The speed of a particle moving in a circle slows down at a rate of `3 m//sec^(2)`. At some instant the magnitude of the total acceleration is `5 m/sec^(2)` and the particle's speed is 12 m/sec. The radius of circle will be :

Options

A. 12 m
B. 24 m
C. 36 m
D. 48 m

Answer

To find the radius of the circle in which a particle is moving, we can follow these steps:

### Step 1: Identify the given values
- The tangential acceleration, \( a_t = -3 \, \text{m/s}^2 \) (negative because the speed is decreasing).
- The total acceleration, \( a = 5 \, \text{m/s}^2 \).
- The speed of the particle, \( v = 12 \, \text{m/s} \).

### Step 2: Use the relationship between total acceleration, tangential acceleration, and centripetal acceleration
The total acceleration \( a \) can be expressed as:
\[ a = \sqrt{a_t^2 + a_c^2} \]
where \( a_c \) is the centripetal acceleration.

### Step 3: Substitute the known values into the equation
We know:
\[ a_t = -3 \, \text{m/s}^2 \quad \text{and} \quad a = 5 \, \text{m/s}^2 \]
Thus, substituting these values gives:
\[ 5 = \sqrt{(-3)^2 + a_c^2} \]
This simplifies to:
\[ 5 = \sqrt{9 + a_c^2} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides results in:
\[ 25 = 9 + a_c^2 \]

### Step 5: Solve for centripetal acceleration \( a_c \)
Rearranging the equation gives:
\[ a_c^2 = 25 - 9 = 16 \]
Taking the square root:
\[ a_c = 4 \, \text{m/s}^2 \]

### Step 6: Relate centripetal acceleration to radius
Centripetal acceleration is given by the formula:
\[ a_c = \frac{v^2}{r} \]
Substituting the known values:
\[ 4 = \frac{12^2}{r} \]
This simplifies to:
\[ 4 = \frac{144}{r} \]

### Step 7: Solve for the radius \( r \)
Rearranging gives:
\[ r = \frac{144}{4} = 36 \, \text{m} \]

### Final Answer
The radius of the circle is \( r = 36 \, \text{m} \).

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