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.
This problem does not really constrain the runners' speeds, but rather the ratio of their speeds. It also does not constrain the distance to R, but rather the approximate time it takes them to get to R.
Call the distance to R: "R" and the three speeds: SA, SB and SC.
Call the time it takes A to get to point R: TRA. Using speed = distance / time:
SA = R/TRA, SB = R/(TRA-5), SC = R/(TRA-10)
When returning,
C meets B at C's meter 64 and B's meter 46. [55 +/- 9]
When returning,
C meets A at C's meter 70 and A's meter 40. [55 +/- 15]
Equating the event times for the passing runners, using time = distance / speed:
64/SC = 46/SB (eq. 1)
64 / [R/(TRA-10)] = 46 / [R/(TRA - 5)]
R drops out and TRA = 410/18
Similarly:
70/SC = 40/SA (eq. 2.)
70 / [R/(TRA-10)] = 40 / [R/TRA]
R drops out and now TRA = 70/3.
These solutions for TRA are different: 410/18 vs 420/18.
So we see that TRA is ill-defined in this problem to a couple of percent (one part in 42). TRA is around 22.8-23.3 sec
Equations (1) and (2) also yield the ratios of the runners' speeds:
SC = (7/4) SA
SB = (23/32) SC
There are 3 "meetings" described in the race: all three runners at R and then C passing each of the others. Playing with TRA near 23 sec we can make one or two of the events match to a half second, but all three meetings cannot be made instantaneous with the positions given. (I think...)
So it is a trick question. Reminds me of this movie clip.
Edited on September 27, 2019, 3:22 am
soln
