Question

A particle starts from rest with a uniform acceleration a. Its velocity after n second will be v. The displacement of the body in the last two second will be
$\begin{aligned} & A.\dfrac{2v\left( n-1 \right)}{n} \\\ & B.\dfrac{v\left( n-1 \right)}{n} \\\ & C.\dfrac{v\left( n+1 \right)}{n} \\\ & D.\dfrac{2v\left( 2n+1 \right)}{n} \\\ \end{aligned}$

Answer 1

Hint: The basic equations of motion will help you to solve this kind of problem.

Formula used: ${{v}^{2}}={{u}^{2}}+2as$
In this equation $v$ is final velocity, u will be the initial velocity, $s$ is the displacement and $a$ will be the acceleration of the body.Substituting the needed values in the appropriate equations of motion will give the answer.

Complete answer:
First of all let us know about the equations of motion in physics. In physics, equations of motion are the equations that explain the nature of a physical system on the basis of its motion as a function of time. In short, the equations of motion explains the characteristics of a physical system as a set of mathematical functions on the basis of changing variables.

As we all know, if
$t=n\sec$
Then

& {{S}_{n}}=\dfrac{1}{2}a{{n}^{2}} \\\ & \\\ \end{aligned}$$ And $${{v}^{2}}=2a{{S}_{n}}$$ Substituting the equation of displacement in this will give $${{v}^{2}}=2a\times \dfrac{1}{2}a{{n}^{2}}$$ Cancel the common terms will give, $${{v}^{2}}={{a}^{2}}{{n}^{2}}$$ Therefore, $$v=an$$ Rearranging the equation will give, $$a=\dfrac{v}{n}$$ Now we can write like this, $${{S}_{n}}-{{S}_{n-2}}=\dfrac{1}{2}a\left( {{n}^{2}}-{{\left( n-2 \right)}^{2}} \right)$$ Which means $$=\dfrac{1}{2}a\left( 4n-4 \right)$$ Taking 2 outside the bracket, $$=2a\left( n-1 \right)$$ Therefore we can write that, $${{S}_{n}}-{{S}_{n-2}}=\dfrac{2v\left( n-1 \right)}{n}$$ **So, the correct answer is “Option A”.** **Note:** Acceleration is the time rate of change of the velocity of an object. Acceleration is a vector quantity. That is it depends on both magnitude and direction. Acceleration of an object is in accordance with is given by the orientation of the net force acting on that object. Acceleration is directly proportional to the force applied. That is as the force increases, acceleration also increases.