For a particle moving along circular path, the radial acceleration `a_x` is proportional to time t. If `a_t` is the tangential acceleration, then which of the following will be independent of time t.

Options

A. `a_t`
B. `a_r.a_t`
C. `a_r/a_t`
D. `a_r(a_t)^2`

Explanation

To solve the problem, we need to analyze the relationship between the radial acceleration \( a_r \) and the tangential acceleration \( a_t \) of a particle moving along a circular path, given that the radial acceleration is proportional to time \( t \).

### Step-by-step Solution:

1. **Understanding Radial Acceleration**:
- The radial acceleration \( a_r \) is given to be proportional to time \( t \). We can express this as:
\[ a_r = k \cdot t \]
where \( k \) is a constant.

2. **Tangential Acceleration**:
- The tangential acceleration \( a_t \) is defined as the rate of change of tangential velocity \( v \):
\[ a_t = \frac{dv}{dt} \]

3. **Finding the Velocity**:
- Since \( a_r = k \cdot t \), we can relate this to the tangential velocity \( v \) using the relationship between radial acceleration and velocity:
\[ a_r = \frac{v^2}{r} \]
- Setting the two expressions for radial acceleration equal gives:
\[ k \cdot t = \frac{v^2}{r} \]
- Rearranging this, we find:
\[ v^2 = k \cdot t \cdot r \]

4. **Differentiating to Find Tangential Acceleration**:
- To find \( a_t \), we differentiate \( v^2 \) with respect to \( t \):
\[ \frac{d(v^2)}{dt} = \frac{d(k \cdot t \cdot r)}{dt} \]
- Using the chain rule:
\[ 2v \frac{dv}{dt} = k \cdot r \]
- Thus, we can express \( a_t \) as:
\[ a_t = \frac{dv}{dt} = \frac{k \cdot r}{2v} \]

5. **Substituting for \( v \)**:
- From the previous step, we have \( v = \sqrt{k \cdot t \cdot r} \). Substituting this into the expression for \( a_t \):
\[ a_t = \frac{k \cdot r}{2\sqrt{k \cdot t \cdot r}} = \frac{k \cdot r}{2\sqrt{k} \cdot \sqrt{t} \cdot \sqrt{r}} = \frac{\sqrt{k} \cdot \sqrt{r}}{2\sqrt{t}} \]
- This shows that \( a_t \) is dependent on \( t \).

6. **Finding \( a_r \cdot a_t^2 \)**:
- Now, we need to check the product \( a_r \cdot a_t^2 \):
\[ a_r \cdot a_t^2 = (k \cdot t) \cdot \left(\frac{\sqrt{k} \cdot \sqrt{r}}{2\sqrt{t}}\right)^2 \]
- Simplifying \( a_t^2 \):
\[ a_t^2 = \left(\frac{\sqrt{k} \cdot \sqrt{r}}{2\sqrt{t}}\right)^2 = \frac{k \cdot r}{4t} \]
- Thus,
\[ a_r \cdot a_t^2 = (k \cdot t) \cdot \left(\frac{k \cdot r}{4t}\right) = \frac{k^2 \cdot r}{4} \]
- This expression is independent of time \( t \).

### Conclusion:
The quantity \( a_r \cdot a_t^2 \) is independent of time \( t \).