The velocity and acceleration vectors of a particle undergoing circular motion are `v = 2 hat(i) m//s and a = 2 hat(i)+4 hat(j)m//s^(2)` respectively at some instant of time. The radius of the circle is

Options

A. 1 m
B. 2 m
C. 3 m
D. 4 m

Answer

To find the radius of the circle for a particle undergoing circular motion with the given velocity and acceleration vectors, we can follow these steps:

### Step 1: Identify the given vectors
The velocity vector is given as:
\[ \mathbf{v} = 2 \hat{i} \, \text{m/s} \]
The acceleration vector is given as:
\[ \mathbf{a} = 2 \hat{i} + 4 \hat{j} \, \text{m/s}^2 \]

### Step 2: Determine the components of acceleration
From the acceleration vector, we can identify the components:
- The x-component of acceleration (tangential acceleration) is \( a_t = 2 \, \text{m/s}^2 \).
- The y-component of acceleration (radial acceleration) is \( a_r = 4 \, \text{m/s}^2 \).

### Step 3: Relate radial acceleration to radius
In circular motion, the radial acceleration \( a_r \) is given by the formula:
\[ a_r = \frac{v^2}{r} \]
where \( v \) is the magnitude of the velocity and \( r \) is the radius of the circle.

### Step 4: Calculate the magnitude of the velocity
The magnitude of the velocity vector is:
\[ v = | \mathbf{v} | = | 2 \hat{i} | = 2 \, \text{m/s} \]

### Step 5: Substitute values into the radial acceleration formula
Now we can substitute the known values into the radial acceleration formula:
\[ 4 = \frac{(2)^2}{r} \]

### Step 6: Solve for the radius \( r \)
Rearranging the equation gives:
\[ r = \frac{(2)^2}{4} = \frac{4}{4} = 1 \, \text{m} \]

### Conclusion
The radius of the circle is:
\[ r = 1 \, \text{m} \]