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A body falling from rest describes $S_{1}$ , $S_{2}$ and $S_{3}$ is the first, second and third seconds of its fall. Then the ratio of $S_{1} : S_{2} : S_{3}$ is

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1 : 1 : 1

1 : 3 : 5

1 : 2 : 3

1 : 4 : 9

answer is B.

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Detailed Solution

A body falling from rest describes

S_{1}

S_{2}

and

S_{3}

is the first, second and third seconds of its fall. Then the ratio of

S_{1} ​ : S_{2} ​ : S_{3}

1 : 3 : 5

.
Given data
Initial velocity,

u = 0

Distance covered in 1st-second

= S_{1}

Distance covered in 2nd-second

= S_{2}

Distance covered in 3rd-second

= S_{3}

Find the ratio of the distance covered. Question Image

Formula used:

S = ut + \frac{1}{2} g t^{2}

Where S is the distance covered,

u

is the initial velocity,

t

is the time taken and

g

is the acceleration due to gravity.
Since we have given body at rest position A, B, and C described.
Therefore,

S = 0 \times t + \frac{1}{2} g t^{2} = \frac{1}{2} g t^{2}

Distance covered in 1st-second at

OA = S_{1} = \frac{1}{2} g \times 1^{2} = \frac{g}{2}

Distance covered in 2nd-second at

S_{2} = OB - OA = \frac{1}{2} g \times 2^{2} - \frac{g}{2} = \frac{3 g}{2}

Distance covered in 3rd-second,

S_{3} = OB - OA = \frac{1}{2} g \times 3^{2} - \frac{1}{2} g \times 2^{2} = \frac{5 g}{2}

Hence, the ratio of

S_{1} ​ : S_{2} ​ : S_{3} = \frac{g}{2} : \frac{3 g}{2} : \frac{5 g}{2} = 1 : 3 : 5

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