Question

A body starts from rest with a uniform acceleration. If its velocity after $n$ second is $v$, then its displacement in the last $2\text{s}$ is

• 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 $\frac{2v(n-1)}{n}$

By using first equation of motion
$v=0+na(\because u=0)$
$\Rightarrow a=\frac{v}{n}$
Distplacement covered in $n$ seconds is
$\Rightarrow S_n=\frac{1}{2}a{n}^2$ and distance travelled in $(n-2)$ seconds is
$S_{n-2}=\frac{1}{2}a(n-2{)}^2$
So, displacement travelled in the last $2\text{s}$ is
$S_n-S_{n-2}=\frac{1}{2}a{n}^2-\frac{1}{2}a(n-2{)}^2$
$=\frac{a}{2}[n^2-(n-2{)}^2]$
$S_n-S_{n-2}=\frac{a}{2}(n^2-n^2-4+4n)=2a(n-1)$
$S_n-S_{n-2}=\frac{2v(n-1)}{n}$