A particle moves in a circle of radius 20 cm. Its linear speed is given by v=2t where t is in s and v in m/s. Then
a) the radial acceleration at `t=2s" is "80ms^(-2)`
b) tangential acceleration at `t-2s" is "2 ms^(-2)`
c)net acceleration at t=2s is greater than `80 ms^(-2)`
d) tangential acceleration remains constant in magnitude.

Options

A. only a,b,c are correct
B. only a,b,d are correct
C. only a,c,d are correct
D. all a,b,c,d are correct

Explanation

To solve the problem, we need to analyze the motion of a particle moving in a circle with a given linear speed. The linear speed is defined as \( v = 2t \), where \( t \) is in seconds and \( v \) is in meters per second. The radius of the circle is given as 20 cm, which we will convert to meters.

### Step-by-Step Solution:

1. **Convert Radius to Meters:**
\[ R = 20 \text{ cm} = \frac{20}{100} \text{ m} = 0.2 \text{ m} \]

**Hint:** Always convert units to the standard SI units for consistency.

2. **Calculate Radial (Centripetal) Acceleration:**
Radial acceleration \( A_R \) is given by the formula:
\[ A_R = \frac{v^2}{R} \]
Substitute \( v = 2t \):
\[ A_R = \frac{(2t)^2}{R} = \frac{4t^2}{0.2} = 20t^2 \]

**Hint:** Remember that radial acceleration depends on the square of the linear speed and the radius.

3. **Evaluate Radial Acceleration at \( t = 2 \) s:**
\[ A_R = 20(2)^2 = 20 \times 4 = 80 \text{ m/s}^2 \]

**Hint:** Substitute the value of \( t \) directly into the formula to find the specific acceleration at that time.

4. **Calculate Tangential Acceleration:**
Tangential acceleration \( A_T \) is the rate of change of linear speed:
\[ A_T = \frac{dv}{dt} \]
Since \( v = 2t \):
\[ A_T = \frac{d(2t)}{dt} = 2 \text{ m/s}^2 \]

**Hint:** The tangential acceleration is constant if the linear speed is a linear function of time.

5. **Evaluate Net Acceleration:**
The net acceleration \( A_{net} \) is the vector sum of radial and tangential accelerations. Since they are perpendicular:
\[ A_{net} = \sqrt{A_R^2 + A_T^2} \]
Substitute the values:
\[ A_{net} = \sqrt{(80)^2 + (2)^2} = \sqrt{6400 + 4} = \sqrt{6404} \approx 80.02 \text{ m/s}^2 \]

**Hint:** Use the Pythagorean theorem to find the resultant acceleration when two components are perpendicular.

6. **Conclusion:**
- a) The radial acceleration at \( t = 2 \) s is \( 80 \text{ m/s}^2 \) (True).
- b) The tangential acceleration at \( t = 2 \) s is \( 2 \text{ m/s}^2 \) (True).
- c) The net acceleration at \( t = 2 \) s is slightly greater than \( 80 \text{ m/s}^2 \) (True).
- d) The tangential acceleration remains constant in magnitude (True).

**Final Answer: All options a, b, c, and d are correct.**