Solution. Let $x$ and $y$ be the speeds (km/h) of the pedestrian and the car. Then $y \cdot 1$ is the total distance the car has traveled, from which $\frac{y-30}{2}$ is the distance the car traveled before meeting the pedestrian, $16-\frac{y-30}{2}=\frac{62-y}{2}$ is the distance the pedestrian traveled before the meeting. Therefore, $\frac{y-30}{2 y}=\frac{62-y}{2 x}$. According to the second condition of the problem, $\frac{5}{y}+2 \frac{2}{3}=\frac{11}{x}$. Thus,

$\left\{\begin{array}{l}\frac{5}{y}+\frac{8}{3}=\frac{11}{x}, \\ \frac{y-30}{y(62-y)}=\frac{1}{x}\end{array} \Rightarrow y=60\right.$.

Answer: 60 km/h.