Kits Triathlon Math Problems!

Problem #1
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On Saturday, June 3, Runner G leaves Kits Pool at 10:30 am, traveling towards the peak of Grouse at a slothful pace of 5:45 min per km. He is exhausted from swimming, has an injured ass, and is very cold. Runner R leaves Kits Pool at 10:35 am, traveling towards the peak of Grouse at a reasonable pace of 5:03 per km. He has an "injured" knee, is somewhat fat and unwell, but runs like a man on fire. Assuming both runners survive, when and where does Runner R overtake Runner G?

Let R1 = Runner G's rate = 1 km/5.75 min = 0.1739 km per min

Let R2 = Runner R's rate = 1 km/5.05 min = 0.1980 km per min

Let T = time for Runner G

Then T-5 = time for Runner R

By hypothesis they cover the same distance, and D=R*T, therefore:

R1 * T = R2 * (T-5)
0.1739 * T = 0.1980 * (T-5)
0.1739T = 0.1980T - 0.9900
0.9900 = 0.0241T
T = 41.08

Eliminating the fraction due to variables such as fatness and injuries, Runner R overtakes Runner G approximately 41 minutes after Runner G starts, namely 11:11 am (G takes 41 mins, R takes only 36 mins)

The distance they have traveled is:

D = R1*T (or R2*(t-5))
D = 0.1739*41.08 (or 0.1980*(41.08 - 5)
D = 7.14 km

So Runner R overtakes Runner G somewhere on the Stanley Park Causeway, just before the Bridge. Hurray! (and thanks to Google Maps).

Problem #2
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It is not likely that Runner G or Runner R will keep up the same "vigorous" pace in the brutal ascent of Capilano Road. Assume Runner R, just after he overtakes Runner G, travels from the start of Stanley Park Bridge to the Grouse parking lot (about 9km) at an average rate of 5:42 min per km. Runner G travels the same distance at an average rate of 6:24 min per km. Assuming both runners survive, how much faster must Runner G be on the Grouse Grind if he is to catch up to Runner R before the finish?

Let R3 = Runner R's rate = 1 km/5.70 min = 0.1754 km per min

Let R4 = Runner G's rate = 1 km/6.40 min = 0.1562 km per min

Let T = Time Runner G will need to make up

Then T = Runner G's run time - Runner R's run time

By hypothesis they cover the same distance (9km), and D=R*T, therefore:

T = D/R4 - D/R3
T = (9/0.1562) - (9/0.1754)
T = 57.62 - 51.31
T = 6.31

Rounding up generously due to Runner G's inevitable cardiac problems somewhere near the Cleveland Dam, we can predict that Runner G will need to make up 7 minutes on the Grouse Grind to catch Runner R.

Problem #3
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Assume you are clever. Predict Runner R's and Runner G's times in the Kits Triathlon!

Here goes:

Runner R:
swim 1.1km - 26:30, T1 - 3:30, run 16km - 1:28:00, T2 - 3:00, Grind 49:00
Total: 2hrs 50mins

Runner G:
swim 1.1km - 22:00, T1 - 3:00, run 16km - 1:38:30, T2 - 3:00, Grind 44:30
Total: 2hrs 51mins

As for Runner K and/or Runner RJ, I have no idea, but they should come in at or around 3 hrs.

Onward!!