A body starts from rest and moves with a constant acceleration. The ratio of distance covered in the nth second to the distance covered in $ \mathrm{n} $ seconds is

Question

A body starts from rest and moves with a constant acceleration. The ratio of distance covered in the nth second to the distance covered in $ \mathrm{n} $ seconds is, $ \frac{2}{n}-\frac{1}{n^{2}} $, $ \frac{1}{n^{2}}-\frac{1}{n} $, $ \frac{2}{n^{2}}-\frac{1}{n} $, $ \frac{2}{n}+\frac{1}{n^{2}} $

MARKS App · Accepted Answer

Given initial velocity u=0 and constant acceleration. -Distance travelled in nth second S n t h = u + 2 n - 1 2 a = 0 + 2 n - 1 2 a = 2 n - 1 2 a Distance travelled in n seconds is S n = u t + 1 2 a t 2 = 0 + 1 2 a n 2 = 1 2 a n 2 ∴ S nth S n = 2 n - 1 a 2 × 2 an 2 = 2 n - 1 n 2

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