Question

A body is moving from rest under constant acceleration and let $S_1$ be the displacement in the first $(p–1)\text{sec}$ and $S_2$ be the displacement in the first $p\text{sec}$. The displacement in $(p^2–p+1{)}^{\text{th}}\text{sec}$ will be

• A. $S_1+S_2$
• B. $S_1{S}_2$
• C. $S_1-S_2$
• D. $\frac{S_1}{S_2}$

Answer 1

The correct option is A $S_1+S_2$

Given, $u=0$, and let the constant acceleration be $a$
From $2^{nd}$ equation of motion we have
$S=ut+\frac{1}{2}a{t}^2$
So, for first $(p-1)$ sec the displacement of the body is given as
$S_1=\frac{1}{2}a(p-1{)}^2$
or, $S_1=\frac{a}{2}(p-1{)}^2$.......(i)
and, for first $\left(p\right)$ sec the displacement of the body is given as
$S_2=\frac{1}{2}a{p}^2$
or, $S_2=\frac{a}{2}{p}^2$........(ii)
Also, we know that the displacement travelled in $n^{th}$ second is given by
$S_n=u+\frac{a}{2}(2n-1)$
$\Rightarrow S_{(p^2-p+1{)}^{\text{th}}}=\frac{a}{2}\left[2\right(p^2-p+1)-1]$
$\Rightarrow S_{(p^2-p+1{)}^{\text{th}}}=\frac{a}{2}[2p^2-2p+1]$
$\Rightarrow S_{(p^2-p+1{)}^{\text{th}}}=\frac{a}{2}(p-1{)}^2+\frac{a}{2}{p}^2$
or, from eq (i) & (ii) we get
$S_{(p^2-p+1{)}^{\text{th}}}=S_1+S_2$