A point moves along a circle with a speed `v=kt` , where `K=0.5m//s^(2)`
Find the total acceleration of the point the momenet when it has covered the `n^(th)` Fraction of the circle after the beginging of motion, where `n=(1)/(10)` .

Explanation

To solve the problem, we need to find the total acceleration of a point moving along a circular path with a speed that varies with time as \( v = kt \), where \( k = 0.5 \, \text{m/s}^2 \). We are to find the total acceleration when the point has covered the \( n^{th} \) fraction of the circle, where \( n = \frac{1}{10} \).

### Step-by-Step Solution:

1. **Understanding the Speed Function**:
The speed of the point is given by:
\[ v = kt \]
where \( k = 0.5 \, \text{m/s}^2 \).

2. **Finding the Displacement**:
The displacement \( s \) can be found by integrating the speed:
\[ s = \int v \, dt = \int kt \, dt = \frac{1}{2} kt^2 \]
Therefore, the displacement after time \( t \) is:
\[ s = \frac{1}{2} k t^2 \]

3. **Relating Displacement to Circle**:
Since the point covers a fraction \( n \) of the circle, we can express the displacement in terms of the radius \( r \) of the circle:
\[ s = 2 \pi r n \]
Setting the two expressions for \( s \) equal gives:
\[ \frac{1}{2} kt^2 = 2 \pi r n \]

4. **Solving for Time \( t \)**:
Rearranging the equation to find \( t^2 \):
\[ t^2 = \frac{4 \pi r n}{k} \]

5. **Finding Tangential Acceleration \( a_t \)**:
The tangential acceleration is the derivative of speed with respect to time:
\[ a_t = \frac{dv}{dt} = k \]
Since \( k = 0.5 \, \text{m/s}^2 \), we have:
\[ a_t = 0.5 \, \text{m/s}^2 \]

6. **Finding Normal Acceleration \( a_n \)**:
The normal (centripetal) acceleration is given by:
\[ a_n = \frac{v^2}{r} \]
Substituting \( v = kt \) gives:
\[ a_n = \frac{(kt)^2}{r} = \frac{k^2 t^2}{r} \]
Now substituting \( t^2 \) from step 4:
\[ a_n = \frac{k^2 \left(\frac{4 \pi r n}{k}\right)}{r} = \frac{4 \pi k n}{1} \]

7. **Calculating Total Acceleration**:
The total acceleration \( a \) is the vector sum of tangential and normal accelerations:
\[ a = \sqrt{a_t^2 + a_n^2} \]
Substituting the values:
\[ a = \sqrt{(0.5)^2 + (4 \pi k n)^2} \]

8. **Substituting Values**:
With \( k = 0.5 \) and \( n = 0.1 \):
\[ a_n = 4 \pi (0.5)(0.1) = 0.2 \pi \]
Now substituting back into the total acceleration:
\[ a = \sqrt{(0.5)^2 + (0.2 \pi)^2} \]

9. **Calculating the Final Result**:
\[ a = \sqrt{0.25 + (0.2 \cdot 3.14)^2} = \sqrt{0.25 + 0.0396} = \sqrt{0.2896} \approx 0.54 \, \text{m/s}^2 \]

### Final Answer:
The total acceleration of the point when it has covered the \( \frac{1}{10} \) fraction of the circle is approximately \( 0.54 \, \text{m/s}^2 \).