Question

A body is moving from rest under constant acceleration and let $S_1$ be the displacement in the first $(p-1)$ seconds and $S_2$ be the displacement in the first $p$ seconds. The displacement in $(p^2-p+1{)}^{th}$ second 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

$S_1+S_2$

From $S=ut+\frac{1}{2}a{t}^2$

$S_1=\frac{1}{2}a(p-1{)}^2$ and $S_2=\frac{1}{2}a{p}^2$ [As u = 0]

We have

$S_1+S_2=\frac{a}{2}[2p^2-2p+1]$

Using the expression for distance traveled in $n^{th}$ second

$S_n=u+\frac{a}{2}(2n-1)$

$S_{(p^2-p+1{)}^{th}}=\frac{a}{2}\left[2\right(p^2-p+1)-1]=\frac{a}{2}[2p^2-2p+1]$

It is clear that $S_{(p^2-p+1{)}^{th}}=S_1+S_2$