Kinematics bike - average speed for three to reach tourist center Three tourists have a bicycle, they must reach the tourist center in the shortest possible time (the time is counted until the last tourist arrives in the center). The bicycle can only carry two people and so the third tourist should go on foot. The cyclist leaves another tourist on the way and returns to pick up what was on foot. Walking speed is $ v_1 $ and biking is $ v_2 $. What is the average speed of tourists in functions $v_1$ and $v_2$?
I thought like this: $ y $ is the distance traveled by A with velocity Vb over a time interval $ ∆t1, $ ie $ y = Vb.∆t2; $
 A: There is a quicker way. The strategy is the following:
First the cyclist goes with another tourist and let him at some point $x$ to finish the travel by himself.
Then the cyclist goes back to pick up the third tourist that had started walking.
Finally, the three of them arrive together.
It suffises the compute $x$, so that they all arrive at the same time. It gives a average speed of:
$$v_a=\frac {v_2(3v_1+v_2)}{v_1+3v_2}.$$
A: Let $d$ be the distance for now. The shortest time is for two of them to bike to the center while the third walks. Then, the bike returns to pick up the third on the road. 
The time taken for the bike to reach the center is 
$$t_1=\frac{d}{v_2}$$
and the time taken for the bike to meet the third tourist on the road is 
$$t_2= \frac{d-v_1 t_1}{v_1+v_2}= \frac {d(v_2-v_1)}{v_2(v_2+v_1)}$$
Then, the total time it takes for all three to reach the center is 
$$t=t_1+2t_2= \frac {d(3v_2-v_1)}{v_2(v_2+v_1)}$$
Thus, the average speed is
$$v_a=\frac dt = \frac {v_2(v_2+v_1)}{3v_2-v_1}$$
