Question

A body is falling freely under gravity. The distances covered by it in the first, second and third seconds of its motion are in the ratio

Answer 1

Given that the body is falling freely under gravity

Initial velocity u = 0

From the equation of motion

$s=ut+0.5a{t}^2$

Considering $a=10m/s^2$

For 1st sec we get,
$s=(0\times 1)+0.5\left(10\right)1^2$

Or, s = 5 m

For 2 sec we get,
$s=0+0.5\left(10\right)2^2$

Or, s = 20 m

So, distance traveled in 2nd second = 20 - 5 = 15 m

For 3 sec we get,
$s=0+0.5\left(10\right)3^2$

Or, s = 45 m

So the distance traveled by the particle in 3rd second = 45 - 20 = 25 m
$s_1$=5
$s_2$=15
$s_3$=25
Therefore, the ratio is in the order of 1:3:5.