A starts from rest and moves with acceleration `a_(1)`. Two seconds later B starts from rest and moves with an acceleration `a_(2)`. If the displacement of A in the `5^(th)` second is the same as that of B in the same interval, the ratio of `a_(1)` to `a_(2)` is

Options

A. `9:5`
B. `5:9`
C. `1:1`
D. `1:3`

Explanation

To solve the problem, we need to find the ratio of the accelerations \( a_1 \) and \( a_2 \) of two bodies A and B, given that their displacements in specific time intervals are equal. Let's break it down step by step.

### Step 1: Understand the Motion of A
- Body A starts from rest with an acceleration \( a_1 \).
- The displacement of A in the \( n^{th} \) second can be calculated using the formula:
\[ S_n = u + \frac{a}{2} (2n - 1) \]
where \( u \) is the initial velocity, \( a \) is the acceleration, and \( n \) is the second.

- Since A starts from rest, \( u = 0 \). Thus, the displacement of A in the 5th second (\( S_A \)) is:
\[ S_A = 0 + \frac{a_1}{2} (2 \cdot 5 - 1) = \frac{a_1}{2} \cdot 9 = \frac{9a_1}{2} \]

### Step 2: Understand the Motion of B
- Body B starts from rest with an acceleration \( a_2 \) but starts 2 seconds later than A.
- Therefore, when A has traveled for 5 seconds, B has traveled for \( 5 - 2 = 3 \) seconds.

- The displacement of B in the 3rd second (\( S_B \)) is given by the same formula:
\[ S_B = 0 + \frac{a_2}{2} (2 \cdot 3 - 1) = \frac{a_2}{2} \cdot 5 = \frac{5a_2}{2} \]

### Step 3: Set the Displacements Equal
- According to the problem, the displacement of A in the 5th second is equal to the displacement of B in the 3rd second:
\[ S_A = S_B \]
\[ \frac{9a_1}{2} = \frac{5a_2}{2} \]

### Step 4: Simplify the Equation
- We can eliminate the \( \frac{1}{2} \) from both sides:
\[ 9a_1 = 5a_2 \]

### Step 5: Find the Ratio of Accelerations
- Rearranging the equation gives us:
\[ \frac{a_1}{a_2} = \frac{5}{9} \]

### Conclusion
- The ratio of the accelerations \( a_1 \) to \( a_2 \) is:
\[ \frac{a_1}{a_2} = \frac{5}{9} \]