13.420 A motorcade of cars with cosmonauts is moving uniformly along the avenue at a speed of $v$ km/h. The length of the motorcade is constantly maintained at $\boldsymbol{m}$ m. A bouquet of flowers, thrown from a window of a house, hit the motorcycle sidecar of a motorcyclist who was riding behind the motorcade. The motorcyclist rode forward, handed the bouquet to the cosmonaut in the first car, and immediately set off back. It took the motorcyclist $t$ min to travel there and back along the moving motorcade. Calculate the speed of the motorcyclist, which was the same throughout the journey.
Solution. Let $x$ be the speed (km/h) of the motorcyclist. Then he will catch up with the beginning of the convoy in $\frac{m}{1000(x-v)}$ hours, and then he will be at the end of the convoy in $\frac{m}{1000(x+v)}$ hours. According to the condition, $\frac{m}{1000(x-v)}+\frac{m}{1000(x+v)}=\frac{t}{60} \Rightarrow 25 t x^{2}-3 m x-25 t v^{2}=0$.

From this

Answer: $x=\frac{3 m+\sqrt{9 m^{2}+2500 t^{2} v^{2}}}{50 t}$ km/h.