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(A) The distance travelled in the $n^{th}$ second is defined as the difference between the total distance covered in $n$ seconds and the total distanc

Vedclass · Accepted Answer

$v_0 + \frac{a}{2}(2n-1)$

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(A) The distance travelled in the $n^{th}$ second is defined as the difference between the total distance covered in $n$ seconds and the total distance covered in $(n-1)$ seconds.Using the kinematic equation $s = v_0 t + \frac{1}{2} a t^2$:$d_n = s_n - s_{n-1}$$d_n = (v_0 n + \frac{1}{2} a n^2) - [v_0 (n-1) + \frac{1}{2} a (n-1)^2]$$d_n = v_0 n + \frac{1}{2} a n^2 - [v_0 n - v_0 + \frac{1}{2} a (n^2 - 2n + 1)]$$d_n = v_0 n + \frac{1}{2} a n^2 - v_0 n + v_0 - \frac{1}{2} a n^2 + a n - \frac{1}{2} a$$d_n = v_0 + a n - \frac{1}{2} a$$d_n = v_0 + \frac{a}{2} (2n - 1)$

$A$ particle with initial velocity $v_0$ moves with constant acceleration $a$ in a straight line. Find the distance travelled in the $n^{th}$ second.

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