Uniform Motion Problem

Two mountain bikers started from the ranger station in the Cascade Mountains traveling in the same direction on the same road. The first biker left 1 hour earlier than the second and traveled 10 mph. The second biker left later and traveled 15 mph. By 4:00 p.m., the second biker had passed the first and was 5 miles ahead of the first biker. What time did the first biker leave the ranger station?

Heres my work...

Velocity Time Distance
First Biker 10 t-1
Second Biker 15 t +5


Equation I came up with...

15t + 5 = 10(t-1)
15t + 5 = 10t - 10
5 = -5t - 10
15 = -5t
t = -3

so the first biker would have left at 1:00 p.m. (4 - 3 = 1)

Does this sound correct or did I set up the formula improperly. If so, can someone tell me where I messed up?

Thank You!

No; not quite. Since speed = distance / time, then distance = speed * time:

1st biker: distance = 10(4 - t)
2nd biker:distance = 15(4 - t - 1) : left an hour later

Since 2nd biker is 5 miles ahead:
10(4 - t) = 15(3 - t) - 5 ; divide by 5:
2(4 - t) = 3(3 - t) - 1
8 - 2t = 9 - 3t - 1
t = 0 : so left at noon

Btw, IF your t = -3 was correct, then 1st biker left at 9am : capish?

Thanks for clarifying this. This makes much more sense to me now!