Q. Let $S_{1}$ and $S_{2}$ be the displacement in the first $\left(\right.p \, - \, 1\left.\right) \, sec$ and $p \, sec$ of a body moving from rest under constant acceleration. Find the displacement of the body in $\left(p^{2} - p + 1\right)^{th} \, sec$ .
NTA AbhyasNTA Abhyas 2022
Solution:
From the second equation of motion we know that
$S=ut+\frac{1}{2}at^{2}$
In $p-1$ seconds
$S_{1}=\frac{1}{2}a\left(P - 1\right)^{2}$
and in $p$ seconds
$S_{2}=\frac{1}{2}aP^{2}$ [As $u=0$ ]
With above equations we know that
$S_{1}+S_{2}=\frac{1}{2}a\left(\right.2p^{2}-2p+1\left.\right)$
From the displacement formula of nth second we know that $S_{n}=u+\frac{a}{2}\left(2 n - 1\right)$
$S_{\left(\right. P^{2} - P + 1 \left.\right)^{th}}=\frac{a}{2}\left[2 \left(P^{2} - P + 1\right) - 1\right]=\frac{a}{2}\left[2 P^{2} - 2 P + 1\right]$
It is clear that $S_{\left(\right. P^{2} - P + 1 \left.\right)^{th}}=S_{1}+S_{2}$
$S=ut+\frac{1}{2}at^{2}$
In $p-1$ seconds
$S_{1}=\frac{1}{2}a\left(P - 1\right)^{2}$
and in $p$ seconds
$S_{2}=\frac{1}{2}aP^{2}$ [As $u=0$ ]
With above equations we know that
$S_{1}+S_{2}=\frac{1}{2}a\left(\right.2p^{2}-2p+1\left.\right)$
$S_{\left(\right. P^{2} - P + 1 \left.\right)^{th}}=\frac{a}{2}\left[2 \left(P^{2} - P + 1\right) - 1\right]=\frac{a}{2}\left[2 P^{2} - 2 P + 1\right]$
It is clear that $S_{\left(\right. P^{2} - P + 1 \left.\right)^{th}}=S_{1}+S_{2}$