P01R2015 – Highway Patrol

To determine when the officer is at the car’s level, we must start by setting up the kinematic equations for both the car and the Officer.

Setting $x=0$ at the initial position of the car, we derive the following equations:

Car:\ \ \ \ \ \ \left\{ \begin{array}{c}
v_1\left(t\right)=v_{0x}+a_xt=v_1 \\
\\
\displaystyle{x_1\left(t\right)=x_0+v_{0x}+\frac{1}{2}a_xt^2=v_1t} \end{array} \ \ \ \ \ \ \ \ \ \ \ \ \ \
\right.

Bike:\ \ \ \ \ \left\{ \begin{array}{c}
v_2\left(t\right)=v_{0x}+a_xt=a_2t \\
\\
\displaystyle{x_2\left(t\right)=\frac{1}{2}a_xt^2+v_{0x}t+x_0=\frac{1}{2}a_2t^2+D} \end{array}
\right.

The Officer is at the car’s level when they have the same position i.e. when $x_1\left(t\right)=x_2\left(t\right)$. We therefore solve:

\begin{aligned}
x_1\left(t\right)=x_2\left(t\right)\ \ \ \ &\Leftrightarrow \ \ \ \ \ v_1t=\frac{1}{2}a_2t^2+D \\
\\
&\Leftrightarrow \ \ \ \ \ \frac{1}{2}a_2t^2-v_1t+D=0 \\
\\
&\Leftrightarrow \ \ \ \ \ 2t^2-30t+100=0
\end{aligned}

The above quadratic has the two following roots

\left\{ \begin{array}{c}
t_1=5\ s\ \ \\
t_2=10\ s \end{array}
\right.

Thus, there are two instants at which the Officer is at the car’s level. At the first time $t_1$, the car will passing up the Officer and at the second time $t_2$ the Officer will have gained enough speed to be at the car’s level again.

Note: a common mistake is to set the velocities equal to each other but this cannot be true if you want the car to catch up with the Officer (or vice versa).

The first time they meet, the speed of each is given by:

\left\{ \begin{array}{c}
v_1\left(t_1\right)=v_1=30\ m/s \\
\\
v_2\left(t_1\right)=a_2t_1=4\cdot 5=20\ m/s \end{array}
\right.

The second time they meet, the speed of each is given by:

\left\{ \begin{array}{c}
v_1\left(t_2\right)=v_1=30\ m/s \\
\\
v_2\left(t_2\right)=a_2t_2=4\cdot 10=40\ m/s \end{array}
\right.

Note that the speed of the Office is different than the speed of the car. Being at the same level means the positions are equal, but does not imply anything about the instantaneous velocities (they do not have to be the same, or else nobody would ever pass anyone up).

Whenever the Officer and the car are at the same level, they share the same position i.e. they are the same distance from the point where the Officer spotted the driver. These positions are given by:

\left\{ \begin{array}{c}
\displaystyle{x_1\left(t_1\right)=x_2\left(t_1\right)=\frac{1}{2}a_2t^2_1+100=\frac{1}{2}\cdot 4\cdot 5^2+100=150\ m} \\
\\
\displaystyle{x_2\left(t_2\right)=x_2\left(t_2\right)=\frac{1}{2}a_2t^2_2+100=\frac{1}{2}\cdot 4\cdot {10}^2+100=300\ m} \end{array}
\right.

They are the same level at distances $d_1=150\ m$ and $d_2=300\ m$ from $x=0$ which means that the distances from the point where the Officer spotted the driver initially are equal to

\left\{ \begin{array}{c}
d_1=x_1\left(t_1\right)-100=50\ m\ \\
\\
d_2=x_2\left(t_2\right)-100=200\ m \end{array}
\right.