A particle moves in a circle of radius 1 cm at a speed given `v = 2t`, where `v` is cm/s and t is in seconds. Total acceleration of the particle at `t=1` second is `2 sqrt(n) cm//s^(2)`. Find value of `n`.

Answer

To solve the problem, we need to find the total acceleration of a particle moving in a circular path with a given speed. The speed of the particle is defined as \( v = 2t \), where \( v \) is in cm/s and \( t \) is in seconds. The radius of the circle is given as \( r = 1 \) cm. We will calculate the total acceleration at \( t = 1 \) second.

### Step-by-Step Solution:

1. **Identify the given parameters:**
- Speed: \( v = 2t \)
- Radius: \( r = 1 \) cm
- Time: \( t = 1 \) s

2. **Calculate the speed at \( t = 1 \) s:**
\[ v = 2t = 2 \times 1 = 2 \text{ cm/s} \]

3. **Calculate the centripetal acceleration (\( a_c \)):**
The formula for centripetal acceleration is:
\[ a_c = \frac{v^2}{r} \]
Substituting the values we have:
\[ a_c = \frac{(2)^2}{1} = \frac{4}{1} = 4 \text{ cm/s}^2 \]

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

5. **Calculate the total acceleration (\( a \)):**
The total acceleration is the vector sum of centripetal and tangential accelerations. We can use the Pythagorean theorem:
\[ a = \sqrt{a_c^2 + a_t^2} \]
Substituting the values:
\[ a = \sqrt{(4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} \text{ cm/s}^2 \]

6. **Express \( \sqrt{20} \) in the required form:**
We can simplify \( \sqrt{20} \):
\[ \sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5} \]
According to the problem, the total acceleration is given as \( 2\sqrt{n} \). Thus, we can equate:
\[ 2\sqrt{n} = 2\sqrt{5} \]

7. **Solve for \( n \):**
Dividing both sides by 2:
\[ \sqrt{n} = \sqrt{5} \]
Squaring both sides gives:
\[ n = 5 \]

### Final Answer:
The value of \( n \) is \( \boxed{5} \).