Question

The velocity of a moving particle along the x-axis is given as $v = {x^2} - 5x + 4$ (in $m.{s^1}$ ), where $x$ denotes the x-coordinate of the particle in meters. Find the magnitude of the acceleration of the particle when the velocity of the particle is zero?
A. $0\,m.{s^{ - 2}}$
B. $2\,m.{s^{ - 2}}$
C. $3\,m.{s^{ - 2}}$
D. None of these

Answer 1

Hint: As per the question, we have the velocity of a moving particle, so we will go through the formula of acceleration in the terms of change in velocity with respect to the time. And the $x$ term in the velocity equation describes that there is no change in velocity because $x$ lies on the x-coordinates of the particle.

Complete step by step answer:
Given that, velocity of a moving particle, $v = {x^2} - 5x + 4$. And, as we know that acceleration is the change in velocity with respect to the time:
$a = \dfrac{{dv}}{{dt}}$
So, now we will differentiate the given velocity of the particle:
$\Rightarrow a = \dfrac{{d({x^2} - 5x + 4)}}{{dt}} \\
\Rightarrow a = 2x\dfrac{{dx}}{{dt}} - 5x\dfrac{{dx}}{{dt}} \\ $
Now, according to the question, the $x$ in the equation of the velocity denotes the x-coordinates of the particles, so we have-
$\because v = \dfrac{{dx}}{{dt}} = 0$
So,
$a = 2x \times 0 - 5 \times 0 \\
\therefore a = 0\,m.{s^{ - 2}} $
Therefore, we got a conclusion here that, when the velocity is zero then the acceleration is also zero.

Hence, the correct option is A.

Note: Acceleration here refers to the rate of change in velocity, thus you'll need the change in velocity. If the question implies that the velocity is always zero, then the acceleration is also zero. However, if the end velocity is zero, the acceleration is negative, resulting in deceleration.