A body is moving from rest under constant acceleration and let S _ 1 be the displacement in the first (P - 1) seconds and S _ 2 be the displacement in the first P seconds. The displacement in (P^ 2 - P + 1) ^ th &

Correct answer: A. S _ 1 + S _ 2 From the equation of motion S = ut + (1)/(2)at^ 2 S _ 1 = 0 + (1)/(2)a (P - 1) ^ 2 And S _ 2 = 0 + (1)/(2)aP^ 2 From S _ n = u + (a)/(2) (2n - 1) S _ (P^ 2 - P + 1) ^ th = (a)/(2) lleft[2(P^ 2 - P + 1) - 1] =(a)/(2)[2P^ 2 - 2P + 1] = S _ 1 + S _ 2

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Motion in One Dimension

NTA Abhyas 2020 · Physics · Motion in One Dimension

A body is moving from rest under constant acceleration and let S _...

A body is moving from rest under constant acceleration and let

{S}_{1}

be the displacement in the first

\left(P - 1\right)

seconds and

{S}_{2}

be the displacement in the first

P

seconds. The displacement in

{\left(P^{2} - P + 1\right)}^{th}

& second will be

Options

A ${S}_{1} + {S}_{2}$
B $\sqrt{{S}_{1}{S}_{2}}$
C ${S}_{1} - {S}_{2}$
DNone of these

Correct answer

A. ${S}_{1} + {S}_{2}$

Step-by-step solution

From the equation of motion

S = ut + \dfrac{1}{2}at^{2}

{S}_{1} = 0 + \dfrac{1}{2}a{\left(P - 1\right)}^{2}

And

{S}_{2} = 0 + \dfrac{1}{2}aP^{2}

From

{S}_{n} = u + \dfrac{a}{2} \left(2n - 1\right)

{S}_{{ \left(P^{2} - P + 1\right)}^{th}} = \dfrac{a}{2}\l

lleft[2\left(P^{2} - P + 1\right)

- 1]

=\dfrac{a}{2}\left[2P^{2} - 2P + 1\right] = {S}_{1} + {S}_{2}

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