The velocity of a particle is v = ( v _ 0 + gt + F t ^ 2 )m s ^ - 1 . Its position is x = 0 at t = 0; then its displacement after time ( t = 1s ) is :

Correct answer: B. v _ 0 + (g)/(2) + (F)/(3) v = v _ 0 + gt + F t ^ 2 (ds)/(dt) = v _ 0 + gt + F t ^ 2 ∫ ds = ∫ _ 0 ^ 1 ( v _ 0 + gt + F t ^ 2 )dt s = [ v _ 0 t + g t ^ 2 2 + F t ^ 3 3 ] _ 0 ^ 1 s = v _ 0 + (g)/(2) + (F)/(3)

JEE Main 202117 MarchEvening Shift

Exam: JEE Main
Year: 2021
Date: 17 March
Shift: Evening Shift
Subject: Physics
Chapter: Motion in One Dimension

JEE Main 2021 · 17 March · Evening Shift · Physics · Motion in One Dimension

The velocity of a particle is v = ( v _ 0 + gt + F t ^ 2 )m s ^ - 1...

The velocity of a particle is

v = \left({v}_{0} + gt + F{t}^{2}\right)m{s}^{- 1}

. Its position is

x = 0

t = 0;

then its displacement after time

( t = 1s )

is :

Options

A ${v}_{0} + g + F$
B ${v}_{0} + \dfrac{g}{2} + \dfrac{F}{3}$
C ${v}_{0} + \dfrac{g}{2} + F$
D ${v}_{0} + 2g + 3F$

Correct answer

B. ${v}_{0} + \dfrac{g}{2} + \dfrac{F}{3}$

Step-by-step solution

$v = {v}_{0} + gt + F{t}^{2}$

$\dfrac{ds}{dt} = {v}_{0} + gt + F{t}^{2}$

$∫ ds = {∫}_{0}^{1}\left({v}_{0} + gt + F{t}^{2}\right)dt$

$s = {\left[{v}_{0}t + \dfrac{g{t}^{2}}{2} + \dfrac{F{t}^{3}}{3}\right]}_{0}^{1}$

$s = {v}_{0} + \dfrac{g}{2} + \dfrac{F}{3}$

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