Question

Prove that the distances traversed during equal intervals of time by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning from unity [namely $1:3:5:\mathrm{7........}$].

Answer 1

Let distance travelled in 1 st n seconds = $S_1$

Let distance travelled in next ' n ' seconds = $S_2$

Let distance in next ' n ' seconds = $S_3$ . . . . . . and so on .

Let distance ' 2n ' seconds = $d_{2n}$

Let distance travelled in '3n' seconds = $d_{3n}$ . . . . . . . . and so on .

So ,

$S_2=d_{2n}-S_1$ ;

$s_3=d_{3n}-d_{2n}$ ;

$S_4=d_{4n}-d_{3n}$ . . . and so on .

As body starts from rest , u = 0 ,

and $ac{c}^n$ = g

So, $s_1=\frac{1}{2g{n}^2}$

$s_2=\frac{1}{2g(2n{)}^2}-\frac{1}{2g{n}^2}$

Now, nth term of the sequence 1 , 2 . 3 , 4 ....is given by (2n - 1)

Hence $s_1:s_2:s_3:s_4:.....:s_n$

= $\frac{1}{2g}\left(n^2\right):\frac{1}{2g}\left(3n^2\right):\frac{1}{2g}\left(5n^2\right):\frac{1}{2g}\left(7n^2\right):....:\frac{1}{2g}(2n-1)n^2$

or $s_1:s_2:s_3:s_4:........:s_n$

= 1 : 3 : 5 : 7 : ....... : (2n - 1)

hence proved