Solution. We solve the problem in a coordinate system associated with the first motorcyclist, denoted as $A$. The positions of the second motorcyclist: at the first moment of time $-B$; at the second moment $-C$; at the third $-D$. Then $A B=$ $40, A C=30, A D=30$.

Since $B C: C D=\frac{46-14}{70-46}=\frac{32}{24}=4: 3=A B: A D$, then $A C$ is the bisector of $\triangle A B D$. By the formula for the length of the bisector, $30^{2}=40 \cdot 30-4 x \cdot 3 x$ (here $B C=4 x ; C D=3 x$). Therefore, $x=5, C D=15$ km. Since the distance $C D$ is covered in 24 minutes, the required relative speed is $\frac{15 \cdot 60}{24}=\frac{75}{2}=37.5$ km/h.
12.1. Two motorcyclists are moving along straight tracks (each on their own) at constant speeds. At 11:14, the distance between them was 40 km, at 11:46 - 30 km, at 12:10 - 30 km. Determine the magnitude of the relative speed of one motorcyclist relative to the other (in kilometers per hour). Provide the answer as a whole number, rounding it if necessary.