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 n seconds is​

Answer 1

Answer:

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Answer 2

Answer:

$(2n-1) : n^{2}$

Explanation:

$s = ut + \frac{1}{2}at^{2}$

Distance covered in n seconds

$= ut + \frac{1}{2}at^{2} \\\\= \frac{1}{2}an^{2}$

Distance covered in the nth second

$= \frac{1}{2}an^{2}$ - distance covered in (n-1) seconds

$= \frac{1}{2}an^{2} - \frac{1}{2}a(n-1)^{2}\\\\= \frac{1}{2}a [n^{2} - (n-1)^{2} ]\\\\= \frac{1}{2}a (n^{2} - n^{2} + 2n -1)\\\\= \frac{1}{2}a (2n-1)$

∴ Required ratio

$= \frac{1}{2}a (2n-1) : \frac{1}{2}an^{2}\\\\= (2n-1) : n^{2}$