Question

A particle starts from rest with uniform acceleration a. Its velocity after n seconds is $v$ . The displacement of the body in the last two seconds is:

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

Answer 1

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

Distance travelled by the particle in the last two seconds will be $=$ Distance travelled by particle in nth second $-$ Distance travelled by particle in $(n-2)$th second

so,

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

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

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

so,

$s=(2n-2).a=2a.(n-1)$ ............................(1)

Now,

Velocity, $v=$ Acceleration $\times$ Time

$v=a\times n$

or

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

so, the displacement in the last two seconds will be

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