A point object moves on a circular path such that distance covered by it is given by function `S=(t^(2)/(2)+2t)` meter ( t in second). The ratio of the magnitude of acceleration at `t = 2 s and t = 5 s`. is `1: 2` then the radius of the circle is

Options

A. 1 m
B. `3sqrt(51) m`
C. `sqrt(51) m`
D. 3 m

Answer

To solve the problem, we need to find the radius of the circular path based on the given distance function and the ratio of accelerations at two different times.

### Step 1: Determine the distance function
The distance covered by the object is given by the function:
\[ S(t) = \frac{t^2}{2} + 2t \]

### Step 2: Find the velocity
To find the velocity \( v(t) \), we differentiate the distance function with respect to time \( t \):
\[ v(t) = \frac{dS}{dt} = \frac{d}{dt}\left(\frac{t^2}{2} + 2t\right) = t + 2 \]

### Step 3: Calculate the tangential acceleration
The tangential acceleration \( a_T \) is the derivative of velocity with respect to time:
\[ a_T = \frac{dv}{dt} = \frac{d}{dt}(t + 2) = 1 \, \text{m/s}^2 \]
This is constant and does not depend on time.

### Step 4: Calculate the centripetal acceleration
The centripetal acceleration \( a_C \) is given by the formula:
\[ a_C = \frac{v^2}{r} \]
We need to calculate \( a_C \) at \( t = 2 \) s and \( t = 5 \) s.

#### For \( t = 2 \) s:
1. Calculate \( v(2) \):
\[ v(2) = 2 + 2 = 4 \, \text{m/s} \]
2. Calculate \( a_C(2) \):
\[ a_C(2) = \frac{(4)^2}{r} = \frac{16}{r} \]

#### For \( t = 5 \) s:
1. Calculate \( v(5) \):
\[ v(5) = 5 + 2 = 7 \, \text{m/s} \]
2. Calculate \( a_C(5) \):
\[ a_C(5) = \frac{(7)^2}{r} = \frac{49}{r} \]

### Step 5: Calculate the total acceleration
The total acceleration \( a \) at any time \( t \) is given by:
\[ a = \sqrt{a_T^2 + a_C^2} \]

#### For \( t = 2 \) s:
\[ a_1 = \sqrt{(1)^2 + \left(\frac{16}{r}\right)^2} = \sqrt{1 + \frac{256}{r^2}} \]

#### For \( t = 5 \) s:
\[ a_2 = \sqrt{(1)^2 + \left(\frac{49}{r}\right)^2} = \sqrt{1 + \frac{2401}{r^2}} \]

### Step 6: Set up the ratio of accelerations
According to the problem, the ratio of the magnitudes of acceleration at \( t = 2 \) s and \( t = 5 \) s is given as:
\[ \frac{a_1}{a_2} = \frac{1}{2} \]

Substituting the expressions for \( a_1 \) and \( a_2 \):
\[ \frac{\sqrt{1 + \frac{256}{r^2}}}{\sqrt{1 + \frac{2401}{r^2}}} = \frac{1}{2} \]

### Step 7: Square both sides
Squaring both sides to eliminate the square roots:
\[ \frac{1 + \frac{256}{r^2}}{1 + \frac{2401}{r^2}} = \frac{1}{4} \]

### Step 8: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4\left(1 + \frac{256}{r^2}\right) = 1 + \frac{2401}{r^2} \]
Expanding and simplifying:
\[ 4 + \frac{1024}{r^2} = 1 + \frac{2401}{r^2} \]
\[ 4 - 1 = \frac{2401}{r^2} - \frac{1024}{r^2} \]
\[ 3 = \frac{1377}{r^2} \]

### Step 9: Solve for \( r^2 \)
Rearranging gives:
\[ r^2 = \frac{1377}{3} \]
\[ r^2 = 459 \]
Thus,
\[ r = \sqrt{459} \]

### Step 10: Final answer
The radius of the circle is:
\[ r \approx 3\sqrt{51} \, \text{meters} \]