X multiplied by Y = why

Many runners enjoy running in groups. Running with company can help in terms of safety, and just extra enjoyment to pass the time.

However, everybody runs at different paces.

I recently experienced a situation where runners set off at different times, running at their own speeds, and hoped to meet up at some point on the same route.

It seemed like the classic high school maths question, “If Train #1 leaves the station travelling at 60km/hr and 30 mins later Train #2 leaves the same station travelling on the same train line at 80km/hr, how long before the trains collide?”.

I’m not good at algebra at the best of times. Let me tell you though, while running I found new levels of hopelessness trying to solve the puzzle of where one runner would catchup to others running at different paces.

Let me document the solution now, for posterity – thanks to http://www.algebra.com/.

The translated problem:

If Runner “Tortoise” leaves at 4:30am travelling at 9km/hr, and Runner “Hare” leaves at 5:00am travelling 12km/hr. At what distance will they meet?

T’s lead = 4.5 kilometres (30 mins @ 9km/hr)

Catching up speed is 3 km/hr (H’s speed 12km/hr – T’s speed 9km/hr)

Solution is to divide the 4.5 kilometre lead, by the speed which H can made up,  3. 

So the answer is…. 1.5 hours.

Let’s test.

The distance which T covers in 1.5 hours is 13.5 kms, then add the 4.5 km head start from beginning 30 mins earlier would mean they would complete 18 kms.

The distance which H covers in 1.5 hours is 18 kms.

The hare after sleeping-in should meet the tortoise after 18kms.