Chapter 11.3, Problem 11.61P

A particle moves in a straight line with a constant acceleration of −4 ft/s² for 6 s, zero acceleration for the next 4 s, and a constant acceleration of +4 ft/s² for the next 4 s. Knowing that the particle starts from the origin and that its velocity is −8 ft/s during the zero acceleration time interval, (a) construct the v−t and x−t curves for 0 ≤ t ≤ 14 s, (b) determine the position and the velocity of the particle and the total distance traveled when t = 14 s.

Fig. P11.61 and P11.62

To determine

Construct the $v-t$ and $x-t$ curves for $0\le t\le 14\sec$.

Explanation of Solution

Given information:

The constant acceleration $\left(a_1\right)$ of particle at 6 sec is $-4{\text{ft/s}}^{\text{2}}$.

The acceleration is zero from 6sec to 10 sec.

From 10 sec to 14 sec the acceleration $\left(a_2\right)$ of the particle is $4{\text{ft/s}}^{\text{2}}$.

The velocity $\left(v_6\right)$ of the particle at 6 sec is $-8\text{ft/s}$.

Calculation:

Show a-t curve of particle that moves in a straight line as in Figure (1).

Calculate the area $\left(A_1\right)$ using the relation:

$A_1=t_1{a}_1$

Substitute 6 sec for $t_1$ and $-4{\text{ft/s}}^{\text{2}}$ for $a_1$.

$\begin{array}{c}{A}_1=\left(6\right)\times \left(-4\right)\\ =-24\text{ft/s}\end{array}$

Calculate the area $\left(A_2\right)$ using the relation:

$A_2=t_3{a}_2$

Substitute 4 sec for $t_3$ and $4{\text{ft/s}}^{\text{2}}$ for $a_2$.

$\begin{array}{c}{A}_2=\left(4\right)\times \left(4\right)\\ =16\text{ft/s}\end{array}$

Calculate the velocity $\left(v_0\right)$ at the origin using the relation:

$v_0=v_6-A_1$

Substitute $-8\text{ft/s}$ for $v_6$ and $-24\text{ft/s}$ for $A_1$.

$\begin{array}{c}{v}_0=-8-\left(-24\right)\\ =16\text{ft/s}\end{array}$

Calculate the velocity $\left(v_{10}\right)$ at 10 sec using the relation;

$v_{10}=v_6$

Substitute $-8\text{ft/s}$ for $v_6$.

$v_{10}=-8\text{ft/s}$

Calculate the velocity $\left(v_{14}\right)$ at 14 sec using the relation:

$v_{14}=v_{10}+A_2$

Substitute $-8\text{ft/s}$ for $v_{10}$ and $16\text{ft/s}$ for $A_2$.

$\begin{array}{c}{v}_{14}=-8+16\\ =8\text{ft/s}\end{array}$

Tabulated the acceleration (a), velocity (v) corresponding to time (t) in Table (1) :

t(s)	$a\left(\text{ft}/{\text{s}}^{\text{2}}\right)$	$v\left(\text{ft}/\text{s}\right)$
0	-4	16
6	0	-8
10	0	-8
14	4	8

Plot the v-t curve of particle that moves in a straight line with areas as in Figure (2).

Calculate the area $\left(A_3\right)$ using the relation:

$A_3=\frac{1}{2}{v}_0{t}_4$

Here, $t_4$ is time between 0 sec to 4 sec.

Substitute 4 sec for $t_4$ and $16\text{ft/s}$ for $v_0$.

$\begin{array}{c}{A}_3=\frac{1}{2}\left(16\right)\left(4\right)\\ =32\text{ft}\end{array}$

Calculate the area $\left(A_4\right)$ using the relation:

$A_4=\frac{1}{2}{v}_6{t}_5$

Here, $t_5$ is time between 4 sec to 6 sec.

Substitute 2 sec for $t_5$ and $-8\text{ft/s}$ for $v_6$.

$\begin{array}{c}{A}_4=\frac{1}{2}\left(-8\right)\left(2\right)\\ =-8\text{ft}\end{array}$

Calculate the area $\left(A_5\right)$ using the relation:

$A_5=t_6{v}_{10}$

Here, $t_6$ is time between 6 sec to 10 sec

Substitute 4 sec for $t_6$ and $-8\text{ft/s}$ for $v_{10}$.

$\begin{array}{c}{A}_5=\left(4\right)\times \left(-8\right)\\ =-32\text{ft}\end{array}$

Calculate the area $\left(A_6\right)$ using the relation:

$A_6=\frac{1}{2}{v}_{10}{t}_7$

Here, $t_7$ is time between 10 sec to 12 sec.

Substitute 2 sec for $t_7$ and $-8\text{ft/s}$ for $v_{10}$.

$\begin{array}{c}{A}_6=\frac{1}{2}\left(-8\right)\left(2\right)\\ =-8\text{ft}\end{array}$

Calculate the area $\left(A_7\right)$ using the relation:

$A_7=\frac{1}{2}{v}_{14}{t}_8$

Here, $t_8$ is time between 12 sec to 14 sec.

Substitute 4 sec for $t_8$ and $8\text{ft/s}$ for $v_{14}$.

$\begin{array}{c}{A}_7=\frac{1}{2}\times 8\times 2\\ =8\text{ft}\end{array}$

Calculate the position $\left(x_0\right)$ of the particle at 0 sec using the relation:

$x_0=0\text{ft}$

Calculate the position $\left(x_4\right)$ of the particle at 4 sec using the relation:

$x_4=x_0+A_3$

Substitute 0 for $x_0$ and $32\text{ft}$ for $A_3$.

$\begin{array}{c}{x}_4=0+32\\ =32\text{ft}\end{array}$

Calculate the position $\left(x_6\right)$ of the particle at 6 sec using the relation:

$x_6=x_4+A_4$

Substitute $32\text{ft}$ for $x_4$ and $-8\text{ft}$ for $A_4$.

$\begin{array}{c}{x}_6=32-8\\ =24\text{ft}\end{array}$

Calculate the position $\left(x_{10}\right)$ of the particle at 10 sec using the relation:

$x_{10}=x_6+A_5$

Substitute $24\text{ft}$ for $x_6$ and $-32\text{ft}$ for $A_5$.

$\begin{array}{c}{x}_{10}=24-32\\ =-8\text{ft}\end{array}$

Calculate the position $\left(x_{12}\right)$ of the particle at 12 sec using the relation:

$x_{12}=x_{10}+A_6$

Substitute $-8\text{ft}$ for $x_{10}$ and $-8\text{ft}$ for $A_6$.

$\begin{array}{c}{x}_{12}=-8-8\\ =-16\text{ft}\end{array}$

Calculate the position $\left(x_{14}\right)$ of the particle at 14 sec using the relation:

$x_{14}=x_{12}+A_7$

Substitute $-16\text{ft}$ for $x_{12}$ and $8\text{ft}$ for $A_7$.

$\begin{array}{c}{x}_{14}=-16+8\\ =-8\text{ft}\end{array}$

Tabulated the position (x) corresponding to time (t) in Table 2:

t (sec)	x (ft)
0	0
4	32
6	24
10	-8
12	-16
14	-8

Plot x-t curve of particle that moves in a straight line with areas as in Figure 3.

To determine

The position, velocity of the particle and the total distance (d) traveled when time (t) 14 sec.

Answer to Problem 11.61P

The total distance (d) traveled when time (t) 14 sec is $\underset{\_}{88\text{ft}}$.

Explanation of Solution

Given information:

The constant acceleration $\left(a_1\right)$ of particle at 6 sec is $-4{\text{ft/s}}^{\text{2}}$.

The acceleration is zero from 6sec to 10 sec.

From 10 sec to 14 sec the acceleration $\left(a_2\right)$ of the particle is $4{\text{ft/s}}^{\text{2}}$.

The velocity $\left(v_6\right)$ of the particle at 6 sec is $-8\text{ft/s}$.

Calculation:

Calculate the area $\left(A_1\right)$ using the relation:

$A_1=t_1{a}_1$

Substitute 6 sec for $t_1$ and $-4{\text{ft/s}}^{\text{2}}$ for $a_1$

$\begin{array}{c}{A}_1=\left(6\right)\times \left(-4\right)\\ =-24\text{ft/s}\end{array}$

Calculate the area $\left(A_2\right)$ using the relation:

$A_2=t_3{a}_2$

Substitute 4 sec for $t_3$ and $4{\text{ft/s}}^{\text{2}}$ for $a_1$

$\begin{array}{c}{A}_2=\left(4\right)\times \left(4\right)\\ =16\text{ft/s}\end{array}$

Calculate the velocity $\left(v_0\right)$ at the origin using the relation:

$v_0=v_6-A_1$

Substitute $-8\text{ft/s}$ for $v_6$ and $-24\text{ft/s}$ for $A_1$.

$\begin{array}{c}{v}_0=-8-\left(-24\right)\\ =16\text{ft/s}\end{array}$

Calculate the velocity $\left(v_{10}\right)$ at 10 sec using the relation;

$v_{10}=v_6$

Substitute $-8\text{ft/s}$ for $v_6$.

$v_{10}=-8\text{ft/s}$

Calculate the velocity $\left(v_{14}\right)$ at 14 sec using the relation:

$v_{14}=v_{10}+A_2$

Substitute $-8\text{ft/s}$ for $v_{10}$ and $16\text{ft/s}$ for $A_2$.

$\begin{array}{c}{v}_{14}=-8+16\\ =8\text{ft/s}\end{array}$

Calculate the area $\left(A_3\right)$ using the relation:

$A_3=\frac{1}{2}{v}_0{t}_4$

Here, $t_4$ is 0 sec to 4 sec.

Substitute 4 sec for $t_4$ and $16\text{ft/s}$ for $v_0$.

$\begin{array}{c}{A}_3=\frac{1}{2}\left(16\right)\left(4\right)\\ =32\text{ft}\end{array}$

Calculate the area $\left(A_4\right)$ using the relation:

$A_4=\frac{1}{2}{v}_6{t}_5$

Here, $t_5$ is 4 sec to 6 sec.

Substitute 2 sec for $t_5$ and $-8\text{ft/s}$ for $v_6$.

$\begin{array}{c}{A}_4=\frac{1}{2}\left(-8\right)\left(2\right)\\ =-8\text{ft}\end{array}$

Calculate the area $\left(A_5\right)$ using the relation:

$A_5=t_6{v}_{10}$

Here, $t_6$ is 6 sec to 10 sec

Substitute 4 sec for $t_6$ and $-8\text{ft/s}$ for $v_{10}$.

$\begin{array}{c}{A}_5=\left(4\right)\times \left(-8\right)\\ =-32\text{ft}\end{array}$

Calculate the area $\left(A_6\right)$ using the relation:

$A_6=\frac{1}{2}{v}_{10}{t}_7$

Here, $t_7$ is 10 sec to 12 sec.

Substitute 2 sec for $t_7$ and $-8\text{ft/s}$ for $v_{10}$.

$\begin{array}{c}{A}_6=\frac{1}{2}\left(-8\right)\left(2\right)\\ =-8\text{ft}\end{array}$

Calculate the area $\left(A_7\right)$ using the relation:

$A_7=\frac{1}{2}{v}_{14}{t}_8$

Here, $t_8$ is 12 sec to 14 sec.

Substitute 4 sec for $t_8$ and $8\text{ft/s}$ for $v_{14}$.

$\begin{array}{c}{A}_7=\frac{1}{2}\times 8\times 2\\ =8\text{ft}\end{array}$

Calculate the position $\left(x_0\right)$ of the particle in 0 sec using the relation:

$x_0=0\text{ft}$

Calculate the position $\left(x_4\right)$ of the particle in 4 sec using the relation:

$x_4=x_0+A_3$

Substitute 0 for $x_0$ and $32\text{ft}$ for $A_3$. 0

$\begin{array}{c}{x}_4=0+32\\ =32\text{ft}\end{array}$

Calculate the position $\left(x_6\right)$ of the particle in 6 sec using the relation:

$x_6=x_4+A_4$

Substitute $32\text{ft}$ for $x_4$ and $-8\text{ft}$ for $A_4$.

$\begin{array}{c}{x}_6=32-8\\ =24\text{ft}\end{array}$

Calculate the position $\left(x_{10}\right)$ of the particle in 10 sec using the relation:

$x_{10}=x_6+A_5$

Substitute $24\text{ft}$ for $x_6$ and $-32\text{ft}$ for $A_5$.

$\begin{array}{c}{x}_{10}=24-32\\ =-8\text{ft}\end{array}$

Calculate the position $\left(x_{12}\right)$ of the particle in 12 sec using the relation:

$x_{12}=x_{10}+A_6$

Substitute $-8\text{ft}$ for $x_{10}$ and $-8\text{ft}$ for $A_6$.

$\begin{array}{c}{x}_{12}=-8-8\\ =-16\text{ft}\end{array}$

Calculate the position $\left(x_{14}\right)$ of the particle in 14 sec using the relation:

$x_{14}=x_{12}+A_7$

Substitute $-16\text{ft}$ for $x_{12}$ and $8\text{ft}$ for $A_7$.

$\begin{array}{c}{x}_{14}=-16+8\\ =-8\text{ft}\end{array}$

Calculate the distance $\left(d_1\right)$ traveled when time is $0\le t\le 4\sec$ using the relation:

$d_1=\left|x_4-x_0\right|$

Substitute 0 for $x_0$ and $32\text{ft}$ for $x_4$.

$\begin{array}{c}{d}_1=\left|32-0\right|\\ =32\text{ft}\end{array}$

Calculate the distance $\left(d_2\right)$ traveled when time is $4\sec \le t\le 12\sec$ using the relation:

$d_2=\left|x_{12}-x_4\right|$

Substitute $32\text{ft}$ for $x_4$ and $-16\text{ft}$ for $x_{12}$.

$\begin{array}{c}{d}_2=\left|-16-32\right|\\ =\left|-48\right|\\ =48\text{ft}\end{array}$

Calculate the distance $\left(d_3\right)$ traveled when time is $12\sec \le t\le 14\sec$ using the relation:

$d_3=\left|x_{14}-x_{12}\right|$

Substitute $-16\text{ft}$ for $x_{12}$ and $-8\text{ft}$ for $x_{14}$.

$\begin{array}{c}{d}_3=\left|-8-\left(-16\right)\right|\\ =8\text{ft}\end{array}$

Calculate the total distance (d) traveled when time (t) is 14 sec

$d=d_1+d_2+d_3$

Substitute $32\text{ft}$ for $d_1$, $48\text{ft}$ for $d_2$ and $8\text{ft}$ for $d_3$.

$\begin{array}{c}d=32+48+8\\ =88\text{ft}\end{array}$

Therefore, the total distance (d) traveled when time (t) 14 sec is $\underset{\_}{88\text{ft}}$.