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. (a)`1/3 s`
B. (b)`2/3s`
C. (c)`5/3s`
D. (d)`4/3s`

Answer

To solve the problem, we need to find the time after which the total acceleration of the particle makes an angle of \(45^\circ\) with the radial acceleration. Here are the steps to arrive at the solution:

### Step 1: Understand the relationship between tangential and radial acceleration
We know that the total acceleration \(A\) of a particle in circular motion can be resolved into two components: tangential acceleration \(a_t\) and radial (centripetal) acceleration \(a_r\). The angle \(\theta\) between the total acceleration and the radial acceleration is given by:

\[ \tan(\theta) = \frac{a_t}{a_r} \]

### Step 2: Set up the equation for the angle
Given that the angle \(\theta = 45^\circ\), we have:

\[ \tan(45^\circ) = 1 \implies a_t = a_r \]

### Step 3: Calculate the radial acceleration
The radial acceleration \(a_r\) is given by the formula:

\[ a_r = \frac{v^2}{r} \]

where \(v\) is the velocity of the particle and \(r\) is the radius of the circular path. We know from the problem that \(r = 4 \, \text{m}\).

### Step 4: Relate tangential acceleration to velocity
We are given that the tangential acceleration \(a_t = 9 \, \text{m/s}^2\). Setting \(a_t\) equal to \(a_r\) gives us:

\[ 9 = \frac{v^2}{4} \]

### Step 5: Solve for velocity \(v\)
From the equation above, we can solve for \(v\):

\[ v^2 = 9 \times 4 = 36 \implies v = \sqrt{36} = 6 \, \text{m/s} \]

### Step 6: Relate tangential acceleration to time
The tangential acceleration is also defined as the rate of change of velocity:

\[ a_t = \frac{dv}{dt} \]

Substituting the known value of \(a_t\):

\[ 9 = \frac{dv}{dt} \]

### Step 7: Integrate to find time \(t\)
To find the time \(t\) taken to reach the velocity of \(6 \, \text{m/s}\):

\[ \int_0^6 dv = \int_0^t 9 \, dt \]

This gives us:

\[ 6 - 0 = 9t - 0 \implies 6 = 9t \]

### Step 8: Solve for \(t\)
Now, solving for \(t\):

\[ t = \frac{6}{9} = \frac{2}{3} \, \text{s} \]

### Conclusion
The time after which the total acceleration of the particle makes an angle of \(45^\circ\) with the radial acceleration is:

\[ \boxed{\frac{2}{3} \, \text{s}} \]