Question

The ratios of the distances covered by a freely falling particle, starting from rest, in the first, second, third.... $n^{th}$ seconds of its motion

• A. form an arithmetic progression
• B. form the series corresponding to the squares of the first $n$ natural numbers
• C. do not form any well-defined series
• D. form a series corresponding to the differences of the squares of the successive whole numbers

Answer 1

Answer: (A), (D)

The correct options are
A form an arithmetic progression
D form a series corresponding to the differences of the squares of the successive whole numbers

Distance covered in $t=1s$,
$u=0$ (Initial velocity is 0 during free fall) $h_1=\frac{1}{2}\times g\times (1{)}^2=\frac{1}{2}\text{g (Using second equation of motion)}$
Distance covered in $t=2s$,
$h_2=\frac{1}{2}\times g(2{)}^2=2g$
Distance covered in $t=3s$,
$h_3=\frac{1}{2}\times g(3{)}^2=\frac{9}{2}g$
Distance covered in $t=4s$,
$h_4=8g$
In 2nd second distance covered
$=h_2-h_1=2g-\frac{1}{2}g=\frac{3}{2}g$
In 3rd second distance covered
$=h_3-h_2=\frac{9}{2}g-2g=\frac{5}{2}g$
$\therefore$ Ratio of distance covered in successive seconds is $1:3:5:7:9$
As there is a common difference between all the successive terms, so it is an A.P. We can also write the following ratios like $1^2:2^2-1^2:3^2-2^2:4^2-3^2$
Thus, it forms a series corresponding to the differences of the squares of the successive whole numbers.