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) $\dfrac{2v\left( n-1 \right)}{n}$
B) $\dfrac{v\left( n-1 \right)}{n}$
C) $\dfrac{v\left( n+1 \right)}{n}$
D) $\dfrac{2v\left( 2n+1 \right)}{n}$

Answer 1

Hint: There are three equations of motion that relate the quantities such as initial and final velocity, acceleration and displacement. When the displacement is asked, and the inputs are time and acceleration, the formula to be used is:
$s=ut+\dfrac{1}{2}\left( a{{t}^{2}} \right)$
where u = initial velocity, a = acceleration and t = time.

Complete step by step answer:
For this problem, the displacement at last two seconds, formula for displacement covered by the particle in n seconds must be applied, by substituting n in place of the quantity t in the equation of motion as demonstrated below –
Displacement covered at ‘n’ seconds –
${{s}_{n}}=un +\dfrac{1}{2}\left( a{{n}^{2}} \right)$
Since the displacement required in the last two seconds is asked in the question, we have to consider the displacement covered in ‘n-2’ seconds.
Displacement covered at ‘n-2’ seconds –
$\Rightarrow {{s}_{n-2}}=u\left( n-2 \right)+\dfrac{1}{2}\left( a{{\left( n-2 \right)}^{2}} \right)$
The displacement covered in the last 2 seconds is equal to the difference in the displacement covered in n seconds and the displacement in $\left( n-2 \right)$ seconds.
$\Rightarrow s={{s}_{n}}-{{s}_{n-2}}$
Substituting the values of displacement in n and $\left( n-2 \right)$ seconds in the above equation, we have –
$\Rightarrow s=un+\dfrac{1}{2}\left( a{{n}^{2}} \right)-\left[ u\left( n-2 \right)+\dfrac{1}{2}\left( a{{\left( n-2 \right)}^{2}} \right) \right]$
Since it is mentioned in the question that the particle starts from rest, it is implied that the initial velocity is zero. Hence, $u=0$
Substituting $u=0$ in the above equation,
$\Rightarrow s=\dfrac{1}{2}\left( a{{n}^{2}} \right)-\left[ \dfrac{1}{2}\left( a{{\left( n-2 \right)}^{2}} \right) \right]$
$\Rightarrow s=\dfrac{1}{2}a\left[ {{n}^{2}}-{{\left( n-2 \right)}^{2}} \right]$
$\Rightarrow s=\dfrac{1}{2}a\left[ {{n}^{2}}-\left( {{n}^{2}}+4-4n \right) \right]$
$\Rightarrow s=\dfrac{1}{2}a\left[ {{n}^{2}}-{{n}^{2}}-4+4n \right]$
$\Rightarrow s=\dfrac{1}{2}a\left[ 4n-4 \right]$
$\Rightarrow s=\dfrac{1}{2}a\times 4\times \left( n-1 \right)$
$\Rightarrow s=2a\left( n-1 \right)$
The velocity of the particle when it covers the distance in n seconds with the acceleration, a is given by –
$\Rightarrow {{v}^{2}}-{{u}^{2}}=2a{{s}_{n}}$
Substituting the value of ${{s}_{n}}$, we get –
$\Rightarrow {{v}^{2}}-{{u}^{2}}=2a\left( un+\dfrac{1}{2}a{{n}^{2}} \right)$
Since the particle starts from rest, it is implied that the initial velocity is zero. Hence, $u=0$.
$\Rightarrow {{v}^{2}}=2a\times \dfrac{1}{2}a{{n}^{2}}$
$\Rightarrow {{v}^{2}}={{a}^{2}}{{n}^{2}}$
$\Rightarrow v=an$
$\Rightarrow a=\dfrac{v}{n}$
Substituting the value of acceleration, a, in the value of displacement equation, we get –
$\Rightarrow s=2a\left( n-1 \right)$
$\therefore s=\dfrac{2v\left( n-1 \right)}{n}$

Hence, the correct option is Option A.

Note: The result of this derivation is very important to memorize as a side note since there could be a question on these lines in competitive exams where instead of ‘n’ seconds, they could ask “velocity of v after 4 seconds” for example, the value of n can be substituted directly, in the expression to obtain the displacement in a single step instead of deriving the expression all over again.