Question

Two Tibetan monks are having a race along a straight path. They both start at the same time, and the race ends in a tie. Prove that at some time during the race the two monks have exactly the same speed.

Answer 1

This is so provided that the velocity changes continuously in which case we can apply the mean value theorem. 
Velocity (v) is the derivative of displacement (x) : 
v = dx/dt 
Monk 1 arrives after a time t* and Monk 2 too. 
Name v1(t) and v2(t) their respective velocities throughout the trajectory. 
Then we know that both average velocities were equal : 
avg1 = avg2 
and avg = integral ( v(t) , t:0->t*) / t* 
so 
integral (v1(t), t:0->t*) = integral (v2(t), t:0->t*) 
which is the same of saying that the covered distances after t* seconds are the same 
=> integral (v1(t) - v2(t) , t:0->t*) = 0 
Thus, name v#(t) = v1(t) - v2(t) , then we obtain 
=> integral ( v#(t) , t:0->t*) = 0 
Name the analytical integral of v#(t) = V(t) , then we have 
=> V(t*) - V(0) = 0 
=> V(t*) = V(0) 
So there exist a c in [0, t*] so that 
V'(c) = (V(t*) - V(0)) / (t* - 0) (mean value theorem) 
We know that V(0) = V(t*) = 0 (covered distances equal at the start and finish), so we get 
V'(c) = v#(c) = v1(c) - v2(c) = 0 
=> v1(c) = v2(c) 
So there exist a point c in [0, t*] so that the velocity of monk 1 equals that of monk 2.