Question

Find the ratio of the distances traveled by a body falling freely from rest in the first, second and third seconds respectively.

• A. $1:4:9$
• B. $1:2:3$
• C. $1:9:25$
• D. $1:3:5$

Answer 1

The correct option is D $1:3:5$

Initial speed of the body is zero i.e. $u=0$

Distance traveled in $t^{th}$ second is given by $S_{t^{th}}=u+\frac{a}{2}(2t-1)$

$⟹S_{t^{th}}=\frac{a}{2}(2t-1)$ $(\because u=0)$

So, distance traveled in first second, $S_1=\frac{a}{2}(2\times 1-1)=\frac{a}{2}$

So, distance traveled in 2nd second, $S_2=\frac{a}{2}(2\times 2-1)=\frac{3a}{2}$

So, distance traveled in 3rd second, $S_3=\frac{a}{2}(2\times 3-1)=\frac{5a}{2}$

$⟹S_1:S_2:S_3=1:3:5$