A particle moves in a circle of radius `20 cm`. Its linear speed is given by `v = 2t` where `t` is in seconds and `v` in `m s^-1`. Then

Options

A. The radial acceleration at `t = 2 s` is `80 m s^-2`.
B. Tangential acceleration at `t = 2 s` is `2 m s^-2`.
C. Net acceleration at `t = 2 s` is greater than `80 m s^-2`.
D. Tangential acceleration remains constant in magnitude.

Explanation

To solve the problem step by step, we will analyze the motion of a particle moving in a circle with a given linear speed and calculate the required accelerations.

### Given Data:
- Radius of the circle, \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Linear speed, \( v = 2t \, \text{m/s} \) (where \( t \) is in seconds)

### Step 1: Calculate Radial (Centripetal) Acceleration
The formula for radial (centripetal) acceleration \( a_r \) is given by:
\[ a_r = \frac{v^2}{r} \]
Substituting the expression for \( v \):
\[ a_r = \frac{(2t)^2}{r} = \frac{4t^2}{0.2} \]
Calculating this gives:
\[ a_r = 20t^2 \, \text{m/s}^2 \]

### Step 2: Find Radial Acceleration at \( t = 2 \) seconds
Now, substituting \( t = 2 \) seconds into the radial acceleration formula:
\[ a_r = 20(2)^2 = 20 \times 4 = 80 \, \text{m/s}^2 \]
Thus, the radial acceleration at \( t = 2 \) seconds is \( 80 \, \text{m/s}^2 \).

### Step 3: Calculate Tangential Acceleration
Tangential acceleration \( a_t \) is given by the derivative of velocity with respect to time:
\[ a_t = \frac{dv}{dt} \]
Since \( v = 2t \):
\[ \frac{dv}{dt} = 2 \, \text{m/s}^2 \]
Thus, the tangential acceleration is constant at \( 2 \, \text{m/s}^2 \).

### Step 4: Calculate Net Acceleration
The net acceleration \( a \) is the vector sum of radial and tangential accelerations:
\[ a = \sqrt{a_t^2 + a_r^2} \]
Substituting the values we found:
\[ a = \sqrt{(2)^2 + (80)^2} = \sqrt{4 + 6400} = \sqrt{6404} \]
Calculating this gives:
\[ a \approx 80.02 \, \text{m/s}^2 \]
Thus, the net acceleration at \( t = 2 \) seconds is greater than \( 80 \, \text{m/s}^2 \).

### Step 5: Conclusion on Tangential Acceleration
Since the tangential acceleration \( a_t = 2 \, \text{m/s}^2 \) is constant, we conclude that the statement regarding tangential acceleration remaining constant in magnitude is correct.

### Summary of Results:
1. Radial acceleration at \( t = 2 \) seconds: \( 80 \, \text{m/s}^2 \)
2. Tangential acceleration: \( 2 \, \text{m/s}^2 \)
3. Net acceleration at \( t = 2 \) seconds: approximately \( 80.02 \, \text{m/s}^2 \) (greater than \( 80 \, \text{m/s}^2 \))
4. Tangential acceleration remains constant.