The position of a particle as a function of time t, is given by x(t) = at + b t ^ 2 - c t ^ 3 where a, b and c are constants. When the particles zero acceleration, then its velocity will be:

Correct answer: A. a + b ^ 2 3c x ( t ) = at + b t ^ 2 - c t ^ 3 velocity v ( t ) = (d)/(dt) ( x ) = (d)/(dt)(at + b t ^ 2 - c t ^ 3 ) = a + 2bt - 3c t ^ 2 Acceleration = (d)/(dt)v ( t ) = 2b - 6tc acceleration = 0 ⇒ 2b - 6tc = 0 t = (b)/(3c) ∴ velocity when t = (b)/(3c), v _ (t = (b)/(3c)) = a + 2b((b)/(3c)) - 3 ((b)/(3c)) ^ 2 = a + 2 b ^ 2 3c - b ^ 2 3c = a + b ^ 2 3c

JEE Main 20199 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 9 April
Shift: Evening Shift
Subject: Physics
Chapter: Motion in One Dimension

JEE Main 2019 · 9 April · Evening Shift · Physics · Motion in One Dimension

The position of a particle as a function of time t, is given by...

The position of a particle as a function of time

t,

is given by

x\left(t\right) = at + b{t}^{2} - c{t}^{3}

where

a, b

and

c

are constants. When the particles zero acceleration, then its velocity will be:

Options

A $a + \dfrac{{b}^{2}}{3c}$
B $a + \dfrac{{b}^{2}}{2c}$
C $a + \dfrac{{b}^{2}}{c}$
D $a + \dfrac{{b}^{2}}{4c}$

Correct answer

A. $a + \dfrac{{b}^{2}}{3c}$

Step-by-step solution

x ( t ) = at + b{t}^{2} - c{t}^{3}

velocity

v ( t ) = \dfrac{d}{dt} ( x ) = \dfrac{d}{dt}\left(at + b{t}^{2} - c{t}^{3}\right)

= a + 2bt - 3c{t}^{2}

Acceleration

= \dfrac{d}{dt}v ( t ) = 2b - 6tc

\text{acceleration} = 0 ⇒ 2b - 6tc = 0

t = \dfrac{b}{3c}

∴

velocity when

t = \dfrac{b}{3c},

{v}_{\left(t = \dfrac{b}{3c}\right)} = a + 2b\left(\dfrac{b}{3c}\right) - 3{\left(\dfrac{b}{3c}\right)}^{2}

= a + \dfrac{2{b}^{2}}{3c} - \dfrac{{b}^{2}}{3c}

= a + \dfrac{{b}^{2}}{3c}

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