A particle moves in a circle of radius `30cm`. Its linear speed is given by `v=2t`, where `t` in second and `v` in `m//s`. Find out its radial and tangential acceleration at `t=3s`.

Options

A. `220m/s^(2),50m//s^(2)`
B. `100m//s^(2),5m//s^(2)`
C. `120m//s^(2)`
D. `110m//s^(2),10m//s^(2)`

Explanation

To solve the problem of finding the radial and tangential acceleration of a particle moving in a circle of radius 30 cm with a linear speed given by \( v = 2t \), we will follow these steps:

### Step 1: Determine the tangential acceleration
The tangential acceleration (\( a_t \)) is given by the derivative of the linear speed with respect to time.

\[ a_t = \frac{dv}{dt} \]

Given \( v = 2t \), we differentiate:

\[ \frac{dv}{dt} = 2 \, \text{m/s}^2 \]

### Step 2: Calculate the linear speed at \( t = 3s \)
Now, we need to find the linear speed at \( t = 3s \):

\[ v = 2t = 2 \times 3 = 6 \, \text{m/s} \]

### Step 3: Determine the radial (centripetal) acceleration
The radial (centripetal) acceleration (\( a_r \)) is given by the formula:

\[ a_r = \frac{v^2}{r} \]

Where \( r \) is the radius of the circle. We convert the radius from cm to meters:

\[ r = 30 \, \text{cm} = 0.3 \, \text{m} \]

Now, substituting the values:

\[ a_r = \frac{(6 \, \text{m/s})^2}{0.3 \, \text{m}} = \frac{36 \, \text{m}^2/\text{s}^2}{0.3 \, \text{m}} = 120 \, \text{m/s}^2 \]

### Step 4: Summarize the results
At \( t = 3s \):
- Tangential acceleration \( a_t = 2 \, \text{m/s}^2 \)
- Radial acceleration \( a_r = 120 \, \text{m/s}^2 \)

### Final Answer
The tangential acceleration is \( 2 \, \text{m/s}^2 \) and the radial acceleration is \( 120 \, \text{m/s}^2 \).

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