A particle is performing circular motion of radius 1 m. Its speed is `v=(2t^(2)) m//s`. What will be magnitude of its acceleration at `t=1s`.

Explanation

To solve the problem, we need to find the magnitude of the acceleration of a particle performing circular motion at a given time \( t = 1 \) second. The speed of the particle is given by the equation \( v = 2t^2 \) m/s, and the radius of the circular motion is \( r = 1 \) m.

### Step-by-Step Solution:

1. **Determine the speed at \( t = 1 \) second:**
\[ v = 2t^2 \]
Substituting \( t = 1 \):
\[ v = 2(1)^2 = 2 \text{ m/s} \]

2. **Calculate the tangential acceleration (\( a_t \)):**
The tangential acceleration is given by the derivative of velocity with respect to time:
\[ a_t = \frac{dv}{dt} \]
First, we differentiate \( v = 2t^2 \):
\[ \frac{dv}{dt} = 4t \]
Now substituting \( t = 1 \):
\[ a_t = 4(1) = 4 \text{ m/s}^2 \]

3. **Calculate the radial (centripetal) acceleration (\( a_r \)):**
The radial acceleration is given by the formula:
\[ a_r = \frac{v^2}{r} \]
We already found \( v = 2 \) m/s and \( r = 1 \) m, so:
\[ a_r = \frac{(2)^2}{1} = \frac{4}{1} = 4 \text{ m/s}^2 \]

4. **Calculate the magnitude of the total acceleration (\( a \)):**
The total acceleration is the vector sum of tangential and radial accelerations:
\[ a = \sqrt{a_t^2 + a_r^2} \]
Substituting the values:
\[ a = \sqrt{(4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \text{ m/s}^2 \]

### Final Answer:
The magnitude of the acceleration at \( t = 1 \) second is \( 4\sqrt{2} \) m/s².

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