Chapter 3.8, Problem 2E

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

$f(t)=\frac{9t}{t^2+9}$

To determine

To find: The velocity at time t.

Answer to Problem 2E

The velocity at time t is $\underset{\_}{f\text{'}\left(t\right)=\frac{-9\left(t^2-9\right)}{{\left(t^2+9\right)}^2}\text{ft/sec}}$.

Explanation of Solution

Given:

The given equation is as below.

$s=f\left(t\right)=\frac{9t}{t^2+9}$ (1)

Calculation:

Calculate the velocity at time $t$.

Differentiate the equation (1) with respect to time by applying $\frac{u}{v}$ method.

$\begin{array}{c}f\text{'}\left(t\right)=\frac{\left(t^2+9\right)\left(9\right)-\left(9t\times \left(2t\right)\right)}{{\left(t^2+9\right)}^2}\\ =\frac{9t^2+81-18t^2}{{\left(t^2+9\right)}^2}\\ =\frac{-9t^2+81}{{\left(t^2+9\right)}^2}\end{array}$

$f\text{'}\left(t\right)=\frac{-9\left(t^2-9\right)}{{\left(t^2+9\right)}^2}$ (2)

Therefore, the velocity at time t is $\underset{\_}{f\text{'}\left(t\right)=\frac{-9\left(t^2-9\right)}{{\left(t^2+9\right)}^2}\text{ft/sec}}$.

To determine

To find: The velocity after 1 second.

Answer to Problem 2E

The velocity after 1 second is $\underset{\_}{v\left(1\right)=0.72\text{ft/sec}}$.

Explanation of Solution

Calculate the velocity after 1 second.

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

$\begin{array}{c}v\left(t\right)=f\text{'}\left(t\right)\\ =\frac{-9\left(t^2-9\right)}{{\left(t^2+9\right)}^2}\\ v\left(1\right)=\frac{-9\left(1-9\right)}{{\left(1+9\right)}^2}\\ =0.72\text{ft/sec}\end{array}$

Therefore, the velocity after 1 second is $\underset{\_}{v\left(1\right)=0.72\text{ft/sec}}$.

To determine

To find: The time when particle at rest.

Answer to Problem 2E

The time when particle at rest is $\underset{\_}{t=3\sec }$.

Explanation of Solution

Calculate the time when particle at rest.

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

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

$\begin{array}{l}v\left(t\right)=\frac{-9\left(t^2-9\right)}{{\left(t^2+9\right)}^2}\\ 0=\frac{-9\left(t^2-9\right)}{{\left(t^2+9\right)}^2}\\ -9\left(t^2-9\right)=0\\ t^2-9=0\\ t=3\sec \end{array}$

Therefore, the time when particle at rest is $\underset{\_}{t=3\sec }$.

To determine

To find: The particle moving in the positive direction.

Answer to Problem 2E

The particle always moves in positive direction when time is within the range $\underset{\_}{0\le t<3}$.

Explanation of Solution

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

If the particle moves in positive direction, the velocity at any time t will be greater than zero.

$\begin{array}{l}v\left(t\right)=\frac{-9\left(t^2-9\right)}{{\left(t^2+9\right)}^2}>0\\ \frac{-9\left(t^2-9\right)}{{\left(t^2+9\right)}^2}>0\\ -9\left(t^2-9\right)>0\\ t^2-9>0\\ t<3\end{array}$

Therefore, the particle moving in the positive direction when $\underset{\_}{0\le t<3}$.

To determine

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

Answer to Problem 2E

The total distance travelled during first 6 second is $\underset{\_}{1.8\text{ft}}$.

Explanation of Solution

Calculate the total distance traveled during first 6 seconds.

Here, the particle moves in positive and negative direction, the total distance traveled should be calculated between the intervals of $\left[0,3\right]$ and $\left[3,6\right]$.

Substitute 3 and 0 for $t$ between interval of $\left[0,3\right]$ in the equation (1) and subtract them.

$\begin{array}{c}\left|f\left(3\right)-f\left(0\right)\right|=\left|\frac{9\left(3\right)}{{\left(3\right)}^2+9}-\frac{9\left(0\right)}{{\left(0\right)}^2+9}\right|\\ =\left|\frac{3}{2}-0\right|\\ =\frac{3}{2}\end{array}$

Substitute 6 and 3 for t between intervals of $\left[3,6\right]$ for $t$ in the equation (1) and subtract them.

$\begin{array}{c}\left|f\left(6\right)-f\left(3\right)\right|=\left|\frac{9\left(6\right)}{{\left(6\right)}^2+9}-\frac{9\left(3\right)}{{\left(3\right)}^2+9}\right|\\ =\left|\frac{6}{5}-\frac{3}{2}\right|\\ =\frac{3}{10}\end{array}$

The total distance travelled is as given below.

$\begin{array}{c}f\left(6\right)=\frac{3}{2}+\frac{3}{10}\\ =1.8\text{ft}\end{array}$

Therefore, the total distance travelled during first 6 second is $\underset{\_}{f\left(6\right)=1.8\text{ft}}$.

To determine

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

Explanation of Solution

Show the diagram to illustrate the motion of the particle as shown below in Figure 1.

Figure 1 shows the movement of particle at different times.

To determine

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

Answer to Problem 2E

The acceleration at time t is $\underset{\_}{f^{"}\left(t\right)=\frac{18t\left(t^2-27\right)}{{\left(t^3+9\right)}^3}}$ and after one second is $\underset{\_}{-0.468{\text{ft/s}}^{\text{2}}}$.

Explanation of Solution

Calculate the acceleration at time t.

Differentiate the equation (2) with respect to t by applying $\frac{u}{v}$ method.

$\begin{array}{c}f\text{'}\left(t\right)=-9\frac{\left(t^2-9\right)}{{\left(t^2+9\right)}^2}\\ f"\left(t\right)=-9\frac{{\left(t^2+9\right)}^2\left(2t\right)-\left[\left(t^2-9\right)\left(2\left(t^2+9\right)\times 2t\right)\right]}{{\left[{\left(t^2+9\right)}^2\right]}^2}\\ =-9\frac{2t\left(t^2+9\right)\left[\left(t^2+9\right)-2\left(t^2-9\right)\right]}{{\left(t^2+9\right)}^4}\end{array}$

On further simplification,

$\begin{array}{c}f"\left(t\right)=\frac{-18t\left[t^2+9-2t^2+18\right]}{{\left(t^2+9\right)}^3}\\ =\frac{-18t\left(-t^2+27\right)}{{\left(t^2+9\right)}^3}\\ =\frac{18t\left(t^2-27\right)}{{\left(t^3+9\right)}^3}\end{array}$

Therefore, the acceleration at time is $\underset{\_}{f"\left(t\right)=\frac{18t\left(t^2-27\right)}{{\left(t^3+9\right)}^3}}$.

Calculate the acceleration after 1 second.

Substitute 1 for $t$ in the above equation.

$\begin{array}{c}f"\left(t\right)=\frac{18t\left(t^2-27\right)}{{\left(t^3+9\right)}^3}\\ f"\left(1\right)=\frac{18\left(1\right)\left({\left(1\right)}^2-27\right)}{{\left(1^3+9\right)}^3}\\ =-0.468\frac{\text{ft}}{{\text{s}}^{\text{2}}}\end{array}$

Therefore, the acceleration after 1 second is $\underset{\_}{f"\left(1\right)=-0.468\frac{\text{ft}}{{\text{s}}^{\text{2}}}}$.

To determine

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

Explanation of Solution

Calculate the position of the particle with respect to time using the expression.

$s\left(t\right)=\frac{9t}{t^2+9}$

Substitute 0 for $t$ in the above equation.

$\begin{array}{c}s\left(0\right)=\frac{9\left(0\right)}{0^2+9}\\ =0\end{array}$

Similarly, calculate the remaining values.

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

$t$	$s\left(t\right)=\frac{9t}{t^2+9}$
0	0
0.5	0.48649
1	0.9
1.5	1.2
2	1.38462
2.5	1.47541
3	1.5
3.5	1.48235
4	1.44
4.5	1.38462
5	1.32353
5.5	1.26115
6	1.2

Calculate the velocity using the formula.

$v\left(t\right)=\frac{-9\left(t^2-9\right)}{{\left(t^2+9\right)}^2}$

Substitute 0 for $t$ in the above equation.

$\begin{array}{c}v\left(0\right)=\frac{-9\left(0^2-9\right)}{{\left(0^2+9\right)}^2}\\ =1\end{array}$

Similarly, calculate the remaining values.

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

$t$	$v\left(t\right)=\frac{-9\left(t^2-9\right)}{{\left(t^2+9\right)}^2}$
0	1
0.5	0.92038
1	0.72
1.5	0.48
2	0.26627
2.5	0.10642
3	0
3.5	-0.0648
4	-0.1008
4.5	-0.1183
5	-0.1246
5.5	-0.1241
6	-0.12

Calculate the acceleration using the formula.

$a\left(t\right)=\frac{18t\left(t^2-27\right)}{{\left(t^3+9\right)}^3}$

Substitute 0 for $t$ in the above equation.

$\begin{array}{c}a\left(0\right)=\frac{18\left(0\right)\left(0^2-27\right)}{{\left(0^3+9\right)}^3}\\ =0\end{array}$

Similarly, calculate the remaining values.

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

$t$	$a\left(t\right)=\frac{18t\left(t^2-27\right)}{{\left(t^3+9\right)}^3}$
0	0.00
0.5	-0.32
1	-0.47
1.5	-0.35
2	-0.17
2.5	-0.06
3	-0.02
3.5	-0.01
4	0.00
4.5	0.00
5	0.00
5.5	0.00
6	0.00

Draw the graph of the position, velocity and acceleration functions as shown in the Figure 2.

To determine

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

Answer to Problem 2E

the acceleration is zero when time t is greater than $3\sqrt{3}\sec$ and the acceleration is negative when time t is less than $3\sqrt{3}\sec$.

Explanation of Solution

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

Substitute 0 for $f"\left(t\right)$ in the equation $f"\left(t\right)=\frac{18t\left(t^2-27\right)}{{\left(t^3+9\right)}^3}$.

$\begin{array}{l}0=\frac{18t\left(t^2-27\right)}{{\left(t^3+9\right)}^3}\\ t^2-27=0\\ t=\sqrt{27}\\ t=3\sqrt{3}\sec \end{array}$

Substitute $1\sec$ for $t$ in the equation $f"\left(t\right)=\frac{18t\left(t^2-27\right)}{{\left(t^3+9\right)}^3}$.

$\begin{array}{c}f"\left(1\right)=\frac{18\times \left(1\right)\left({\left(1\right)}^2-27\right)}{{\left({\left(1\right)}^3+9\right)}^3}\\ =-0.468<0\end{array}$

Substitute $4\sec$ for $t$ in the equation $f"\left(t\right)=\frac{18t\left(t^2-27\right)}{{\left(t^3+9\right)}^3}$.

$\begin{array}{c}{f}^{"}\left(4\right)=\frac{18\times \left(4\right)\left({\left(4\right)}^2-27\right)}{{\left({\left(4\right)}^3+9\right)}^3}\\ =0\end{array}$

Therefore, the acceleration is zero when time t is greater than $3\sqrt{3}\sec$ and the acceleration is negative when time t is less than $3\sqrt{3}\sec$.