A point moves along a circle with speed v=at. The total acceleration of the point at a time when it has traced 1//8th of the circumference is:

Options

A. `(v)/(8a)`
B. `2asqrt(4+pi^(2))`
C. a
D. `(a)/(2)sqrt(4+pi^(2))`

Answer

To solve the problem step by step, we will analyze the motion of a point moving along a circular path with a speed given by \( v = at \). We need to find the total acceleration of the point after it has traced \( \frac{1}{8} \) of the circumference.

### Step 1: Determine the distance covered
The circumference of a circle is given by \( 2\pi r \). Therefore, the distance covered when the point has traced \( \frac{1}{8} \) of the circumference is:
\[ s = \frac{2\pi r}{8} = \frac{\pi r}{4} \]

**Hint:** Remember that the total distance covered is a fraction of the circumference based on the fraction of the circle traced.

### Step 2: Use the kinematic equation
Since the initial velocity \( u = 0 \) (the point starts from rest) and the speed varies with time as \( v = at \), we can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting \( u = 0 \):
\[ \frac{\pi r}{4} = \frac{1}{2} a t^2 \]
From this, we can solve for \( t^2 \):
\[ t^2 = \frac{2\pi r}{4a} = \frac{\pi r}{2a} \]

**Hint:** This equation relates distance, acceleration, and time. Make sure to isolate \( t^2 \) correctly.

### Step 3: Find the speed at that time
Using \( v = at \), we can express \( v \) in terms of \( t \):
\[ v = a \sqrt{t^2} = a \sqrt{\frac{\pi r}{2a}} = \sqrt{a \cdot \frac{\pi r}{2}} = \sqrt{\frac{\pi a r}{2}} \]

**Hint:** The speed is derived from the acceleration and time, so ensure you substitute \( t \) properly.

### Step 4: Calculate centripetal acceleration
Centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{r} \]
Substituting for \( v \):
\[ a_c = \frac{\left(\sqrt{\frac{\pi a r}{2}}\right)^2}{r} = \frac{\frac{\pi a r}{2}}{r} = \frac{\pi a}{2} \]

**Hint:** Remember that centripetal acceleration depends on the square of the speed divided by the radius.

### Step 5: Determine tangential acceleration
The tangential acceleration \( a_t \) is constant and equal to \( a \) since \( v = at \) implies a constant acceleration:
\[ a_t = a \]

**Hint:** Tangential acceleration is related to the change in speed along the circular path.

### Step 6: Calculate the total acceleration
The total acceleration \( a_{total} \) is the vector sum of the tangential and centripetal accelerations. Since they are perpendicular, we can use the Pythagorean theorem:
\[ a_{total} = \sqrt{a_t^2 + a_c^2} = \sqrt{a^2 + \left(\frac{\pi a}{2}\right)^2} \]
\[ = \sqrt{a^2 + \frac{\pi^2 a^2}{4}} = \sqrt{a^2 \left(1 + \frac{\pi^2}{4}\right)} = a \sqrt{1 + \frac{\pi^2}{4}} \]

**Hint:** The total acceleration is found using the Pythagorean theorem because the components are perpendicular.

### Final Answer
Thus, the total acceleration of the point when it has traced \( \frac{1}{8} \) of the circumference is:
\[ a_{total} = a \sqrt{1 + \frac{\pi^2}{4}} \]