Chapter 3.8, Problem 1E

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet.

(a) Find the velocity at time t.

(b) What is the velocity after 1 second?

(c) When is the particle at rest?

(d) When is the particle moving in the positive direction?

(e) Find the total distance traveled during the first 6 seconds.

(f) Draw a diagram like Figure 2 to illustrate the motion of the particle.

(g) Find the acceleration at time t and after 1 second.

(h) Graph the position, velocity, and acceleration functions for 0 ≤ t ≤ 6.

(i) When is the particle speeding up? When is it slowing down?

FIGURE 2

f(t) = t³ – 8t² + 24t

To determine

To find: The velocity at time t.

Answer to Problem 1E

The velocity at time is $\underset{\_}{3t^2-16t+24\text{ft/sec}}$.

Explanation of Solution

Given:

The given equation is as below.

$s=f\left(t\right)=t^3-8t^2+24t$ (1)

Calculation:

Calculate the velocity at time $t$.

Differentiate the equation (1) with respect to time.

$v\left(t\right)=f\text{'}\left(t\right)=3t^2-16t+24$ (2)

Therefore, the velocity at time $t$ is $\underset{\_}{3t^2-16t+24\text{ft/sec}}$.

To determine

To find: The velocity after 1 second.

Answer to Problem 1E

The velocity after 1 second is $\underset{\_}{v\left(1\right)=11\text{ft/sec}}$.

Explanation of Solution

Calculate the velocity after 1 second.

Substitute 1 for $t$ in the equation (2).

$\begin{array}{c}v\left(t\right)=f\text{'}\left(t\right)=3t^2-16t+24\\ v\left(1\right)=f\text{'}\left(1\right)=3{\left(1\right)}^2-16\left(1\right)+24\\ v\left(1\right)=11\text{ft/sec}\end{array}$

Therefore, the velocity after 1 second is $\underset{\_}{v\left(1\right)=11\text{ft/sec}}$.

To determine

To find: The time when particle at rest.

Answer to Problem 1E

The particle never is at rest.

Explanation of Solution

Calculate the time when particle will be at rest.

The velocity will be zero, when the particle is at rest.

Substitute 0 for $v\left(t\right)$ in the equation (2).

$\begin{array}{c}v\left(t\right)=3t^2-16t+24\\ 0=3t^2-16t+24\\ t=\frac{16t\pm \sqrt{{\left(16\right)}^2-4\left(3\right)\left(24\right)}}{2\left(3\right)}\\ =\frac{16\pm \sqrt{-32}}{6}\end{array}$

From the above equation, the value of time t doesn’t exist. Therefore, the particle never is at rest.

To determine

To find: The particle moving in the positive direction.

Answer to Problem 1E

The velocity of particle always moves in positive direction.

Explanation of Solution

Calculate the time at which the particle will be moving in the positive direction.

If speed is positive, the particle moves in positive direction whereas the speed is negative, the particle moves in negative direction.

Substitute 0 for $t$ in the equation (2).

$\begin{array}{c}v\left(t\right)=3t^2-16t+24\\ v\left(0\right)=3{\left(0\right)}^2-16\left(0\right)+24\\ =24\end{array}$

Therefore, the velocity at $t=0$ is positive and the velocity never get zero since the value of $t$ will be positive. Hence, the particle will always move in positive direction.

To determine

To find: The total distance traveled during the first 6 seconds.

Answer to Problem 1E

The total distance travelled during first 6 seconds is $\underset{\_}{f\left(6\right)=72\text{ft}}$.

Explanation of Solution

Calculate the total distance traveled during first 6 seconds.

Substitute 0 for $t$ in the equation (1).

$\begin{array}{c}f\left(t\right)=t^3-8t^2+24t\\ f\left(0\right)=\left(0^3\right)-8{\left(0\right)}^2+24\left(0\right)\\ f\left(0\right)=0\end{array}$

Substitute 6 for $t$ in the equation (1).

$\begin{array}{c}f\left(t\right)=t^3-8t^2+24t\\ f\left(6\right)=\left(6^3\right)-8{\left(6\right)}^2+24\left(6\right)\\ f\left(6\right)=72\end{array}$

Therefore, the total distance travelled during first 6 seconds is $\underset{\_}{f\left(6\right)=72\text{ft}}$.

To determine

To find: The diagram to illustrate the motion of the particle.

Answer to Problem 1E

The diagram is shown in the figure (1).

Explanation of Solution

Show the diagram to illustrate the motion of the particle as shown below in figure (1).

To determine

To find: The acceleration at time t and after 1 second.

Answer to Problem 1E

The acceleration at time is $\underset{\_}{f"\left(t\right)=6t-16}$ and after one second is $\underset{\_}{-10{\text{ft/s}}^2}$.

Explanation of Solution

Calculate the acceleration at time t.

Differentiate the equation (2) with respect to t.

$\begin{array}{l}f\text{'}\left(t\right)=3t^2-16t+24\\ f"\left(t\right)=6t-16\end{array}$

Therefore, the acceleration at time is $\underset{\_}{\left(6t-16\right){\text{ft/s}}^2}$.

Calculate the acceleration after 1 second.

Substitute 1 for $t$ in the above equation.

$\begin{array}{c}f"\left(t\right)=6t-16\\ f"\left(1\right)=6\left(1\right)-16\\ =-10\raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\end{array}$

Therefore, the acceleration after 1 second is $\underset{\_}{-10{\text{ft/s}}^2}$

To determine

To sketch: The graph the position, velocity, and acceleration function for $0\le t\le 6$.

Explanation of Solution

Calculate the position using the formula.

$f\left(t\right)=t^3-8t^2+24t$

Substitute 0 for $t$ in the above equation.

$\begin{array}{c}f\left(t\right)=t^3-8t^2+24t\\ f\left(0\right)={\left(0\right)}^3-8{\left(0\right)}^2+24\left(0\right)\\ =0\end{array}$

Similarly, calculate the remaining values.

Calculate the value of $t$ and $f\left(t\right)$ as shown in the table (1).

$t$	$f\left(t\right)=t^3-8t^2+24t$
0	0
0.5	10.125
1	17
1.5	21.375
2	24
2.5	25.625
3	27
3.5	28.875
4	32
4.5	37.125
5	45
5.5	56.375
6	72

Calculate the velocity using the expression.

$v\left(t\right)=3t^2-16t+24$

Substitute 0 for $t$ in the above equation.

$\begin{array}{c}v\left(t\right)=3t^2-16t+24\\ v\left(0\right)=3{\left(0\right)}^2-16\left(0\right)+24\\ =24\end{array}$

Similarly, calculate the remaining values.

Calculate the value of $t$ and $v\left(t\right)$ as shown in the table (2).

$t$	$v\left(t\right)=t^3-8t^2+24t$
0	24
0.5	16.75
1	11
1.5	6.75
2	4
2.5	2.75
3	3
3.5	4.75
4	8
4.5	12.75
5	19
5.5	26.75
6	36

Calculate the acceleration using the formula.

$v\left(t\right)=6t-16$

Substitute 0 for $t$ in the above equation.

$\begin{array}{c}v\left(t\right)=6t-16\\ v\left(0\right)=6\left(0\right)-16\\ =-16\end{array}$

Similarly, calculate the remaining values.

Calculate the value of $t$ and $a\left(t\right)$ as shown in the table (3).

$t$	$a\left(t\right)=t^3-8t^2+24t$
0	-16
0.5	Z-13
1	-10
1.5	-7
2	-4
2.5	-1
3	2
3.5	5
4	8
4.5	11
5	14
5.5	17
6	20

Draw the position as a function of time curve as shown in the figure (1).

Draw the speed as a function of time curve as shown in the figure (2).

Draw the acceleration as a function of time curve as shown in the figure (3).

To determine

To find: The time when the particle is speeding up and slowing down.

Answer to Problem 1E

The time when the particle is speeding up is when time t is greater than $\frac{8}{3}\text{seconds}$, and the time when the particle is slowing down is when time is lesser than $\frac{8}{3}\text{seconds}$.

Explanation of Solution

Calculate the time when particle is speeding up and slowing down.

Substitute 0 for $f"\left(t\right)$ in the equation $f"\left(t\right)=6t-16$.

$\begin{array}{l}f"\left(t\right)=6t-16\\ 0=6t-16\\ 6t=16\\ t=\frac{16}{6}\\ t=\frac{8}{3}\sec \end{array}$

Substitute $\frac{9}{3}\sec$ for $t$ in the equation $f"\left(t\right)=6t-16$.

$\begin{array}{c}f"\left(t\right)=6t-16\\ f"\left(\frac{9}{3}\right)=6\left(\frac{9}{3}\right)-16\\ =2\end{array}$

Substitute $\frac{7}{3}\sec$ for $t$ in the equation $f"\left(t\right)=6t-16$.

$\begin{array}{c}f"\left(t\right)=6t-16\\ f"\left(\frac{7}{3}\right)=6\left(\frac{7}{3}\right)-16\\ =-2\end{array}$

Therefore, the acceleration is positive when the value of time is $t>\frac{8}{3}\sec$ and this will be speeding up and the acceleration is negative when the value of time is $t<\frac{8}{3}\sec$ and this will be slowing down.