The distance of a particle moving on a circle of radius 12 m measured from a fixed point on the circle and measured along the circle is given by `s=2t^(3)` (in meters). The ratio of its tangential to centripetal acceleration at t = 2s is

Options

A. `1:1`
B. `1:2`
C. `2:1`
D. `3:1`

Explanation

To solve the problem, we need to find the ratio of tangential acceleration (\(a_t\)) to centripetal acceleration (\(a_c\)) at \(t = 2\) seconds for a particle moving along a circular path of radius \(r = 12\) m, where the distance \(s\) traveled along the circle is given by the equation \(s = 2t^3\).

### Step-by-Step Solution:

1. **Find the expression for tangential acceleration (\(a_t\))**:
- The tangential acceleration is defined as the rate of change of tangential velocity:
\[ a_t = \frac{dv}{dt} \]
- The tangential velocity \(v\) can be found by differentiating \(s\) with respect to \(t\):
\[ v = \frac{ds}{dt} = \frac{d(2t^3)}{dt} = 6t^2 \]
- Now, differentiate \(v\) to find \(a_t\):
\[ a_t = \frac{dv}{dt} = \frac{d(6t^2)}{dt} = 12t \]

2. **Calculate \(a_t\) at \(t = 2\) seconds**:
- Substitute \(t = 2\) into the expression for \(a_t\):
\[ a_t = 12 \times 2 = 24 \, \text{m/s}^2 \]

3. **Find the expression for centripetal acceleration (\(a_c\))**:
- The centripetal acceleration is given by the formula:
\[ a_c = \frac{v^2}{r} \]
- We already found \(v = 6t^2\). Now, substitute \(t = 2\) to find \(v\):
\[ v = 6 \times (2^2) = 6 \times 4 = 24 \, \text{m/s} \]
- Now, calculate \(a_c\):
\[ a_c = \frac{(24)^2}{12} = \frac{576}{12} = 48 \, \text{m/s}^2 \]

4. **Calculate the ratio of tangential to centripetal acceleration**:
- Now that we have both accelerations, we can find the ratio:
\[ \frac{a_t}{a_c} = \frac{24}{48} = \frac{1}{2} \]

### Final Answer:
The ratio of tangential to centripetal acceleration at \(t = 2\) seconds is \(1:2\).

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