Take the hint, then f(t) is continuous on [0,T] (where T is the time they finish),

and assume that s1 and s2 are differentiable on (0,T), then so is f and by the

mean value theorem there is some point c in (0,T) such that:

f'(c)=(f(0)-f(T))/(0-T)=0

(as f(0)=f(T)=0).

But f'(t)=s1'(t)-s2'(t), so at some point c in (0,T) s1'(c)=s2'(c) as required.

RonL