Chapter 11.3, Problem 11.64P

A particle moves in a straight line with the velocity shown in the figure. Knowing that x = −540 m at t = 0, (a) construct the a−t and x−t curves for 0 < t < 50 s, and determine (b) the maximum value of the position coordinate of the particle, (c) the values of t for which the particle is at x = 100 m.

Fig. P11.63 and P11.64

To determine

Construct the $a-t$ and $x-t$ curves for $0\le t\le 50\sec$.

Explanation of Solution

Given information:

The position $\left(x_0\right)$ at time is 0 sec is $-540\text{m}$.

Calculation:

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

Calculate the acceleration $\left(a_0\right)$ when time (t) is $0\sec \le t\le 10\sec$:

Refer Figure (1), the slope of the v-t curve is constant from 0 sec to 10 sec. Hence, the acceleration $\left(a_0\right)$ of the particle is $0{\text{m/s}}^{\text{2}}$.

Calculate the acceleration $\left(a_{10}\right)$ when time $10\sec \le t\le 26\sec$ using the relation:

$a_{10}=\frac{v_{26}-v_{10}}{t_{26}-t_{10}}$

Here, $v_{10}$ is velocity of particle when time is 10 sec, $v_{26}$ is velocity of particle when time is 26 sec, $t_{26}$ is time coordinates, and $t_{10}$ is time coordinates.

Substitute $-20\text{m}/\text{s}$ for $v_{26}$, $60\text{m}/\text{s}$ for $v_{10}$, 26s for $t_{26}$, and 10s for $t_{10}$.

$\begin{array}{c}{a}_{10}=\frac{-20-60}{26-10}\\ =\frac{-80}{16}\\ =-5\text{m}/\text{s}\end{array}$

Calculate the acceleration $\left(a_{26}\right)$ of the particle when time $26\sec \le t\le 41\sec$:

Refer Figure (1), the slope of the v-t curve is constant from 26 sec to 41 sec. Hence, the acceleration $\left(a_{26}\right)$ of the particle is $0{\text{m/s}}^{\text{2}}$.

Calculate the acceleration $\left(a_{41}\right)$ when time $41\sec \le t\le 46\sec$ using the relation:

$a_{41}=\frac{v_{46}-v_{41}}{t_{46}-t_{41}}$

Here, $v_{41}$ is velocity of particle when time is 41 sec, $v_{46}$ is velocity of particle when time is 46 sec, $t_{46}$ is time coordinate and $t_{41}$ is time coordinate.

Substitute $-5\text{m}/\text{s}$ for $v_{46}$, $-20\text{m}/\text{s}$ for $v_{41}$, 46s for $t_{46}$, and 41s for $t_{41}$.

$\begin{array}{c}{a}_{41}=\frac{-5-\left(-20\right)}{46-41}\\ =\frac{15}{5}\\ =3\text{m}/{\text{s}}^2\end{array}$

Calculate the acceleration $\left(a_{46}\right)$ of the particle when time $46\sec$:

Refer Figure (1), the slope of the v-t curve is constant from 46 to end of the graph. Hence the acceleration $\left(a_{46}\right)$ of the particle is $0{\text{m/s}}^{\text{2}}$.

Tabulated the acceleration (a) corresponding to time (t) as in Table 1:

t (sec)	$a\left({\text{m/s}}^{\text{2}}\right)$
0	0
10	–5
26	0
41	3
46	0

Plot the a-t curve of the motion of the particle as in Figure (2).

Calculate the position $\left(x_{10}\right)$ when time is 10 sec using the relation:

$x_{10}=x_0+\left(v_{\left(10\right)}{t}_1\right)$

Here, $x_0$ is position of the particle when time is 0 sec and $v_{\left(10\right)}$ is velocity of particle at 10 sec.

Substitute $-540\text{m}$ for $x_0$, 10 sec for $t_1$, and $60\text{m/s}$ for $a_{10}$.

$\begin{array}{c}{x}_{10}=-540+\left(10\times 60\right)\\ =60\text{m}\end{array}$

Calculate the time $\left(t_f\right)$ when velocity is equal to zero.

$\begin{array}{c}{t}_f-t_0=\frac{v-v_0}{a}\\ t_f=t_0+\frac{v-v_0}{a}\end{array}$

Here, v is final velocity of particle, $v_0$ is initial velocity of particle and $t_0$ is time corresponding to velocity $v_0$.

Substitute 10s for $t_0$, 0 for v, $60\text{m}/\text{s}$ for $v_0$, and $-5{\text{m}/\text{s}}^2$ for a.

$\begin{array}{c}{t}_f=10+\frac{0-60}{\left(-5\right)}\\ =10+12\\ =22\text{s}\end{array}$

Calculate the position $\left(x_{22}\right)$ when time is 22 sec using the relation:

$x_{22}=x_{10}+\left(\frac{1}{2}{v}_{\left(10\right)}{t}_2\right)$

Substitute $60\text{m}$ for $x_0$, 12 sec for $t_2$, and $60\text{m/s}$ for $v_{\left(10\right)}$.

$\begin{array}{c}{x}_{22}=60+\left(\frac{1}{2}\times 60\times 12\right)\\ =60+360\\ =420\text{m}\end{array}$

Calculate the position $\left(x_{26}\right)$ when time is 26 sec using the relation:

$x_{26}=x_{22}+\left(\frac{1}{2}{v}_{\left(26\right)}{t}_3\right)$

Here, $v_{\left(26\right)}$ is velocity of particle at 26 sec.

Substitute $420\text{m}$ for $x_{22}$, 4 sec for $t_2$, and $-20\text{m/s}$ for $v_{\left(26\right)}$.

$\begin{array}{c}{x}_{26}=420+\left(\frac{1}{2}\times -20\times 4\right)\\ =420-40\\ =380\text{m}\end{array}$

Calculate the position $\left(x_{41}\right)$ when time is 41 sec using the relation:

$x_{41}=x_{26}+\left(-v_{\left(41\right)}{t}_4\right)$

Here, $v_{\left(41\right)}$ is velocity of particle at 41 sec.

Substitute $380\text{m}$ for $x_{26}$, 15 sec for $t_4$, and $-20\text{m/s}$ for $v_{\left(41\right)}$.

$\begin{array}{c}{x}_{41}=380+\left(-20\times 15\right)\\ =380-300\\ =80\text{m}\end{array}$

Calculate the position $\left(x_{46}\right)$ when time is 46 sec using the relation:

$x_{46}=x_{41}+t_5\left(\frac{v_{\left(41\right)}+v_{46}}{2}\right)$

Here, $v_{\left(46\right)}$ is velocity of particle at 46 sec.

Substitute $80\text{m}$ for $x_{41}$, 5 sec for $t_5$, $-5\text{m/s}$ for $v_{46}$, and $-20\text{m/s}$ for $v_{\left(41\right)}$.

$\begin{array}{c}{x}_{46}=80+5\left(\frac{-20+\left(-5\right)}{2}\right)\\ =80-62.5\\ =17.5\text{m}\end{array}$

Calculate the position $\left(x_{50}\right)$ when time is 50 sec using the relation:

$x_{50}=x_{46}+\left(v_{\left(50\right)}{t}_6\right)$

Here, $v_{\left(50\right)}$ is velocity of particle at 50 sec .

Substitute $17.5\text{m}$ for $x_{46}$, 4 sec for $t_6$, and $-5\text{m/s}$ for $v_{\left(50\right)}$.

$\begin{array}{c}{x}_{50}=17.5+\left(-5\times 4\right)\\ =17.5-20\\ =2.5\text{m}\end{array}$

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

t (s)	x(m)
0	–540
10	60
22	420
26	380
41	80
46	17.5
50	–2.5

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

To determine

The maximum value of position coordinate $\left(x_{\max }\right)$ of the particle.

Answer to Problem 11.64P

The maximum value of position coordinate $\left(x_{\max }\right)$ of the particle is $\underset{\_}{420\text{m}}$.

Explanation of Solution

Given information:

The position $\left(x_0\right)$ at time is 0 sec is $-540\text{m}$.

Calculation:

Refer Figure (3) for maximum value of position coordinate $\left(x_{\max }\right)$ of the particle.

Therefore, the maximum value of position coordinate $\left(x_{\max }\right)$ of the particle is $\underset{\_}{420\text{m}}$.

To determine

The value of ${\left(t_{x=100}\right)}_1$ and ${\left(t_{x=100}\right)}_2$ for which the particle is at distance (x) $100\text{m}$.

Answer to Problem 11.64P

The value of ${\left(t_{x=100}\right)}_1$ and ${\left(t_{x=100}\right)}_2$ for which the particle is at distance (x) $100\text{m}$ are $\underset{\_}{10.69\sec }$ and $\underset{\_}{40\sec }$.

Explanation of Solution

Given information:

The position $\left(x_0\right)$ at time is 0 sec is $-540\text{m}$.

Calculation:

Show the v-t curve between 10 to 22 sec in Figure (4).

Calculate velocity $\left(v_1\right)$ using the relation:

$x=x_{\max }-\frac{1}{2}\left(t_f-{\left(t_{x=100}\right)}_1\right)v_1$

Substitute $100\text{m}$ for x, $420\text{m}$ for $x_{\max }$ and $22\sec$ for $t_f$.

$\begin{array}{l}100=420-\frac{1}{2}\left(22-{\left(t_{x=100}\right)}_1\right)v_1\\ \left(100-420\right)\times 2=-\left(22-{\left(t_{x=100}\right)}_1\right)v_1\\ \left(22-{\left(t_{x=100}\right)}_1\right)v_1=640\\ v_1=\frac{640}{\left(22-{\left(t_{x=100}\right)}_1\right)}\end{array}$

Calculate the time ${\left(t_{x=0}\right)}_1$ using similar triangle:

$\frac{v_1}{t_f-{\left(t_{x=100}\right)}_1}=\frac{v_{10}}{t_f-t_1}$

Substitute $22\sec$ for $t_f$, $60\text{m/s}$ for $v_{10}$, 10 sec for $t_1$ and $\frac{640}{\left(22-{\left(t_{x=100}\right)}_1\right)}$ for $v_1$.

$\begin{array}{l}\frac{\left(\frac{640}{\left(22-{\left(t_{x=100}\right)}_1\right)}\right)}{22-{\left(t_{x=100}\right)}_1}=\frac{60}{22-10}\\ 640=\left[5\left(22-{\left(t_{x=100}\right)}_1\right)\right]\left(22-{\left(t_{x=100}\right)}_1\right)\\ 640=\left(110-5{\left(t_{x=100}\right)}_1\right)\left(22-{\left(t_{x=100}\right)}_1\right)\end{array}$

$\begin{array}{l}640=2,420-110{\left(t_{x=100}\right)}_1-110{\left(t_{x=100}\right)}_1+5{\left(t_{x=100}\right)}_1^2\\ -2,420+640=-220{\left(t_{x=100}\right)}_1+5{\left(t_{x=100}\right)}_1^2\\ 0=5{\left(t_{x=100}\right)}_1^2-220{\left(t_{x=100}\right)}_1+1780\end{array}$

Solve the above quadratic equation for the roots:

The roots ${\left(t_{x=100}\right)}_1$ are 33.3 sec and 10.69 sec. Take the value 10.69 sec because the time (t) is $10\sec <{\left(t_{x=100}\right)}_1<22\sec$.

Show the x-t curve between 26 sec to 41 sec in Figure (5).

Calculate time ${\left(t_{x=100}\right)}_2$ using similar triangle:

$\frac{t_{41}-{\left(t_{x=100}\right)}_2}{\left(x_{100}-x_{41}\right)}=\frac{t_{41}-t_f}{\left(x_{26}-x_{41}\right)}$

Here, $t_{41}$ is time at 41 sec.

Substitute $22\sec$ for $t_f$, 41 sec for $t_{41}$, $100\text{m}$ for $x_{100}$, $80\text{m}$ for $x_{41}$ and $380\text{m}$ for $x_{26}$.

$\begin{array}{l}\frac{41-{\left(t_{x=100}\right)}_2}{\left(100-80\right)}=\frac{41-22}{\left(380-80\right)}\\ 12,300-300{\left(t_{x=100}\right)}_2=380\\ {\left(t_{x=100}\right)}_2=\frac{-11,920}{-300}\\ {\left(t_{x=100}\right)}_2=40\sec \end{array}$

Therefore, the value of ${\left(t_{x=100}\right)}_1$ and ${\left(t_{x=100}\right)}_2$ for which the particle is at distance (x) $100\text{m}$ are $\underset{\_}{10.69\sec }$ and $\underset{\_}{40\sec }$.