Question

The position of a particle is given by the function $x=2t^3-6t^2+12$m, where t is in s. a. At what time does the particle reach its minimum velocity? What is ${\left(V_x\right)}_{min}$? b. At what time is the acceleration zero?

Short answer

(a)The particle reaches its minimum velocity at $t=1s$. The minimum velocity of the particle is $-6\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

(b) The acceleration of the particle is zero at$t=1s$.

Step-by-step solution

Given information

The position of the particle is given by$x=2t^3-6t^2+12$
.

The velocity of the particle.

The velocity is given by $v_x=\frac{dx}{dt}$.

Differentiate the given function with respect to $t$.

$$\begin{gathered} v_x=\frac{d}{dt}\left(2t^3-6t^2+12\right) \\ =6t^2-12t \end{gathered}$$

Acceleration of the particle.

To find the acceleration, differentiate $v_x=6t^2-12t$with respect to $t$.

$$\begin{gathered} a_x=\frac{d}{dt}\left(6t^2-12t\right) \\ =12t-12 \end{gathered}$$

Substitute 0 for $a_x$.

$$\begin{gathered} 0=12t-12 \\ 12t=12 \\ t=1s \end{gathered}$$

Part (a)

The velocity of the particle at $t=1s$is

$$\begin{gathered} v_x=6{\left(1\right)}^2-12\left(1\right) \\ =6-12 \\ =-6\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

The particle reaches its minimum velocity at $t=1s$. The minimum velocity of the particle is $-6\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Part (b)

The acceleration of the particle is zero at$t=1s$.