Question

A body starts from rest with an uniform acceleration. If its velocity after $n$ seconds is $v$, then its displacement in the last $2s$ 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

Answer: (D)

As v = 0 + na $\Rightarrow$ a = $\frac{v}{n}$
$\Rightarrow S_n=\frac{1}{2}a{n}^2$ and distance travelled in (n-2) second is
$S_{n-2}=\frac{1}{2}a(n-2{)}^2$
So distance travelled in the last 2 s is
$S_n-S_{n-2}=\frac{1}{2}a{n}^2-\frac{1}{2}a(n-2{)}^2$
$=\frac{a}{2}[n^2-(n-2{)}^2]$
$\frac{a}{2}[n+(n-2)][n-(n-2)]$
$=\frac{2v(n-1)}{n}$