Chapter 2.4, Problem 22E

a.

To determine

To calculate: The instantaneous velocity.

Answer to Problem 22E

The instantaneous velocity is $3t^2-6t+8$ .

Explanation of Solution

Given information:

The function is $s\left(t\right)=t^3-6t^2+8t+2$ and time is $t$

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

Calculation:

The function is $s\left(t\right)=t^3-6t^2+8t+2$ .

Differentiate the function with respect to time $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left[t^3-6t^2+8t+2\right]\\ =\frac{d}{dt}\left(t^3\right)-6\frac{d}{dt}\left(t^2\right)+8\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(2\right)\\ =3t^{3-1}-6\left(2t^{2-1}\right)+8t^{1-1}+0\\ =3t^2-12t+8\end{array}$

Conclusion: The instantaneous velocity of the particle is. $3t^2-6t+8$ .

b.

To determine

To calculate: The acceleration of the particle.

Answer to Problem 22E

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

Explanation of Solution

Given information:

The function is $s\left(t\right)=t^3-6t^2+8t+2$ and time is $t$

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

Calculation:

The function is $s\left(t\right)=t^3-6t^2+8t+2$ .

Differentiate the function with respect to time $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left[t^3-6t^2+8t+2\right]\\ =\frac{d}{dt}\left(t^3\right)-6\frac{d}{dt}\left(t^2\right)+8\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(2\right)\\ =3t^{3-1}-6\left(2t^{2-1}\right)+8t^{1-1}+0\\ =3t^2-12t+8\end{array}$

Differentiate the function again with respect to time $t$ .

  $\begin{array}{c}\frac{d}{dt}\left(\frac{ds}{dt}\right)=\frac{d}{dt}\left(3t^2-12t+8\right)\\ =\frac{d}{dt}\left(3t^2\right)-12\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(8\right)\\ =3\left(2t^{2-1}\right)-12.1-0\\ =6t-12\end{array}$

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

c.

To determine

To calculate: The time at which the particle is at rest.

Answer to Problem 22E

The times when particle is at rest are $t=3.2$ , $t=0.9$ .

Explanation of Solution

Given information:

The function is $s\left(t\right)=t^3-6t^2+8t+2$ and time is $t$ .

Concept used:

The particle will be in rest when the velocity of the particle becomes zero, quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

The function is $s\left(t\right)=t^3-6t^2+8t+2$ .

Differentiate the function with respect to time $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left[t^3-6t^2+8t+2\right]\\ =\frac{d}{dt}\left(t^3\right)-6\frac{d}{dt}\left(t^2\right)+8\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(2\right)\\ =3t^{3-1}-6\left(2t^{2-1}\right)+8t^{1-1}+0\\ =3t^2-12t+8\end{array}$

Equate the velocity to zero.

  $\begin{array}{c}\frac{ds}{dt}=0\\ 3t^2-12t+8=0\\ t=\frac{-\left(-12\right)\pm \sqrt{144-4\times \left(3\right)\times \left(8\right)}}{6}\\ t=\frac{12\pm \sqrt{144-96}}{6}\end{array}$

Solve further.

  $\begin{array}{c}t=\frac{12\pm 4\sqrt{3}}{6}\\ t=\frac{12}{6}+\frac{4\sqrt{3}}{6},\frac{12}{6}-\frac{4\sqrt{3}}{6}\\ =\frac{12+4\sqrt{3}}{6},\frac{12-4\sqrt{3}}{6}\\ =3.2,0.9\end{array}$

Conclusion: The particle will be in rest when $t=3.2$ , $t=0.9$ .

d.

To determine

To calculate: The time when particle changes its direction.

Explanation of Solution

Given information:

The function is $s\left(t\right)=t^3-6t^2+8t+2$ and time is $t$ .

Calculation:

The function is $s\left(t\right)=t^3-6t^2+8t+2$ .

The graph of the function is shown below.

  

The motion of the particle can be divided into six phases

(a) The particle starts when $t=0$ , where $s=2$ .

(b) The particle changes direction and continues its motion until at $t=0.85$ where $s=5.07$

(c) The particle stops at $t=2.5$ when the distance that is $s=0$

(d) The particle changes direction and continues its motion until at $t=3.21$ where $s=-1.1$

(e) The particle stops at $t=3.7$ when the distance that is $s=0$

(f) Then it stops and continues its motion.