P01G2008 – Passing a Car

First we setup the kinematic equations for each car. The first car moves at constant speed $v_1$ $\left(a_1=0\right)$ and its initial position is $x_0=D$. Thus, the kinematic equations describing the motion of the first car are

\left\{ \begin{array}{c}
x_1\left(t\right)=v_1t+D \\
\\
v_1\left(t\right)=v_1 \ \ \ \ \ \ \ \ \ \end{array}
\right.

The second car starts at rest $\left(v_0=0\right)$ with an acceleration $a_2$ and an initial position $x_0=0$. Thus, the kinematic equations describing the motion of the second car are

\left\{ \begin{array}{c}
\displaystyle{x_2\left(t\right)=\frac{1}{2}a_2t^2} \\
\\
v_2\left(t\right)=a_2t \ \ \ \ \end{array}
\right.

The second car catches the first in a time $t_1$ that satisfies $x_1\left(t_1\right)=x_2\left(t_1\right)$ and we solve for $t_1$ as follows

\begin{aligned}
x_1\left(t_1\right)=x_2\left(t_1\right)\ \ \ \ &\Rightarrow \ \ \ \ \ v_1t_1+D=\frac{1}{2}a_2t^2_1 \\
\\
&\Rightarrow \ \ \ \ \ 0.75t^2_1-15t_1-20=0
\end{aligned}

This quadratic equation has two roots

\left\{ \begin{array}{c}
t_1=-1.25\ s \\
t_1=21.25\ s \ \end{array}
\right.

and we keep the positive solution (the only one that makes physical sense) and conclude that the time $t_1$ it takes for the second car to catch up with the first is equal to

\boxed{t_1=21.25\ s}

The speed of the second car at $t=t_1$ is equal to

\boxed{v_2\left(t_1\right)=a_2t_1=1.5\cdot 21.25\approx 31.9\ m/s}

The distance covered by the second car at $t=t_1$ is equal to

\boxed{x_2\left(t_1\right)=\frac{1}{2}a_2t^2_1=\frac{1}{2}\cdot 1.5\cdot {21.25}^2\approx 339\ m}