Question

A body starts from rest with uniform acceleration. If its velocity after n second is v, then its displacement in last two seconds 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 $\frac{2v(n-1)}{n}$

The distance traveled by the particle in the last 2 seconds = distance traveled by particle in nth second- distance traveled by particle in (n-2)th second

So,

$s=\frac{1}{2}a{n}^2-\frac{1}{2}a(n-2{)}^2$

$s=\frac{1}{2}a[n^2-n^2+4n-4]$

$s=\frac{a}{2}[4n-4]$

$s=(2n-2)a=2a(n-1).........\left(1\right)$

Now,

Velocity, $v=acceleration\times time$

$v=a\times n$

$a=\frac{v}{n}...........................\left(2\right)$

Substitute value of 'a' from (2) to (1) we get,

$s=\frac{2v(n-1)}{n}$