Chapter 11.3, Problem 11.76P

Car A is traveling at 40 mi/h when it enters a 30 mi/h speed zone. The driver of car A decelerates at a rate of 16 ft/s² until reaching a speed of 30 mi/h, which she then maintains. When car B, which was initially 60 ft behind car A and traveling at a constant speed of 45 mi/h, enters the speed zone, its driver decelerates at a rate of 20 ft/s² until reaching a speed of 28 mi/h. Knowing that the driver of car B maintains a speed of 28 mi/h, determine (a) the closest that car B comes to car A, (b) the time at which car A is 70 ft in front of car B.

  

To determine

(a)

The closest that car B comes to car A.

Answer to Problem 11.76P

${\left(x_{A/B}\right)}_{\min }=34.623ft$

Explanation of Solution

Given information:

${\left(v_A\right)}_0=40mi/h=58.667ft/s$ For $30mi/h<v_A<40mi/h$

$a_A=-16ft/s^2$ For $v_A=30mi/h=44ft/s$

$a_A=0$

${\left(v_B\right)}_0=45mi/h=66ft/s$

${\left(x_{A/B}\right)}_0=60ft$

When, $x_B=0,$

$a_B=20ft/s^2$ and,

For, $v_B=28mi/h=41.067ft/s,$

$a_B=0$

The v-t curve of two cars is as shown:

At, $t=0$ : car A enters the speed zone.

$t={\left(t_B\right)}_1$ : Car B enters the speed zone.

$t=t_A$ : Car A reaches its final speed.

$t=t_{\min }$ : $v_A=v_B$

$t={\left(t_B\right)}_2$ : Car B reaches its final speed.

Acceleration of A, $a_A=\frac{{\left(v_A\right)}_{final}-{\left(v_A\right)}_0}{t_A}$

$\begin{array}{l}-16=\frac{44-58.667}{t_A}\\ t_A=0.91669s\end{array}$

And, ${\left(x_{A/B}\right)}_0={\left(t_B\right)}_1{\left(v_B\right)}_0$

$\begin{array}{l}60={\left(t_B\right)}_1\times 66\\ {\left(t_B\right)}_1=\frac{60}{66}=0.90909s\end{array}$

Again, similarly acceleration of B, $a_B=\frac{{\left(v_B\right)}_{final}-{\left(v_B\right)}_0}{{\left(t_B\right)}_2-{\left(t_B\right)}_1}$

$-20=\frac{41.067-66}{{\left(t_B\right)}_2-0.90909}$

Car B will continue to overtake car A while $v_B>v_A.$ Therefore, ${\left(x_{A/B}\right)}_{\min }$ will occur when $v_A=v_B,$ which occurs for,

${\left(t_B\right)}_1<t_{\min }<{\left(t_B\right)}_2$

For his time interval, $v_A=44ft/s$

$v_B={\left(v_B\right)}_0+a_B\left[t-{\left(t_B\right)}_1\right]$

Then at, $t=t_{\min }$

$\begin{array}{l}44=66+\left(-20\right)\left[t_{\min }-0.90909\right]or,\\ t_{\min }=2.009s\end{array}$

Thus, ${\left(x_{A/B}\right)}_{\min }={\left(x_A\right)}_{t_{\min }}-{\left(x_B\right)}_{t_{\min }}$

$\begin{array}{l}{\left(x_{A/B}\right)}_{\min }=\left\{t_A\left[\frac{{\left(v_B\right)}_{final}+{\left(v_A\right)}_0}{2}\right]+\left(t_{\min }-t_A\right){\left(v_A\right)}_{final}\right\}\\ -\left\{{\left(x_B\right)}_0+{\left(t_B\right)}_1{\left(v_B\right)}_0+\left[t_{\min }-{\left(t_B\right)}_1\right]\left[\frac{{\left(v_A\right)}_{final}+{\left(v_B\right)}_0}{2}\right]\right\}\end{array}$

$\begin{array}{l}{\left(x_{A/B}\right)}_{\min }=\left\{\left(0.91669\right)\left[\frac{58.667+44}{2}\right]+\left(2.009-0.91669\right)\left(44\right)\right\}\\ -\left\{60+\left(0.90909\right)\left(66\right)+\left[2.009-0.90909\right]\left[\frac{66+44}{2}\right]\right\}\\ {\left(x_{A/B}\right)}_{\min }=\left(47.057+48.066\right)-\left(-60+60+60.500\right)\\ {\left(x_{A/B}\right)}_{\min }=34.623ft\end{array}$

To determine

(b)

The time at which car A is 70 feet in front of car B.

Answer to Problem 11.76P

$t=14.144s$

Explanation of Solution

Given information:

${\left(v_A\right)}_0=40mi/h=58.667ft/s$ For $30mi/h<v_A<40mi/h$

$a_A=-16ft/s^2$ For $v_A=30mi/h=44ft/s$

$a_A=0$

${\left(v_B\right)}_0=45mi/h=66ft/s$

${\left(x_{A/B}\right)}_0=60ft$

When, $x_B=0,$

$a_B=20ft/s^2$ and,

For, $v_B=28mi/h=41.067ft/s,$

$a_B=0$

The v-t curve of two cars is as shown:

At, $t=0$ : car A enters the speed zone.

$t={\left(t_B\right)}_1$ : Car B enters the speed zone.

$t=t_A$ : Car A reaches its final speed.

$t=t_{\min }$ : $v_A=v_B$

$t={\left(t_B\right)}_2$ : Car B reaches its final speed.

Acceleration of A, $a_A=\frac{{\left(v_A\right)}_{final}-{\left(v_A\right)}_0}{t_A}$

$\begin{array}{l}-16=\frac{44-58.667}{t_A}\\ t_A=0.91669s\end{array}$

And, ${\left(x_{A/B}\right)}_0={\left(t_B\right)}_1{\left(v_B\right)}_0$

$\begin{array}{l}60={\left(t_B\right)}_1\times 66\\ {\left(t_B\right)}_1=\frac{60}{66}=0.90909s\end{array}$

Again, similarly acceleration of B, $a_B=\frac{{\left(v_B\right)}_{final}-{\left(v_B\right)}_0}{{\left(t_B\right)}_2-{\left(t_B\right)}_1}$

$-20=\frac{41.067-66}{{\left(t_B\right)}_2-0.90909}$

Car B will continue to overtake car A while $v_B>v_A.$ Therefore, ${\left(x_{A/B}\right)}_{\min }$ will occur when $v_A=v_B,$ which occurs for,

${\left(t_B\right)}_1<t_{\min }<{\left(t_B\right)}_2$

For his time interval, $v_A=44ft/s$

$v_B={\left(v_B\right)}_0+a_B\left[t-{\left(t_B\right)}_1\right]$

Then at, $t=t_{\min }$

$\begin{array}{l}44=66+\left(-20\right)\left[t_{\min }-0.90909\right]or,\\ t_{\min }=2.009s\end{array}$

Thus, ${\left(x_{A/B}\right)}_{\min }={\left(x_A\right)}_{t_{\min }}-{\left(x_B\right)}_{t_{\min }}$

${\left(x_{A/B}\right)}_{\min }=\left\{t_A\left[\frac{{\left(v_B\right)}_{final}+{\left(v_A\right)}_0}{2}\right]+\left(t_{\min }-t_A\right){\left(v_A\right)}_{final}\right\}-\left\{{\left(x_B\right)}_0+{\left(t_B\right)}_1{\left(v_B\right)}_0+\left[t_{\min }-{\left(t_B\right)}_1\right]\left[\frac{{\left(v_A\right)}_{final}+{\left(v_B\right)}_0}{2}\right]\right\}$

$\begin{array}{l}{\left(x_{A/B}\right)}_{\min }=\left\{\left(0.91669\right)\left[\frac{58.667+44}{2}\right]+\left(2.009-0.91669\right)\left(44\right)\right\}\\ -\left\{60+\left(0.90909\right)\left(66\right)+\left[2.009-0.90909\right]\left[\frac{66+44}{2}\right]\right\}\\ {\left(x_{A/B}\right)}_{\min }=\left(47.057+48.066\right)-\left(-60+60+60.500\right)\\ {\left(x_{A/B}\right)}_{\min }=34.623ft\end{array}$

Since, ${\left(x_{A/B}\right)}_{\min }\le 60$ for $t\le t_{\min },$ it follows that $\left(x_{A/B}\right)=70ft$ for $t>{\left(t_B\right)}_2$.

Then for, $t>{\left(t_B\right)}_2$

$x_{A/B}={\left(x_{A/B}\right)}_{\min }+\left[\left(t-t_{\min }\right){\left(v_A\right)}_{final}\right]-\left\{\left[{\left(t_B\right)}_2-\left(t_{\min }\right)\right]\left[\frac{{\left(v_A\right)}_{final}+{\left(v_B\right)}_{final}}{2}\right]+\left(t-{\left(t_B\right)}_2\right){\left(v_B\right)}_{final}\right\}$

$\begin{array}{l}70=34.623+\left[\left(t-2.009\right)\times 44\right]-\left\{\left[2.15574-\left(2.009\right)\right]\left[\frac{44+41.06}{2}\right]+\left(t-2.15574\right)\left(41.067\right)\right\}\\ 70=34.623+44t-88.396-\left\{0.14674\times 42.53+41.067t-88.5297746\right\}\\ 70=34.623+44t-88.396-\left\{6.241+41.067t-88.5297746\right\}\\ 70=34.623+44t-88.396-6.241-41.067t+88.5297746\\ 70=28.5156+2.933t\\ 2.933t=70-28.5156\\ 2.933t=41.4844\\ t=14.144s\end{array}$