Question

A body starts from rest with uniform acceleration and remains in motion for $n$ seconds. If its final velocity after $n$ second is $v$, then its displacement in the last two seconds will be

• A. $\frac{2v(n+1)}{n}$
• B. $\frac{v(n+1)}{n}$
• C. $\frac{v(n-1)}{n}$
• D. $\frac{2v(n-1)}{n}$

Answer 1

The correct option is D $\frac{2v(n-1)}{n}$

The distance travelled by the particle in the last $2$ seconds $=$ distance travelled by $n^{th}$ second $-$distance travelled by $(n-2)th$ second.

$\because v=0+na\Rightarrow a=\frac{v}{n}$
$\therefore$ displacement in last two seconds,
$=S_n-S_{n-2}=\frac{1}{2}a{n}^2-\frac{1}{2}a(n-2{)}^2$
$=\frac{a}{2}[n^2-(n-2{)}^2]=\frac{a}{2}[n+(n-2\left)\right][n-(n-2\left)\right]$
$=a(2n-2)=\frac{v}{n}(2n-2)=\frac{2v(n-1)}{n}$.