Question

A body is falling freely under gravity. At time t=0 s, velocity = 0 m/s. The distances covered by it in the first, second and third seconds of its motion are in the ratio of

• A. 1 : 3 :5
• B. 1 : 3 : 9
• C. 1 : 2 : 3
• D. 5 : 3 :1

Answer 1

The correct option is A 1 : 3 :5

We know, $s=ut+\left(1/2\right)a{t}^2$
Considering $t_0=0,t_1=1,t_2=2and{t}_3=3sec$, and given $u=0$
we can write ,
$s_1-s_0=\left(1/2\right)g({t_1}^2-{t_0}^2)=g/2$
$s_2-s_1=\left(1/2\right)g({t_2}^2-{t_1}^2)=3g/2$
$s_3-s_2=\left(1/2\right)g({t_3}^2-{t_2}^2)=5g/2$
So, the ratio of distances covered in 1st, 2nd and 3rd seconds is 1:3:5