A particle is moving on a circle of radius A .At an instant its speed and tangential acceleration are B and C respectively.Total acceleration of the particle at that instant is

Options

A. `sqrt(A^2/B^2 +C^2)`
B. `sqrt(A^4/B^2 +C^2)`
C. `sqrt(B^4/A^2 +C^2)`
D. `B^2/A +C`

Answer

To find the total acceleration of a particle moving in a circle, we need to consider both the tangential acceleration and the centripetal acceleration. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Radius of the circle, \( A \)
- Speed of the particle, \( B \)
- Tangential acceleration, \( C \)

### Step 2: Calculate the centripetal acceleration
Centripetal acceleration (\( a_c \)) is given by the formula:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the speed and \( r \) is the radius of the circle.

Substituting the values:
\[ a_c = \frac{B^2}{A} \]

### Step 3: Write the expression for total acceleration
The total acceleration (\( a \)) of the particle is the vector sum of the tangential acceleration and the centripetal acceleration. The relationship can be expressed as:
\[ a = \sqrt{a_t^2 + a_c^2} \]
where \( a_t \) is the tangential acceleration and \( a_c \) is the centripetal acceleration.

### Step 4: Substitute the values into the total acceleration formula
Substituting \( a_t = C \) and \( a_c = \frac{B^2}{A} \):
\[ a = \sqrt{C^2 + \left(\frac{B^2}{A}\right)^2} \]

### Step 5: Simplify the expression
Now, we can simplify the expression:
\[ a = \sqrt{C^2 + \frac{B^4}{A^2}} \]

### Final Answer
Thus, the total acceleration of the particle at that instant is:
\[ a = \sqrt{C^2 + \frac{B^4}{A^2}} \]