A particle moves in a circular path of radius R with an angualr velocity `omega=a-bt`, where a and b are positive constants and t is time. The magnitude of the acceleration of the particle after time `(2a)/(b)` is

Options

A. `(a)/(R)`
B. `a^(2)R`
C. `R(a^(2)+b)`
D. `Rsqrt(a^(4)+b^(2))`

Explanation

To solve the problem, we need to find the magnitude of the acceleration of a particle moving in a circular path with a given angular velocity. The angular velocity is defined as \(\omega = a - bt\), where \(a\) and \(b\) are positive constants.

### Step-by-Step Solution:

1. **Determine Angular Acceleration (\(\alpha\))**:
The angular acceleration \(\alpha\) is the time derivative of angular velocity \(\omega\):
\[ \alpha = \frac{d\omega}{dt} = \frac{d}{dt}(a - bt) = -b \]

2. **Calculate the Tangential Acceleration (\(a_t\))**:
The tangential acceleration is given by:
\[ a_t = R \alpha = R(-b) = -Rb \]
The magnitude of the tangential acceleration is:
\[ |a_t| = Rb \]

3. **Calculate the Angular Velocity at \(t = \frac{2a}{b}\)**:
Substitute \(t = \frac{2a}{b}\) into the expression for \(\omega\):
\[ \omega = a - b\left(\frac{2a}{b}\right) = a - 2a = -a \]
The magnitude of the angular velocity is:
\[ |\omega| = a \]

4. **Calculate the Centripetal Acceleration (\(a_c\))**:
The centripetal acceleration is given by:
\[ a_c = R \omega^2 = R(-a)^2 = Ra^2 \]

5. **Calculate the Total Acceleration (\(a\))**:
The total acceleration \(a\) is the vector sum of tangential and centripetal accelerations. Since they are perpendicular to each other, we can use the Pythagorean theorem:
\[ |a| = \sqrt{a_t^2 + a_c^2} = \sqrt{(Rb)^2 + (Ra^2)^2} \]
\[ |a| = R\sqrt{b^2 + a^4} \]

6. **Final Expression**:
Thus, the magnitude of the acceleration of the particle after time \(t = \frac{2a}{b}\) is:
\[ |a| = R\sqrt{b^2 + a^4} \]

### Answer:
The magnitude of the acceleration of the particle after time \(\frac{2a}{b}\) is \(R\sqrt{b^2 + a^4}\).