If the velocity (in `ms^(-1)` ) of a particle is given by , `4.0 hati + 5.0 t hatj` then the magnitude of its acceleration (in `m s^(-2`) is

Options

A. 4
B. 9
C. 1
D. 5

Explanation

To find the magnitude of the acceleration of a particle whose velocity is given by the vector equation \( \vec{v} = 4.0 \hat{i} + 5.0 \hat{j} \), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the velocity vector**:
The velocity of the particle is given as:
\[ \vec{v} = 4.0 \hat{i} + 5.0 \hat{j} \]
Here, \( 4.0 \hat{i} \) is the component in the x-direction and \( 5.0 \hat{j} \) is the component in the y-direction.

2. **Differentiate the velocity to find acceleration**:
The acceleration \( \vec{a} \) is the time derivative of the velocity \( \vec{v} \):
\[ \vec{a} = \frac{d\vec{v}}{dt} \]
Since the velocity vector has a constant component in the x-direction and a time-dependent component in the y-direction, we differentiate each component:
- The x-component \( 4.0 \hat{i} \) is constant, so its derivative is:
\[ \frac{d(4.0 \hat{i})}{dt} = 0 \hat{i} \]
- The y-component \( 5.0 \hat{j} \) does not depend on time, so its derivative is:
\[ \frac{d(5.0 \hat{j})}{dt} = 0 \hat{j} \]

Therefore, the acceleration vector becomes:
\[ \vec{a} = 0 \hat{i} + 0 \hat{j} \]

3. **Magnitude of the acceleration**:
The magnitude of the acceleration \( |\vec{a}| \) can be calculated using the formula:
\[ |\vec{a}| = \sqrt{(a_x)^2 + (a_y)^2 + (a_z)^2} \]
In our case:
- \( a_x = 0 \)
- \( a_y = 0 \)
- There is no z-component, so \( a_z = 0 \)

Thus, the magnitude of the acceleration is:
\[ |\vec{a}| = \sqrt{(0)^2 + (0)^2 + (0)^2} = \sqrt{0} = 0 \]

### Final Answer:
The magnitude of the acceleration is \( 0 \, \text{m/s}^2 \).