P01R2020 – Coming at Each Other
Coming at Each Other
In this problem, two cars (treated as particles) move along a horizontal road. The car on the left, labeled 1, is initially at rest and accelerates with $a_1=2\ m/s^2$. The car on the right, labeled 2, moves with constant speed $8\ m/s$ throughout the problem. Initially, the distance between the two cars is $20\ m$.
1. Write the kinematic equations for each car. Make sure you explain your work.
View answerHide answerCar 1 is initially at rest, therefore $v_{01}=0$ and we set $x=0$ at the initial location of car 1 which means that $x_{01}=0$.
Car 2 moves to the left and therefore its initial velocity — which remains constant — is equal to $v_{02}=-8\ m/s$ while its acceleration is zero. Its initial position is $x_{02}=20\ m$.
These quantities are summarized in the figure below for convenience.
Thus, the kinematic equation for both cars are
\boxed{
car\ 1:\ \ \left\{ \begin{array}{c}
\displaystyle{x_1\left(t\right)=\frac{1}{2}a_1t^2} \\
\\
v_1\left(t\right)=a_1t \end{array}
\right.\ \ \ \ \ \ \ and\ \ \ \ \ \ \ car\ 2:\ \ \left\{ \begin{array}{c}
x_2\left(t\right)=v_{02}t+x_{02} \\
\\
v_2\left(t\right)=v_{02} \end{array}
\right.
}2. Show that at $t=2\ s$ the cars pass each other.
View answerHide answerWhen the cars pass each other, they have the same position and $x_1\left(t\right)=x_2\left(t\right)$. Thus, solving this equation we derive the time as follows
\begin{aligned}
x_1\left(t\right)=x_2\left(t\right)\ \ \ \ &\Rightarrow \ \ \ \ \frac{1}{2}a_1t^2=v_{02}t+x_{02} \\
\\
&\Rightarrow \ \ \ \ \ t^2=-8t+20 \\
\\
&\Rightarrow \ \ \ \ \ t^2+8t-20=0 \\
\\
&\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{t=\frac{-8-\sqrt{8^2+4\cdot 20}}{2}=\frac{-8-\sqrt{144}}{2}=-10\ s} \\
\\
or \\
\\
\displaystyle{t=\frac{-8+\sqrt{8^2-4\cdot 20}}{2}=\frac{-8+\sqrt{144}}{2}=2\ s \ \ \ \ \ \ } \end{array}
\right.
\end{aligned}We rule out the negative root and it is physically irrelevant and thus conclude that, indeed, the cars pass each other at time $t=2\ s.$
3. What is the speed of each car when they pass each other?
View answerHide answerThe speed of car 1 at time $t=2\ s$ is equal to
\boxed{v_1\left(2\right)=2\cdot 2=4\ m/s}The speed of car 2 at time $t=2\ s$ is equal to
\boxed{\left|v_2\left(2\right)\right|=\left|-8\right|=8\ m/s}4. How far apart are the cars at time $t=4\ s$?
View answerHide answerThe position of car 1 at time $t=4\ s$ is equal to
x_1\left(4\right)=4^2=16\ m
The position of car 2 at time $t=4\ s$ is equal to
x_2\left(4\right)=-8\cdot 4+20=-12\ m
Thus, the distance between the two cars at time $t=4\ s$ is equal to
\boxed{d=x_1\left(4\right)-x_2\left(4\right)=16-\left(-12\right)=28\ m}5. When do the cars have the same speed?
View answerHide answerTo derive the time when the cars have the same speed, we solve
\begin{aligned}
\left|v_1\left(t\right)\right|=\left|v_2\left(t\right)\right|\ \ \ \ &\Rightarrow \ \ \ \ \ a_1t=\left|v_{02}\right| \\
\\
&\Rightarrow \ \ \ \ \ t=\frac{\left|v_{02}\right|}{a_1}
\end{aligned}Therefore, the cars have the same speed at time $t$ equal to
\boxed{t=\frac{\left|v_{02}\right|}{a_1}=\frac{8}{2}=4\ s}