Question

A body starts from rest, what is the ratio of the distance travelled by the body during the $4th$ and $3rd$ second?

• A. $\frac{7}{5}$
• B. $\frac{5}{7}$
• C. $\frac{7}{3}$
• D. $\frac{3}{7}$

Answer 1

Answer: (A)

$\frac{7}{5}$

Distance traveled by a free a free falling body in $nth$ second $=$

distance traveled in n secs$-$distance traveled in $n-1$ secs.

$=0\ast n+\frac{1}{2}g{n}^2-0\ast (n-1)+\frac{1}{2}g{(n-1)}^2=\frac{g}{2}(n^2-(n^2+1-2n\left)\right)=\frac{g}{2}(2n-1)$

ratio of distance traveled in 4th sec. to 3rd sec $=\frac{2\times 4-1}{2\times 3-1}=\frac{7}{5}$