Question

Derive an equation for the distance covered by a uniformly accelerated body in $n^{th}$ second of its motion. A body travels half its total path in the last second of its fall from rest. Calculate the time of its fall.

Answer 1

Consider a particle moving with uniform acceleration ‘a’.Let u be the initial velocity of the particle.

Distance travelled during nth second = Distance travelled in ‘n’ seconds – Distance travelled in (n-1) seconds

$s_n=un+1/2a{n}^2–\left[u\right(n–1)+1/2a(n–1{)}^2]$

$=un+1/2a{n}^2–[un–u+1/2a(n^2–2n+1\left)\right]$

$=un+1/2a{n}^2–[un–u+1/2an²–an+1/2a]$

$=un+1/2a{n}^2–un+u–1/2a{n}^2+an–1/2a$

$=u+an–1/2a$

$s_n=u+a(n–1/2)$

$s_n=4+a(n–1/2)$

Given $u=0,a=g$

$s=1/2g{t}^2=1/2g{n}^2$

By the given condition $1/2s=s_n$

$\Rightarrow \frac{g{n}^2}{4}=g(n–1/2)$

$\Rightarrow n^2–4n+2=0$

$\Rightarrow n=\frac{4\pm \sqrt{16-8}}{2}$

$\Rightarrow n=2\pm \sqrt{2}$

∴ Time of the body’s fall $=2\pm \sqrt{2}$ s