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 after a specific time. Let's break this down step by step.

### Step 1: Understand the given parameters
The particle is moving in a circular path of radius \( R \) with an angular velocity defined as:
\[ \omega = a - bt \]
where \( a \) and \( b \) are positive constants, and \( t \) is time.

### Step 2: Calculate the angular velocity at time \( t = \frac{2a}{b} \)
Substituting \( t = \frac{2a}{b} \) into the angular velocity equation:
\[ \omega = a - b\left(\frac{2a}{b}\right) = a - 2a = -a \]
Thus, at \( t = \frac{2a}{b} \), the angular velocity \( \omega \) is \( -a \).

### Step 3: Calculate centripetal acceleration \( A_C \)
Centripetal acceleration is given by the formula:
\[ A_C = \omega^2 R \]
Substituting \( \omega = -a \):
\[ A_C = (-a)^2 R = a^2 R \]

### Step 4: Calculate angular acceleration \( \alpha \)
Angular acceleration is defined as the time derivative of angular velocity:
\[ \alpha = \frac{d\omega}{dt} \]
Differentiating \( \omega = a - bt \):
\[ \alpha = 0 - b = -b \]

### Step 5: Calculate tangential acceleration \( A_T \)
Tangential acceleration is given by:
\[ A_T = \alpha R \]
Substituting \( \alpha = -b \):
\[ A_T = -b R \]

### Step 6: Calculate the total acceleration
The total acceleration \( A \) is the vector sum of centripetal and tangential accelerations. Since these two accelerations are perpendicular to each other, we can use the Pythagorean theorem:
\[ A = \sqrt{A_T^2 + A_C^2} \]
Substituting the values we calculated:
\[ A = \sqrt{(-b R)^2 + (a^2 R)^2} \]
\[ A = \sqrt{b^2 R^2 + a^4 R^2} \]
Factoring out \( R^2 \):
\[ A = R \sqrt{b^2 + a^4} \]

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