Optimizing Travel Time for Three Brothers Exhibiting Their Vehicles

[ceptimus]

Three brothers decide to exhibit their two vehicles at a vintage car and motorcycle show twelve miles from their house. They jointly own a single-seat car and a very early single-seat motorcycle. They are discussing how to get themselves and their vehicles to the show.

"Our car goes at 60mph and our bike at 15mph", says Archie. "Whoever drives the car will only take twelve minutes to get there, the bike will take forty-eight minutes. One of us will have to walk - at four miles an hour that will take three hours."

"Why not share the driving?", says Brian. "If one of us drives the car for four miles, then parks it, and another of us does the same with the bike, then we can swap over, and park them again after another four miles. If we work it out right we can each walk, ride and drive for one third of the distance - that will be fair and we'll all arrive together and faster than one of us having to walk the whole way."

"I see what you mean", says Charlie, the youngest and smartest brother. "But there will be some waiting around involved because the bike won't be at the final swapping point early enough. I have a better idea..."

What is Charlie's idea and how quickly can all three brothers arrive at the show if they all set off together and share the driving, riding and walking?

 

[mabs239]

Is it possible that they don't wait for the late comers and just leave the vehicle where one have to leave it? The first person to reach first 4mile line could leave the car there and moves on foot. Then bike rider comes at this point and travels on car for next 4mi and parks after another 4miles. The person who starts by walking, takes the bike then Car.
The sequence being
Walk-Bike-Car
Bike-Car-Walk
Car-Walk-Bike

Their avg speed = (60+15+4)/3

 

[ceptimus]

The problem with that approach is that one brother has to wait at the eight mile mark for the motorcycle to arrive:

Brother one: Walks four miles (60 minutes), then rides the motorcycle for four miles (16 minutes), takes car for final four miles (4 minutes) = 80 minutes total.

Brother two: Rides motorcycle four miles (16 minutes), takes car for four miles (4 minutes), walks remaining 4 miles (60 minutes) = 80 minutes total.

Brother three: Drives car four miles (4 minutes), walks four miles (60 minutes), now has to wait for 12 minutes for Brother one to arrive with the motorcycle before he can ride the final four miles (16 minutes) = 92 minutes total.

 

[mabs239]

I made mistake in writing the avg speed formula. Avg time should by 80 minutes in my proposed strategy, which is of course not correct.

This problem has really confused. If all of them have to be simultaneously set foot in the show, then considering when any of them would have started.

Let us say no one has changed his vehicle or velocity since a constant time x. We may suppose 10 minutes(Just to start) before reaching the show. That means car has been traveling for 6miles, bike for 1.5 miles and walk for 0.4 miles.

Now the person who is walking could not have either a car or a bike before this last 0.4 miles as both are engaged and are left behind. He is already 9.6miles ahead of car and 1.1miles ahead of car.

The person on bike could not have a car before because car is already left behind 4.5miles.

There is no restriction on car driver, he could have been walking or on bike before.

I think from here we could move backwards until all times and distances are equal. This problem has moved my attention to Dynamic Programming, Multi tasking and Lagrange functions at the same time.

Good puzzle any way. I am still thinking...

 

[mabs239]

I hope this time I have done it.

The plan goes like this:

Spoiler

All three start at the same instant.

Bike rider goes for 36.4 minutes and travels 9.1 miles. Leaved the motor cycle there and covers the rest of the distance on foot, 2.9 miles in 43.6 minutes.

Car rider travels for 4.86 minutes covering 4.86 kilometers. Then he leaves the car there and covers 4.23 miles in 63.5 minutes. There he founds the motorcycle left by first brother and covers rest of the distance, 2.9 miles in 11.64 minutes.

The third brother goes on foot for 72.86 minutes and covers 4.86 miles. There he finds the car by second brother and covers 7.14 miles in 7.14 minutes.

For accuracy (minutes, miles): (400/11, 100/11), (1020/14, 34/7), (34/7, 34/7), (752/11, 100/11)

I am feeling my self guilty as this problem has stolen a lot of time from my home work. I couldn’t help spending my time here. Anyway thanks for a good problem…