In a circular motion of a particle , the tangential acceleration of the particle is given by `a_t = 9 m s^(-2)` . The radius of the circle is 4m . The particle was initially at rest. Time after which total acceleration of the particle makes an angle of `45^@` with the radial acceleration is

Options

A. `1/3 s`
B. `2/3s`
C. `5/3s`
D. `4/3s`

Answer

To solve the problem step-by-step, we will analyze the circular motion of the particle and use the information given to find the time after which the total acceleration makes an angle of 45 degrees with the radial acceleration.

### Step 1: Understand the Components of Acceleration
In circular motion, the total acceleration \( A \) of a particle can be broken down into two components:
- **Tangential Acceleration \( a_t \)**: This is the acceleration along the tangent to the circular path, which is given as \( a_t = 9 \, \text{m/s}^2 \).
- **Radial (Centripetal) Acceleration \( a_r \)**: This is directed towards the center of the circular path and is given by the formula:
\[ a_r = \frac{v^2}{r} \]
where \( v \) is the tangential velocity and \( r \) is the radius of the circular path.

### Step 2: Set Up the Condition for the Angle
We are given that the total acceleration makes an angle of \( 45^\circ \) with the radial acceleration. This means:
\[ \tan(45^\circ) = 1 = \frac{a_t}{a_r} \]
Thus, we can write:
\[ a_t = a_r \]

### Step 3: Calculate the Radial Acceleration
Substituting the expression for radial acceleration:
\[ a_t = \frac{v^2}{r} \]
Given \( a_t = 9 \, \text{m/s}^2 \) and \( r = 4 \, \text{m} \), we have:
\[ 9 = \frac{v^2}{4} \]
Multiplying both sides by 4 gives:
\[ v^2 = 36 \]
Taking the square root, we find:
\[ v = 6 \, \text{m/s} \]

### Step 4: Relate Velocity to Time
The tangential acceleration is constant, and it is related to the change in velocity over time:
\[ a_t = \frac{dv}{dt} \]
Substituting \( a_t = 9 \, \text{m/s}^2 \):
\[ 9 = \frac{dv}{dt} \]
Integrating this from the initial velocity (which is 0, since the particle starts from rest) to the final velocity \( v \):
\[ \int_0^v dv = \int_0^t 9 \, dt \]
This gives:
\[ v = 9t \]
Substituting \( v = 6 \, \text{m/s} \):
\[ 6 = 9t \]
Solving for \( t \):
\[ t = \frac{6}{9} = \frac{2}{3} \, \text{s} \]

### Step 5: Conclusion
The time after which the total acceleration of the particle makes an angle of \( 45^\circ \) with the radial acceleration is:
\[ \boxed{\frac{2}{3} \, \text{s}} \]