The speed of a particle moving in a circle of radius `r=2m` varies with time `t` as `v=t^(2)`, where `t` is in second and `v` in `m//s`. Find the radial, tangential and net acceleration at `t=2s`.

Options

A. ` a=sqrt(80)m//s^(2)`
B. ` a=sqrt(30)m//s^(2)`
C. ` a=sqrt(20)m//s^(2)`
D. None of the above

Explanation

To solve the problem, we need to find the radial, tangential, and net acceleration of a particle moving in a circle of radius \( r = 2 \, \text{m} \) with a speed that varies with time as \( v = t^2 \). We will evaluate these accelerations at \( t = 2 \, \text{s} \).

### Step 1: Calculate the velocity at \( t = 2 \, \text{s} \)

Given:
\[ v = t^2 \]
At \( t = 2 \, \text{s} \):
\[ v = (2)^2 = 4 \, \text{m/s} \]

### Step 2: Calculate the radial acceleration \( a_r \)

Radial acceleration is given by the formula:
\[ a_r = \frac{v^2}{r} \]
Substituting the values:
\[ a_r = \frac{(4 \, \text{m/s})^2}{2 \, \text{m}} = \frac{16}{2} = 8 \, \text{m/s}^2 \]

### Step 3: Calculate the tangential acceleration \( a_t \)

Tangential acceleration is given by the derivative of velocity with respect to time:
\[ a_t = \frac{dv}{dt} \]
First, we differentiate \( v = t^2 \):
\[ \frac{dv}{dt} = 2t \]
Now, substituting \( t = 2 \, \text{s} \):
\[ a_t = 2 \times 2 = 4 \, \text{m/s}^2 \]

### Step 4: Calculate the net acceleration \( a_n \)

The net acceleration is the vector sum of radial and tangential accelerations, calculated using the Pythagorean theorem:
\[ a_n = \sqrt{a_r^2 + a_t^2} \]
Substituting the values:
\[ a_n = \sqrt{(8 \, \text{m/s}^2)^2 + (4 \, \text{m/s}^2)^2} = \sqrt{64 + 16} = \sqrt{80} \, \text{m/s}^2 \]
Calculating \( \sqrt{80} \):
\[ a_n = \sqrt{16 \times 5} = 4\sqrt{5} \, \text{m/s}^2 \approx 8.94 \, \text{m/s}^2 \]

### Summary of Results
- Radial acceleration \( a_r = 8 \, \text{m/s}^2 \)
- Tangential acceleration \( a_t = 4 \, \text{m/s}^2 \)
- Net acceleration \( a_n \approx 8.94 \, \text{m/s}^2 \)