Chapter 4.3, Problem 26E

a.

To determine

To find the velocity of the particle.

Answer to Problem 26E

The velocity is $-2t-2$ .

Explanation of Solution

Given:

The given function is $x(t)=-t^2-2t+6$ .

Calculation:

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

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

b.

To determine

To find the acceleration of the particle.

Answer to Problem 26E

The acceleration of the particle is $-2$ .

Explanation of Solution

Given:

The given function is $x(t)=-t^2-2t+6$

Calculation:

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

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

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

C.

To determine

To describe the motion of the particle

Answer to Problem 26E

The negative direction is $x=6$ .

Explanation of Solution

Given:

The given function is $x(t)=t^2-4t+3$

Calculation:

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

  $\begin{array}{l}v(t)=\frac{dx(t)}{dt}\\ v(t)=\frac{d}{dx}(t^2-4t+3\\ v(t)=2t-4\end{array}$

  $\begin{array}{l}a(t)=\frac{dv(t)}{dt}\\ a(t)=\frac{d}{dt}(2t-4)\\ a(t)=2\end{array}$

It begins at position $x=6$ moving in a negative direction.