Chapter 2.4, Problem 21E

a.

To determine

To calculate: The instantaneous velocity.

Answer to Problem 21E

The instantaneous velocity is $3t^2-16t+20$ .

Explanation of Solution

Given information:

The function is $s\left(t\right)={\left(t-2\right)}^2\left(t-4\right)$ and time is $t$

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ , product rule is $\frac{d}{dx}\left(u.v\right)=u.\frac{d}{dx}\left(v\right)+v\frac{d}{dx}\left(u\right)$ .

Calculation:

The function is $s\left(t\right)={\left(t-2\right)}^2\left(t-4\right)$ .

Differentiate the function with respect to time $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left[{\left(t-2\right)}^2\left(t-4\right)\right]\\ ={\left(t-2\right)}^2\frac{d}{dt}\left(t-4\right)+\left(t-4\right)\frac{d}{dt}{\left(t-2\right)}^2\\ ={\left(t-2\right)}^2.1+\left(t-4\right).2\left(t-2\right)\\ =t^2+4-4t+2t^2-12t+16\\ =3t^2-16t+20\end{array}$

Conclusion: The instantaneous velocity of the particle is $3t^2-16t+20$ .

b.

To determine

To calculate: The acceleration of the particle.

Answer to Problem 21E

The acceleration of the particle is $6t-16$ .

Explanation of Solution

Given information:

The function is $s\left(t\right)={\left(t-2\right)}^2\left(t-4\right)$ 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 $s\left(t\right)={\left(t-2\right)}^2\left(t-4\right)$ .

Differentiate the function with respect to time $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left[{\left(t-2\right)}^2\left(t-4\right)\right]\\ ={\left(t-2\right)}^2\frac{d}{dt}\left(t-4\right)+\left(t-4\right)\frac{d}{dt}{\left(t-2\right)}^2\\ ={\left(t-2\right)}^2.1+\left(t-4\right).2\left(t-2\right)\\ =t^2+4-4t+2t^2-12t+16\\ =3t^2-16t+20\end{array}$

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

  $\begin{array}{c}\frac{d}{dt}\left(\frac{ds}{dt}\right)=\frac{d}{dt}\left(3t^2-16t+20\right)\\ =\frac{d}{dt}\left(3t^2\right)-16\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(20\right)\\ =6t-16+0\\ =6t-16\end{array}$

Conclusion: The acceleration of the particle is $6t-16$ .

c.

To determine

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

Answer to Problem 21E

The times when particle is at rest are $t=\frac{10}{3}$ , $t=2$ .

Explanation of Solution

Given information:

The function is $s\left(t\right)={\left(t-2\right)}^2\left(t-4\right)$ and time is $t$ .

Concept used:

The particle will be in rest when the velocity of the particle becomes zero.

Calculation:

The function is $s\left(t\right)={\left(t-2\right)}^2\left(t-4\right)$ .

Differentiate the function with respect to time $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left[{\left(t-2\right)}^2\left(t-4\right)\right]\\ ={\left(t-2\right)}^2\frac{d}{dt}\left(t-4\right)+\left(t-4\right)\frac{d}{dt}{\left(t-2\right)}^2\\ ={\left(t-2\right)}^2.1+\left(t-4\right).2\left(t-2\right)\\ =t^2+4-4t+2t^2-12t+16\\ =3t^2-16t+20\end{array}$

Equate the velocity to zero.

  $\begin{array}{c}\frac{ds}{dt}=0\\ 3t^2-16t+20=0\\ 3t^2-6t-10t+20=0\\ 3t\left(t-2\right)-10\left(t-2\right)=0\end{array}$

Solve further.

  $\begin{array}{l}\left(3t-10\right)\left(t-2\right)=0\\ t=\frac{10}{3},2\end{array}$

Conclusion: The particle will be in rest when $t=\frac{10}{3}$ , $t=2$ .

d.

To determine

To calculate: The time when particle changes its direction.

Explanation of Solution

Given information:

The function is $s\left(t\right)={\left(t-2\right)}^2\left(t-4\right)$ and time is $t$ .

Calculation:

The function is $s\left(t\right)={\left(t-2\right)}^2\left(t-4\right)$ .

The graph of the function is shown below.

  

The motion of the particle can be divided into four phases

(a) The particle starts when $t=0$ , where $s=-16$ .

(b) The particle stops at $t=2$ when the distance that is $s=0$

(c) The particle changes direction and continues its motion until at $t=3.33$ where $s=-1.185$

(d) Then it stops and continues its motion.