Chapter 4.3, Problem 28E

a.

To determine

To find the velocity of the particle.

Answer to Problem 28E

The velocity is $6t-6t$ .

Explanation of Solution

Given:

The given function is $x(t)=3t^2-2t^3$

Calculation:

Let $x(t)$ is displacement then $v(t)$ is the velocity of the particle.

  $\begin{array}{l}v(t)=\frac{dx(t)}{dt}\\ v(t)=\frac{d}{dx}(3t^2-2t^3)\\ v(t)=6t-6t^2\end{array}$

b.

To determine

To find the acceleration of the particle.

Answer to Problem 28E

The acceleration of the particle is $6-12t$ .

Explanation of Solution

Given:

The given function is $x(t)=3t^2-2t^3$

Calculation:

Let $x(t)$ is displacement then $v(t)$ is the velocity of the particle and $a(t)$ is the acceleration of the particle.

  $\begin{array}{l}v(t)=\frac{dx(t)}{dt}\\ v(t)=\frac{d}{dx}(3t^2-2t^3)\\ v(t)=6t-6t^2\end{array}$

  $\begin{array}{l}a(t)=\frac{dv(t)}{dt}\\ a(t)=\frac{d}{dt}(6t-6t^2)\\ a(t)=6-12t\end{array}$

C.

To determine

To describe the motion of the particle

Answer to Problem 28E

The negative direction is $x=1$ . The positive direction is $x=0$ at $t=1$ .

Explanation of Solution

Given:

The given function is $x(t)=3t^2-2t^3$

Calculation:

Let $x(t)$ is displacement then $v(t)$ is the velocity of the particle and $a(t)$ is the acceleration of the particle.

  $\begin{array}{l}v(t)=\frac{dx(t)}{dt}\\ v(t)=\frac{d}{dx}(t^3-3t+3)\\ v(t)=3t^2-3\end{array}$

  $\begin{array}{l}a(t)=\frac{dv(t)}{dt}\\ a(t)=\frac{d}{dt}(3t^2-3)\\ a(t)=6t\end{array}$

It begins at position $x=0$ . It starts moving in the position direction until it reaches position $x=1$ when $t=1$ , and then changes direction, moving in a negative direction thereafter.