Chapter 3.8, Problem 4E

(a)

To determine

To find: The velocity at time t.

Answer to Problem 4E

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

Explanation of Solution

Given:

The given equation is as below.

$s=f\left(t\right)=t^2{e}^{-t}$ (1)

Calculation:

Calculate the velocity at time $t$.

Differentiate the equation (1) with respect to time.

$\begin{array}{c}v\left(t\right)=f\text{'}\left(t\right)=t^2\left(-e^{-t}\right)+e^{-t}\left(2t\right)\\ =-e^{-t}{t}^2+2t{e}^{-t}\end{array}$

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

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

(b)

To determine

To find: The velocity after 1 second.

Answer to Problem 4E

The velocity after 1 second is $\underset{\_}{v\left(1\right)=0.3678\text{ft/sec}}$.

Explanation of Solution

Calculate the velocity after 1 second.

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

$\begin{array}{c}f\text{'}\left(t\right)=t{e}^{-t}\left(2-t\right)\\ f\text{'}\left(1\right)=1e^{-1}\left(2-1\right)\\ =0.3678\text{ft}/\text{s}\end{array}$

Therefore, the velocity after 1 second is $\underset{\_}{v\left(1\right)=0.3678\text{ft/sec}}$.

(c)

To determine

To find: The time when particle at rest.

Answer to Problem 4E

The particle never is at rest when $\underset{\_}{0<t<2\sec }$

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).

$\begin{array}{c}v\left(t\right)=t{e}^{-t}\left(2-t\right)\\ 0=t{e}^{-t}\left(2-t\right)\\ t=0\sec \\ 2-t=0\\ t=2\sec \end{array}$

Therefore, the particle at rest when $\underset{\_}{t=0\sec }$ and $\underset{\_}{t=2\sec }$.

(d)

To determine

To find: The particle moving in the positive direction.

Answer to Problem 4E

The particle will be moving in the positive direction within the time limits $0<t<2\sec$ and $t>2\sec$.

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.

$\begin{array}{c}v\left(t\right)=t{e}^{-t}\left(2-t\right)>0\\ 0<t<2;t>2\end{array}$

Therefore, the particle will be moving in the positive direction within the time limits $0<t<2\sec$ and $t>2\sec$.

(e)

To determine

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

Answer to Problem 4E

The total distance travelled during first 6 second is $\underset{\_}{f\left(6\right)=0.0892\text{ft}}$.

Explanation of Solution

Calculate the total distance traveled during first 6 seconds.

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

$\begin{array}{c}f\left(t\right)=t^2{e}^{-t}\\ f\left(6\right)=\left(6^2\right)e^{-6}\\ f\left(6\right)=0.0892\text{ft}\end{array}$

Therefore, the total distance travelled during first 6 seconds is $\underset{\_}{f\left(6\right)=0.0892\text{ft}}$.

(f)

To determine

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

Answer to Problem 4E

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).

(g)

To determine

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

Answer to Problem 4E

The acceleration at time is $\underset{\_}{e^{-t}\left(2-4t+t^2\right){\text{ft/s}}^2}$ and after one second is $\underset{\_}{-0.368{\text{ft/s}}^2}$.

Explanation of Solution

Calculate the acceleration at time t.

Differentiate the equation (2) with respect to t.

$\begin{array}{c}f"\left(t\right)=2e^{-t}-4e^{-t}t+e^{-t}{t}^2\\ f"\left(t\right)=e^{-t}\left(2-4t+t^2\right)\end{array}$

Therefore, the acceleration at time is $\underset{\_}{e^{-t}\left(2-4t+t^2\right){\text{ft/s}}^2}$.

Calculate the acceleration after 1 second.

Substitute 1 for $t$ in the above equation.

$\begin{array}{c}f"\left(t\right)=e^{-t}\left(2-4t+t^2\right)\\ f"\left(1\right)=e^{-\left(1\right)}\left(2-4\left(1\right)+{\left(1\right)}^2\right)\\ =0.368\left(-1\right)\end{array}$

$f^{"}\left(1\right)=-0.368\text{ft}/{\text{s}}^{\text{2}}$

Therefore, the acceleration after 1 second is $\underset{\_}{-0.368{\text{ft/s}}^2}$.

(h)

To determine

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

Answer to Problem 4E

The position, velocity, and acceleration functions are plotted for time limits $0\le t\le 6$ in figures (2), (3), and (4).

Explanation of Solution

Calculate the position using the formula.

$s=f\left(t\right)=t^2{e}^{-t}$

Substitute 0 for $t$ in the above equation.

$\begin{array}{c}f\left(t\right)=t^2{e}^{-t}\\ f\left(0\right)={\left(0\right)}^2{e}^{-0}\\ =0\end{array}$

Similarly, calculate the remaining values.

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

$t$	$s=f\left(t\right)=t^2{e}^{-t}$
0	0
0.5	0.15163
1	0.36788
1.5	0.50204
2	0.54134
2.5	0.51303
3	0.44808
3.5	0.36992
4	0.29305
4.5	0.22496
5	0.16845
5.5	0.12362
6	0.08924

Calculate the velocity using the expression.

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

Substitute 0 for $t$ in the above equation.

$\begin{array}{c}f\text{'}\left(t\right)=t{e}^{-t}\left(2-t\right)\\ f\text{'}\left(0\right)=\left(0\right)e^{-0}\left(2-0\right)\\ =0\end{array}$

Similarly, calculate the remaining values.

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

$t$	$f\text{'}\left(t\right)=t{e}^{-t}\left(2-t\right)$
0	0
0.5	0.4549
1	0.36788
1.5	0.16735
2	0
2.5	-0.1026
3	-0.1494
3.5	-0.1585
4	-0.1465
4.5	-0.125
5	-0.1011
5.5	-0.0787
6	-0.0595

Calculate the acceleration using the formula.

$f"\left(t\right)=e^{-t}\left(2-4t+t^2\right)$

Substitute 0 for $t$ in the above equation.

$\begin{array}{c}f"\left(t\right)=e^{-t}\left(2-4t+t^2\right)\\ f"\left(0\right)=e^{-0}\left(2-4\left(0\right)+{\left(0\right)}^2\right)\\ =2\end{array}$

Similarly, calculate the remaining values.

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

$t$	$f"\left(t\right)=e^{-t}\left(2-4t+t^2\right)$
0	2
0.5	0.15163
1	-0.3679
1.5	-0.3905
2	-0.2707
2.5	-0.1436
3	-0.0498
3.5	0.00755
4	0.03663
4.5	0.04721
5	0.04717
5.5	0.04189
6	0.0347

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

Draw the speed as a function of time curve as shown in the Figure (3).

Draw the acceleration as a function of time curve as shown in the Figure (4).

(i)

To determine

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

Answer to Problem 4E

The acceleration is positive when the value of time $t$ is greater than $1.30\sec$ and till 3.41 seconds. This will be increasing in speed. The acceleration is negative when the value of time is lesser than $1.3\sec$ and this will be slowing down. 3.41 seconds. The acceleration is positive when the value of time $t$ is greater than $3.41\sec$ and till 4.5 seconds. This will be increasing in speed.

Explanation of Solution

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

Substitute 0 for $f^{"}\left(t\right)$ in the equation $f"\left(t\right)=e^{-t}\left(2-4t+t^2\right)$.

$\begin{array}{l}0=e^{-t}\left(2-4t+t^2\right)\\ t=3.41\sec \end{array}$

Substitute $1\sec$ for $t$ in the equation $f"\left(t\right)=e^{-t}\left(2-4t+t^2\right)$.

$\begin{array}{c}{f}^{"}\left(1\right)=e^{-1}\left(2-4\times 1+1^2\right)\\ =-0.368<0\end{array}$

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

$\begin{array}{c}{f}^{"}\left(4\right)=e^{-4}\left(2-4\times 4+4^2\right)\\ =0.0366>0\end{array}$

Therefore, the acceleration is positive when the value of time $t$ is greater than $1.30\sec$ and till 3.41 seconds. This will be increasing in speed. The acceleration is negative when the value of time is lesser than $1.3\sec$ and this will be slowing down. 3.41 seconds. The acceleration is positive when the value of time $t$ is greater than $3.41\sec$ and till 4.5 seconds. This will be increasing in speed.