"Simple" Math question.

We were presented with this question in my math 111 class yesterday.

Sally and bob leave for each others house at sun rise. They pass each other at noon and at 4pm sally arrives at bobs house, at 9pm bob arrives at sallys house. What time did the sun rise?

This is an extra credit question because the professor couldn't solve it after working on it for 30 minutes. So can any one help show us how to do this by using math 111 equivalent procedures?

I imagine their speeds are constant. Say sunrise occurred at time t=s, Bob's speed is v, and Sally's speed is u. Also suppose Bob's position is 0 at time s, and Sally's is 1 at time s. Bob's position is then B(t) = (t-s)v, and Sally's is S(t) = (s-t)u + 1. We have the following data:

B(12) = S(12)
S(16) = 0
B(21) = 1

Explicitly,
(12-s)v = (s-12)u + 1
(s-16)u + 1 = 0
(21-s)v = 1

This is a non-linear system of equations, though it's not terrible. Having a computer algebra system at my disposal, the solutions are

s = 6 and u = 1/10 and v = 1/15
s = 18 and u = -1/2 and v = 1/3

where we can discard the second solution since in that case Bob is walking away from Sally's house. That is, sunrise occurred at 6am.


Solving this system by hand isn't horrible, either. We can rewrite the first equation in terms of s alone using the second two equations:
u = -1/(s-16) = 1/(16-s)
v = 1/(21-s)
=> (12-s)/(21-s) = (s-12)/(16-s) + 1
=> (12-s)(16-s) = (s-12)(21-s) + (16-s)(21-s)
=> s^2 - 24s + 108 = 0
=> s = 6 or 18

(a) Let R = the time of sunrise in hours, and let D = the distance between the two houses.

(b) At 4 PM (16 hours), Sally (S) has traveled D and at 9 PM (5 hours later, at 21 hours),

Bob (B) has also traveled D.

(c) So, Sally's speed of travel SS = D/(16 - R) and Bob's speed BS = D/(21 - R)