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.
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.