A body travels $102.5 \mathrm{~m}$ in $\mathrm{n}^{\text {th }}$ second and $115.0 \mathrm{~m}$ in $(\mathrm{n}+2)^{\text {th }}$ second. The acceleration is :

<p>The distance covered by a body in the $n^{\text{th}}$ second can be found using the equation:</p> <p>$S_{n} = u + \dfrac{1}{2}a(2n-1)$</p> <p>where,</p> <ul> <li>$S_{n}$ is the distance covered in the $n^{\text{th}}$ second,</li> <li>$u$ is the initial velocity,</li> <li>$a$ is the acceleration, and</li> <li>$n$ is the nth second.</li> </ul> <p>The distance covered in the $n^{\text{th}}$ second is given as $102.5 \, \text{m}$, so we have:</p> <p>$102.5 = u + \dfrac{1}{2}a(2n-1)$ ---- (1)</p> <p>For the $(n + 2)^{\text{th}}$ second, the distance covered is:</p> <p>$S_{n+2} = u + \dfrac{1}{2}a(2(n+2)-1)$</p> <p>Substituting $n + 2$ in place of $n$, we get:</p> <p>$115.0 = u + \dfrac{1}{2}a(2n+3)$ ---- (2)</p> <p>Subtracting equation (1) from equation (2), we get:</p> <p>$115.0 - 102.5 = \dfrac{1}{2}a(2n + 3 - 2n + 1)$</p> <p>$12.5 = \dfrac{1}{2}a(4)$</p> <p>So, solving for $a$ gives:</p> <p>$a = \dfrac{12.5 \times 2}{4} = \dfrac{25}{4} = 6.25 \, \text{m/s}^2$</p> <p>Therefore, the acceleration of the body is:</p> <p>$6.25 \, \text{m/s}^2$</p> <p>Which corresponds to Option A: $6.25 \, \text{m/s}^2$.</p>