1. 

![](https://cdn.mathpix.com/cropped/2024_05_06_508ac94edeef1bbe30b1g-3.jpg?height=65&width=956&top_left_y=293&top_left_x=413)

Notations:

$C$ - the point on the highway where Kolya got off his bicycle, $A C=10 t$

$D$ - the point on the highway where Vasya was at time $t, A D=5 t$

$E$ - the point where Petya picked up Vasya on the bicycle

According to the problem:

On the segment $D C, |D C|=10 t-5 t=5 t$, Vasya and Petya approached each other at a speed of 20 km/h and met at point $E$ at time $t+\frac{10 t-5 t}{20}=\frac{5 t}{4}$.

where $D E=5 \cdot \frac{t}{4}, E C=15 \cdot \frac{t}{4}, C B=20-10 t, B E=\frac{15 t}{4}+20-10 t=20-\frac{25 t}{4}$

The time of the end of the trip for Vasya and Petya

$T_{1}=t+\frac{t}{4}+\frac{20-\frac{25 t}{4}}{10}=2+\frac{5 t}{8}$

The time of the end of the trip for Kolya

$T_{2}=t+\frac{20-10 t}{5}=4-t$

$T=T(t)=\max \left(T_{1}, T_{2}\right)=\left\{\begin{array}{c}T_{1}=2+\frac{5 t}{8}, t \geq \frac{16}{13} \\ T_{2}=4-t, 0 \leq t<\frac{16}{13}\end{array}\right.$

The graph of the function $T(t)$ is shown in the figure

![](https://cdn.mathpix.com/cropped/2024_05_06_508ac94edeef1bbe30b1g-3.jpg?height=439&width=443&top_left_y=1551&top_left_x=584)
Answer: 1) $t=\frac{16}{13}$ hour; 2) $T_{\min }=\frac{38}{13}$ hour