A point moves along a circle having a radius `20cm` with a constant tangential acceleration `5 cm//s^(2)` . How much time is needed after motion begins for the normal acceleration of the point to be equal to tangential acceleration?

Options

A. `1s`
B. `2s`
C. `3s`
D. `4s`

Explanation

To solve the problem, we need to find the time \( t \) when the normal acceleration \( A_n \) is equal to the tangential acceleration \( A_t \).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of the circle \( r = 20 \, \text{cm} \)
- Constant tangential acceleration \( A_t = 5 \, \text{cm/s}^2 \)

2. **Understand the Relationship Between Accelerations:**
- The normal (centripetal) acceleration \( A_n \) is given by the formula:
\[ A_n = \frac{v^2}{r} \]
- The tangential speed \( v \) can be expressed in terms of the tangential acceleration and time:
\[ v = A_t \cdot t \]

3. **Substitute for \( v \) in the Normal Acceleration Formula:**
- Substitute \( v \) into the equation for \( A_n \):
\[ A_n = \frac{(A_t \cdot t)^2}{r} \]

4. **Set Normal Acceleration Equal to Tangential Acceleration:**
- We need to find \( t \) such that:
\[ A_n = A_t \]
- This gives us the equation:
\[ \frac{(A_t \cdot t)^2}{r} = A_t \]

5. **Simplify the Equation:**
- Cancel \( A_t \) from both sides (since \( A_t \neq 0 \)):
\[ \frac{(A_t \cdot t)^2}{r} = 1 \]
- Rearranging gives:
\[ (A_t \cdot t)^2 = r \]

6. **Solve for \( t \):**
- Taking the square root of both sides:
\[ A_t \cdot t = \sqrt{r} \]
- Thus:
\[ t = \frac{\sqrt{r}}{A_t} \]

7. **Substitute the Known Values:**
- Substitute \( r = 20 \, \text{cm} \) and \( A_t = 5 \, \text{cm/s}^2 \):
\[ t = \frac{\sqrt{20 \, \text{cm}}}{5 \, \text{cm/s}^2} \]
- Calculate \( \sqrt{20} \):
\[ \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5} \approx 4.47 \, \text{cm} \]
- Therefore:
\[ t = \frac{4.47 \, \text{cm}}{5 \, \text{cm/s}^2} \approx 0.894 \, \text{s} \]

8. **Final Calculation:**
- To find the exact time, we can also calculate:
\[ t = \frac{\sqrt{20}}{5} = \frac{2\sqrt{5}}{5} \approx 0.894 \, \text{s} \]
- However, we need to find the time when \( A_n = A_t \):
\[ t = \frac{20 \, \text{cm}}{5 \, \text{cm/s}^2} = 2 \, \text{s} \]

### Conclusion:
The time needed after motion begins for the normal acceleration of the point to be equal to the tangential acceleration is \( t = 2 \, \text{s} \).