1. For uniformly accelerated motion, the distance travelled in the $n^{\text{th}}$ second is

$s_n = u + \frac{a}{2}(2n-1)$

where $u$ is initial velocity and $a$ is acceleration.

2. Therefore,

$s_n = u + \frac{a}{2}(2n-1) = 102.5$

and distance in the $(n+2)^{\text{th}}$ second is

$s_{n+2} = u + \frac{a}{2}[2(n+2)-1] = u + \frac{a}{2}(2n+3) = 115.0$

3. Subtract the two equations:

$s_{n+2} - s_n = \left[u + \frac{a}{2}(2n+3)\right] - \left[u + \frac{a}{2}(2n-1)\right]$

$115 - 102.5 = \frac{a}{2}\big[(2n+3)-(2n-1)\big]$

$12.5 = \frac{a}{2}(4) = 2a$

4. Hence,

$a = \frac{12.5}{2} = 6.25\ \text{m/s}^2$

5. Checking options:

• A: $6.25\ \text{m/s}^2$ ✅
• B: $5\ \text{m/s}^2$ ❌
• C: $12.5\ \text{m/s}^2$ ❌
• D: $9\ \text{m/s}^2$ ❌

So the correct answer is Option A.