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 of a point object moving along it, given the distance function \( S(t) = \frac{t^2}{2} + 2t \) and the ratio of the magnitudes of acceleration at \( t = 2 \, \text{s} \) and \( t = 5 \, \text{s} \) is \( 1:2 \).

### Step 1: Find the velocity function
The velocity \( v(t) \) is the first derivative of the displacement function \( S(t) \).

\[ v(t) = \frac{dS}{dt} = \frac{d}{dt} \left( \frac{t^2}{2} + 2t \right) = t + 2 \]

### Step 2: Find the tangential acceleration
The tangential acceleration \( a_t \) is the derivative of the velocity function.

\[ a_t = \frac{dv}{dt} = \frac{d}{dt}(t + 2) = 1 \, \text{m/s}^2 \]

### Step 3: Find the centripetal acceleration at \( t = 2 \, \text{s} \) and \( t = 5 \, \text{s} \)
The centripetal acceleration \( a_c \) is given by the formula \( a_c = \frac{v^2}{r} \).

1. **At \( t = 2 \, \text{s} \)**:
\[ v(2) = 2 + 2 = 4 \, \text{m/s} \]
\[ a_c(2) = \frac{v(2)^2}{r} = \frac{4^2}{r} = \frac{16}{r} \]

2. **At \( t = 5 \, \text{s} \)**:
\[ v(5) = 5 + 2 = 7 \, \text{m/s} \]
\[ a_c(5) = \frac{v(5)^2}{r} = \frac{7^2}{r} = \frac{49}{r} \]

### Step 4: Find the total acceleration at \( t = 2 \, \text{s} \) and \( t = 5 \, \text{s} \)
The total acceleration \( a \) is the vector sum of the tangential and centripetal accelerations.

1. **At \( t = 2 \, \text{s} \)**:
\[ a(2) = \sqrt{a_c(2)^2 + a_t^2} = \sqrt{\left(\frac{16}{r}\right)^2 + 1^2} = \sqrt{\frac{256}{r^2} + 1} \]

2. **At \( t = 5 \, \text{s} \)**:
\[ a(5) = \sqrt{a_c(5)^2 + a_t^2} = \sqrt{\left(\frac{49}{r}\right)^2 + 1^2} = \sqrt{\frac{2401}{r^2} + 1} \]

### Step 5: Set up the ratio of the accelerations
Given that the ratio of the magnitudes of acceleration at \( t = 2 \, \text{s} \) and \( t = 5 \, \text{s} \) is \( 1:2 \):

\[ \frac{a(2)}{a(5)} = \frac{1}{2} \]

Substituting the expressions for \( a(2) \) and \( a(5) \):

\[ \frac{\sqrt{\frac{256}{r^2} + 1}}{\sqrt{\frac{2401}{r^2} + 1}} = \frac{1}{2} \]

### Step 6: Square both sides and cross-multiply
\[ \frac{256}{r^2} + 1 = \frac{1}{4} \left( \frac{2401}{r^2} + 1 \right) \]

Multiplying both sides by \( 4r^2 \):

\[ 4 \left( 256 + r^2 \right) = 2401 + r^2 \]

### Step 7: Simplify and solve for \( r^2 \)
\[ 1024 + 4r^2 = 2401 + r^2 \]
\[ 3r^2 = 1377 \]
\[ r^2 = \frac{1377}{3} \]

### Step 8: Find \( r \)
\[ r = \sqrt{\frac{1377}{3}} = \frac{\sqrt{1377}}{\sqrt{3}} = \frac{\sqrt{51 \times 27}}{\sqrt{3}} = 3\sqrt{51} \]

### Final Answer
The radius of the circle is \( 3\sqrt{51} \, \text{meters} \).

---