13.079. 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 $56 \mathrm{km} / \mathrm{h}$. What distance will they cover before the second tourist catches up with the first?
## Solution.

The motion graph is shown in Fig. 13.3. Let $t$ be the time (in hours) it takes for the second tourist to catch up with the first. Since $v_{\text {bike }}=16 \mathrm{km} / \mathbf{h}$, $\mathrm{v}_{\text {motor }}=56 \mathrm{km} /$ h, then $S_{\text {bike }}=(2.5+t) 16, S_{\text {motor }}=56 t$. Therefore, $56 t=16 t+40$, i.e., $t=1$ (h). Thus, $S=56$ km.

Answer: 56 km.