve the problem, we need to determine the direction of the change in velocity of the particle as it moves from an initial velocity towards the east to a final velocity towards the north. Here’s a step-by-step solution:

### Step 1: Understand the Initial and Final Velocities
- The initial velocity \( \vec{v_i} \) is \( 5 \, \text{m/s} \) towards the east.
- The final velocity \( \vec{v_f} \) is \( 5 \, \text{m/s} \) towards the north.

### Step 2: Represent the Velocities as Vectors
- We can represent the initial velocity vector \( \vec{v_i} \) as:
\[ \vec{v_i} = 5 \hat{i} \quad \text{(where } \hat{i} \text{ is the unit vector in the east direction)} \]
- The final velocity vector \( \vec{v_f} \) can be represented as:
\[ \vec{v_f} = 5 \hat{j} \quad \text{(where } \hat{j} \text{ is the unit vector in the north direction)} \]

### Step 3: Calculate the Change in Velocity
- The change in velocity \( \Delta \vec{v} \) is given by:
\[ \Delta \vec{v} = \vec{v_f} - \vec{v_i} \]
- Substituting the values:
\[ \Delta \vec{v} = 5 \hat{j} - 5 \hat{i} = -5 \hat{i} + 5 \hat{j} \]

### Step 4: Determine the Direction of the Change in Velocity
- The change in velocity vector \( \Delta \vec{v} = -5 \hat{i} + 5 \hat{j} \) can be visualized as a vector pointing from the origin (0,0) to the point (-5,5) in the Cartesian plane.
- This forms a right triangle where:
- The horizontal leg is \( -5 \) (westward)
- The vertical leg is \( 5 \) (northward)

### Step 5: Calculate the Angle of the Change in Velocity
- To find the angle \( \theta \) that \( \Delta \vec{v} \) makes with the east direction (positive x-axis), we can use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{5} = 1 \]
- Therefore, \( \theta = 45^\circ \).

### Step 6: Determine the Direction Relative to East
- Since the change in velocity is directed towards the northwest (as it is moving from east to north), we find the angle relative to the east:
\[ \text{Angle from east} = 180^\circ - 45^\circ = 135^\circ \]

### Conclusion
The change in velocity is directed at \( 135^\circ \) to the east.