Three sprinters A, B, and C had to sprint from points P to Q, which are 55 meters apart, and back again (starting in that order). The time interval between their starting times was 5 seconds each. C started 10 seconds after A, while B started 5 seconds after A. They passed a certain point R, which is somewhere between P and Q, simultaneously (none of them having reached point Q yet). Having reached Q and reversed the direction, the third sprinter met the second one 9m short of Q and met the first sprinter 15m short of Q.
Find the speed of the three sprinters.
(In reply to
re: soln by Kenny M)
Thanks so much Kenny M! Yes, I got my time formulae wrong. Fixing these changes all the results because R does not now drop out. Again, TRA is the time A takes to get to R.
As you point out, the correct equations for the (C,B) and (C,A) crossings include delay times and I give the new equations below. Here I use "clock times" where A starts sprinting at clock time 0:
Time when C crosses B:
64 / [R/(TRA-10)] + 10 = 46 / [R/(TRA-5)] + 5
Time when C crosses A:
70 / [R/(TRA - 10)] + 10 = 40 / (R/TRA)
-----
Solving these gives R = 10m and TRA = 20s
So:
Speed A = 10/20 = 0.5 m/s
Speed B = 10 /(20-5) = 0.667 m/s
Speed C = 10/(20-10) = 1.0 m/s
(BTW, Most runners can sprint at many times these speeds)
Anyway, I am sorry I suggested the problem was a trick problem. It is a very nice problem in actuality.
Edited on September 27, 2019, 4:55 pm
re(2): soln - Correction!
