3.19 The first tourist, after riding a bicycle for 1.5 hours at a speed of $16 \mathrm{km} / \mathrm{h}$, makes a stop for 1.5 hours, and then continues the journey at the initial speed. After 4 hours from the departure of the first tourist, the second tourist sets off in pursuit on a motorcycle at a speed of 56 km/h. What distance will they have traveled before the second tourist catches up with the first?
3.19 An approximate graph of the movement is shown in Fig. 3.4.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-099.jpg?height=398&width=373&top_left_y=1293&top_left_x=167)

Fig. 3.4

Let $t$ be the time (in hours) it takes for the second tourist to catch up with the first. Since

$$
\begin{aligned}
& v_{\text {bike }}=16 \mathrm{km} / \mathrm{h} \\
& v_{\text {motor }}=56 \mathrm{km} / \mathrm{h}, \text { then } \\
& s_{\text {bike }}=(2.5+t) \cdot 16
\end{aligned}
$$

$$
s_{\text {motor }}=56 t \text {. }
$$

From this, $56 t=16 t+40$, i.e., $t=1$ (hour). Therefore, $s=56$ km.