In a circular motion of a particle the tangential acceleration of the particle is given by `a_(1)= 2m//s^(2)`. The radius of the circle described is 4 m. The particle is initially at rest. Time after which total acceleration of the particle makes `45^(@)` with radial acceleration is :

Options

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

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the Given Information
- Tangential acceleration, \( a_t = 2 \, \text{m/s}^2 \)
- Radius of the circular path, \( r = 4 \, \text{m} \)
- Initial velocity, \( u = 0 \, \text{m/s} \)

### Step 2: Determine the Condition for 45 Degrees
For the total acceleration \( a \) to make an angle of \( 45^\circ \) with the radial acceleration \( a_c \), the magnitudes of the tangential acceleration \( a_t \) and centripetal acceleration \( a_c \) must be equal:
\[ a_t = a_c \]

### Step 3: Express Centripetal Acceleration
Centripetal acceleration is given by:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the linear velocity of the particle.

### Step 4: Set Up the Equation
From the condition \( a_t = a_c \):
\[ 2 = \frac{v^2}{4} \]

### Step 5: Solve for Velocity \( v \)
Rearranging the equation gives:
\[ v^2 = 2 \times 4 = 8 \]
\[ v = \sqrt{8} = 2\sqrt{2} \, \text{m/s} \]

### Step 6: Use the Equation of Motion to Find Time
We will use the equation of motion to find the time \( t \):
\[ v = u + a_t \cdot t \]
Substituting the known values:
\[ 2\sqrt{2} = 0 + 2 \cdot t \]
\[ t = \frac{2\sqrt{2}}{2} = \sqrt{2} \, \text{s} \]

### Final Answer
The time after which the total acceleration of the particle makes \( 45^\circ \) with the radial acceleration is:
\[ \boxed{\sqrt{2} \, \text{s}} \]