A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5$^{th}$ $\sec$ to that covered in $5\ sec$ is

Question

Dr. Abubacker · Accepted Answer

(a) Distance covered in $5^{th}$ second
$S_{5^{th}} = u + \frac{a}{2}(2n - 1) = 0 + \frac{a}{2}(2 	imes 5 - 1) = \frac{9a}{2}$
and distance covered in $5\ second$,
$S_{5} = ut + \frac{1}{2}at^{2} = 0 + \frac{1}{2} 	imes a 	imes 25 = \frac{25a}{2}$
$	herefore\frac{S_{5^{th}}}{S_{5}} = \frac{9}{25}$

Q. A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5$^{th}$ $\sec$ to that covered in $5\ sec$ is

Explanation

Reviewed by Dr. Abubacker

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