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JEE PYQ: Motion in a Straight Line - Question ID e0f9e0ad9245 (JEE Main 2019)

ID: e0f9e0ad9245JEE Main 2019Single Correct MCQ

The position of a particle as a function of time t, is given by
x(t) = at + bt² – ct³
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

Select Option

Step-by-step Explanation

Core Formula & Concept:

In kinematics, the motion of a particle along a straight line is described by its position function $x(t)$ . The velocity $v(t)$ and acceleration $a(t)$ are obtained by successive time derivatives of $x(t)$ :

Velocity: $v(t) = \frac{dx}{dt}$
Acceleration: $a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

When the acceleration becomes zero, the particle’s velocity at that instant is what we must find.

Step-by-Step Derivation:

Given position function:

x(t) = a\,t + b\,t^2 - c\,t^3

Step 1 – Compute velocity $v(t)$ :

v(t) = \frac{dx}{dt} = \frac{d}{dt}\bigl(a\,t + b\,t^2 - c\,t^3\bigr)

v(t) = a + 2b\,t - 3c\,t^2

Step 2 – Compute acceleration $a(t)$ :

a(t) = \frac{dv}{dt} = \frac{d}{dt}\bigl(a + 2b\,t - 3c\,t^2\bigr)

a(t) = 2b - 6c\,t

Step 3 – Find the time $t_0$ when acceleration is zero:

Set

a(t_0) = 0

2b - 6c\,t_0 = 0

6c\,t_0 = 2b

t_0 = \frac{2b}{6c} = \frac{b}{3c}

Step 4 – Compute velocity at $t = t_0$ :

Substitute

t_0 = \frac{b}{3c}

into

v(t)

v(t_0) = a + 2b\Bigl(\frac{b}{3c}\Bigr) - 3c\Bigl(\frac{b}{3c}\Bigr)^2

Simplify each term:

$2b \cdot \frac{b}{3c} = \frac{2b^2}{3c}$
$3c \cdot \frac{b^2}{9c^2} = \frac{b^2}{3c}$

Hence

v(t_0) = a + \frac{2b^2}{3c} - \frac{b^2}{3c} = a + \frac{b^2}{3c}

Conclusion:

The velocity when acceleration is zero is

a + \frac{b^2}{3c}

, which corresponds to option C. Common Traps & Exam Tip:

Many students mistakenly stop after finding $t_0$ and forget to substitute back into the velocity expression. Others confuse the order of derivatives, leading to incorrect expressions for $v(t)$ or $a(t)$ . Always double-check each derivative and substitution step.

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