Question

A particle moves along a straight line such that its displacement at any time t is givem by $\text{s}={\text{t}}^3-6{\text{t}}^2+3\text{t}+4$ The velocity, when its acceleration is zero, is

• A. $-9\text{m/s}$
• B. $42\text{m/s}$
• C. $-12\text{m/s}$
• D. $3\text{m/s}$

Answer 1

The correct option is A $-9\text{m/s}$

Find the velocity of the particle.
Formula Used: $\upsilon =\frac{\text{ds}}{\text{dt}}$
Given, displacement of the equation
$\text{s}={\text{t}}^3-6{\text{t}}^2+3\text{t}+4$
Velocity,
$\upsilon =\frac{\text{ds}}{\text{dt}}=3{\text{t}}^2-6\times 2\text{t}+3\times 1+0$
$\upsilon =3{\text{t}}^2-12\text{t}+3\dots \left(i\right)$
Find the acceleration of the particle.
Formula Used: $\text{a}=\frac{\text{ds}}{\text{dt}}$
Acceleration
$\text{a}=\frac{\text{dv}}{\text{dt}}=3\times 2\text{t}\times 1+0$
$a=6\text{t}-12\dots \left(ii\right)$
Find the velocity when acceleration of the particle is zero
Acceleration is zero at time 𝑡 given by from equation $\left(ii\right)$
$6\text{t}-12=0$
$\Rightarrow \text{t}=2\text{seconds}$
$\therefore$ Velocity $\upsilon$ at $\text{t}=2$ seconds is
$\upsilon =(3{\text{t}}^2-12\text{t}+3{)}_{\text{t}=2s}$
$=3\times \left(2^2\right)-12\times 2+3$
$\upsilon =-9\text{m/s}$