The ratio of displacement in n seconds and in | vedclass

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Question

(D) For a particle starting from rest $(u = 0)$ under constant acceleration $(a)$:The displacement in $n$ seconds is given by the kinematic equation:$

Vedclass · Accepted Answer

$\frac{n^2}{2n - 1}$

Vedclass · Answer

(D) For a particle starting from rest $(u = 0)$ under constant acceleration $(a)$:The displacement in $n$ seconds is given by the kinematic equation:$s_n = ut + \frac{1}{2}an^2 = 0 + \frac{1}{2}an^2 = \frac{1}{2}an^2$The displacement in the $n^{th}$ second is given by:$s_{nth} = u + \frac{a}{2}(2n - 1) = 0 + \frac{a}{2}(2n - 1) = \frac{a}{2}(2n - 1)$Taking the ratio of displacement in $n$ seconds to the displacement in the $n^{th}$ second:$\frac{s_n}{s_{nth}} = \frac{\frac{1}{2}an^2}{\frac{a}{2}(2n - 1)} = \frac{n^2}{2n - 1}$Thus,the correct option is $D$.

The ratio of displacement in $n$ seconds and in the $n^{th}$ second for a particle moving in a straight line under constant acceleration starting from rest is:

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