The Great Chase Burger

Cyclist $A$ has speed $55$ km/h, cyclist $B$ has speed $50$ km/h, cyclist $C$ has speed $45$ km/h, and the van has speed $40$ km/h. The race is $240$ km long; only cyclists $A$ and $B$ meet the van.

To work all this out, let the race length be $s$ . It takes the van $6$ hours to cover this distance, so its speed is $\frac{s}{6}$ (from now on all speeds will be in km/h and all times in hours).

The van starts an hour before the cyclists, so has a headstart of $\frac{s}{6}$ km.

Consider a cyclist riding at speed $v$ . The speed of the cyclist relative to the van is $v-\frac{s}{6}$ , so it will take the cyclist a time of $t_1=\frac{\frac{s}{6}}{v-\frac{s}{6}}=\frac{s}{6v-s}$ to catch the van. This happens at a distance $\frac{sv}{6v-s}$ from point $E$ . After this, the cyclist's speed reduces by $20$ ; there is a distance of $s-\frac{sv}{6v-s}$ remaining in the race, which takes the cyclist $t_2=\frac{s-\frac{sv}{6v-s}}{v-20}$ to complete. The total time for the cyclist to complete the race is therefore $t_1+t_2=\frac{s}{6v-s}+\frac{s-\frac{sv}{6v-s}}{v-20}=\frac{s(s-6v+20)}{(v-20)(s-6v)}$

If any two of the cyclists don't reach the van, they must take a different time to finish the race; so at least two of them do reach the van.

It's easy to check there's no solution if all three cyclists reach the van; the only remaining possibility is that cyclists $A$ and $B$ reach the van during the race, and $C$ (the slowest) does not.

If cyclist $A$ has velocity $v_A$ , and so on, then we have to solve $T=\frac{s(s-6v_A+20)}{(v_A-20)(s-6v_A)}=\frac{s(s-6v_B+20)}{(v_B-20)(s-6v_B)}=\frac{s}{v_C}$

subject to $v_A-v_B=v_B-v_C=5$ . Solving this leads to the results above.

Note that even though cyclist $C$ is faster than the van, it would take $8$ hours for them to meet - longer than the race itself.