Race of the century

Question

Four drivers face off in the race of the century. The path of the race follows this layout:

• 100 KM road with a sharp turn every 5 KM
• 50 KM straight road
• 50 KM straight river
• 40 KM upstream river
• 40 KM downstream river
• 40 KM uphill road with a sharp turn every 2 km
• 40 KM uphill straight road
• 40 KM straight river
• 30 KM downhill road with a sharp turn every 5 km
• 50 KM dowhill straight road

The four drivers build their vehicles with special attributes and specific values, they always follow these setting and never cheat, they always accelerate if they can until they are at their maximum velocity. They only come to a full stop when they need to change modes and they only slow down at turns if they are going to fast to actualy make that turn. All four vehicles have two modes: normal and amphibious. The vehicles behave as follows:

Driver 1: Java Gee Roe

Maximum speed: 1000 m/s
Acceleration: 2 m/ss, minimum 1 m/ss
Breaking speed: 25 m/ss
Maximum turn speed: 350 m/s
Time to change modes: 50 s
Special: Every 10,000 m, the vehicle's acceleration is increased by 15 m/ss for 12 seconds.

Driver 2: Frames Wist

Maximum speed: 600 m/s
Acceleration: 1 m/ss, minimum 0.5 m/ss
Breaking speed: 20 m/ss
Maximum turn speed: 300 m/s
Time to change modes: 0
Special: This vehicle requires no time to change modes, it doesn't need to stop or even slow down to go from road->water or water->road.
It also gains +2 m/ss to acceleration while in water.

Driver 3: Pupa Noblest

Maximum speed: infinite
Acceleration: 1 m/ss, minimum 0.5 m/ss
Breaking speed: 50 m/ss
Maximum turn speed: 550 m/s
Time to change modes: 70 s
Special: This vehicle gains 0.02 m/ss acceleration every second.
Acceleration is reset back to 1 m/ss when the vehicle needs to slow down or stop.

Driver 4: Bight Fuller

Maximum speed: 1200 m/s
Acceleration: 1.5 m/ss, minimum 0.5 m/ss
Breaking speed: 60 m/ss
Maximum turn speed: 500 m/s
Time to change modes: 30 s
Special: Every time a vehicle is infront of this one, this vehicle gains a acceleration bonus depending on distance between the two:

• 0-2 km -> 0.05 m/ss
• 2-4 km -> 0.04 m/ss
• 4-6 km -> 0.03 m/ss
• 6-8 km -> 0.02 m/ss
• 8-10 km -> 0.01 m/ss
• > 10 km -> 0.005 m/ss

Acceleration is reset back to 1.5 m/ss every time this vehicle overtakes another one.

On every uphill/upstream road part, the vehicle's acceleration is reduced by 1 m/ss it can't go below their minimum acceleration. At downhill/downstream parts the vehicles gain 1 m/ss.

At what place do each of the drivers finish the race?

Answer 1

And their off...

Advancing into the first turn at 5KM, JavaGeeRoe's superior speed at a full 50CM/ss showing off. Poor Wist isn't even half-way to the turn yet! JavaGee in the lead, Pupa and Bight fighting for 2nd.

On to the turns, Java's gotta stomp on those breaks to keep control, but Pupa just laughs and continues at full speed. I gotta get some binoculars to see Wist way in the distance not even half-way done with the turns. Completing the turns is Pupa at 100KM, Java & Bight neck and neck at 84/83KM Wist back at the 44M mark.

A quick stop to reconfigure at the river, Pupa's still got the lead at 150KM, Java and Bight still sparring over 2nd at 132/130, Wist still working his way through the turns back at 70KM.

River Stretch Completed.

What's this? Pupa seems to be having mechanical difficulties. His infinite accelerator has ground to a halt in the water. Caulk's cheap, you'd think he'd have waterproofed it! Meanwhile Java's nitrous works just fine and he's left all the competition in the dust. Java Gee Whiz stopping to reconfigure just before the uphill hairpins at 280KM, Pupa at 216KM working his way upstream followed by Bight and Wist still on the flat part of the river. Wist is travelling much faster than anyone else, he didn't even slow-down for river, just kept accelerating! Still, Java's 60KM+ lead is over 10% of the track can he be caught?

Checking in from the mini-lake at 360KM, the hairpins didn't seem to slow Java down, that nitrous boost of 15m/ss letting him maintain a fairly high average speed in the turns, Wist passing Bight, Pupa, leaving them behind at 313KM, he's about to enter the hairpins. Bight has just executed a high-speed draft pass on Pupa as they tie at 285KM. But, drafting only lets you keep second, that'll never win here. In fact, I'd say it only worked because Bight tossed one of his oily rags straight into Pupa's accelerator while they were reconfiguring!

Lake Hop completed

Java's leaving the river at 400KM. What's he doing? He's actually stopped to wash his car! What a show-off. It may cost him though, Wist isn't far behind. Java may have forgotten that Wist won't stop at the edge. Pupa's left Bight behind at the turns and is now at 355KM, Bight at 318KM.

Watching from the finish line, It's Wist coming out of the quick turns at 430KM, Java not even a single KM behind. Pupa and Bight meanwhile have just finished swapping to water gear to enter the last river. Pupa at 363KM, bight not far behind at 360KM.

Finish Line

FINISH And the prize goes to Java. Race completed in just over 22min 10secs. Right behind is FramesWist at 23min, 7secs. Java's top speed took home the prize. 7min later, for those of you who stuck around, Bight and Pupa. Bight executed a well-timed manuver passing Pupa just before the finish line to end in 27min, 57secs compared to Pupa's 28min.

Closing Banter

Java could cut his nitrous by 10% and still win. Wist states he'd have won if not for the hairpins, just 10% faster through the pins would have done it. Bight points to Pupa's habit of throwing rocks at him when he passes and says that if Pupa just has a slightly faster car, they wouldn't be stuck so far behind. Pupa claims Bight is being ridiculous and that he had the fastest car of them all if not for those rivers. Who ever heard of a race with a lake?

Completed via a Java program, may have inaccuracies.

Answer 2

Here is a first stab... please continue if you think it is appropriate

First off, I noticed the racers names were anagrams.

1. Average Joe
2. Swim Faster
3. Unstoppable
4. Bull Fighter

Before we start, here is a handy formula for calculating distance travelled under constant acceleration.

$$d=v_i\times t+\frac{a\times t^2}{2}$$

100 km

The fastest out of the gate is 1 followed by 4. 3 starts off similar to 2 but his acceleration increases. This makes 2 the slowest of the 4 for this stretch, and the easiest to calculate for this section.

2. Swim Faster

First 9 turns - up to 45km

His maximum turn speed is $v=300m/s$, which at an acceleration of $1m/s^2$ will be achieved after $t=300s$. The distance travelled in this time is $d=\frac{(300s)^2}{2}=45,000m$. This is conveniently right at turn 9, so he accelerates all the way up to this point.

Next 10 turns - up to 95km

For the next 10 turns he will accelerate at $a_a=1m/s^2$ until just before the turn when he will start breaking at $a_b=-20m/s^2$. Since the starting and ending velocities are the same, we know that whatever speed is gained during acceleration must be lost during deceleration. Thus, we get the following:

$$v_{max}=v_i+a_a\times t_a$$ $$v_i=v_{max}+a_d\times t_d$$

Thus: $$v_i=v_i+a_a\times t_a + a_d\times t_d$$ $$a_a\times t_a = -a_d\times t_d$$ $$t_a = 20 \times t_d$$

As for distances, we know that $$d_a=v_i \times t_a + \frac{a_a \times t_a^2}{2} = 300m/s \times t_a + \frac{1 m/s^2 \times t_a^2}{2} = t_a\frac{600 + t_a}{2}$$ $$d_d=v_{max} \times t_d + \frac{a_d \times t_d^2}{2} = v_{max} \times t_d + \frac{-20m/s^2 \times t_d^2}{2} = t_d\frac{2v_{max} - 20t_d}{2}$$ $$d_a + d_d = 5,000$$

Substituting in, we get: $$d_a=t_a\frac{600 + t_a}{2}=20t_d \frac{600 + 20t_d}{2}$$ $$d_d=t_d\frac{2v_{max} - 20t_d}{2}= t_d\frac{2(v_i+a_a\times t_a) - 20t_d}{2} = t_d\frac{2(300+20t_d) - 20t_d}{2}=t_d\frac{600+20t_d}{2}$$ $$5,000=d_a + d_d = 20t_d \frac{600 + 20t_d}{2} + t_d\frac{600+20t_d}{2}=21t_d\frac{600+20t_d}{2}$$ $$420t_d^2 + 12,600t_d - 10,000 = 0$$ $$21t_d^2 + 630t_d - 500 = 0$$

Solving this equation gives us $t_d=5\sqrt{\frac{209}{21}}-15~=0.77s$.

We can verify that this is the answer by seeing that $t_a~=15.4s$ and the distance travelled during that interval is:

$$d_a=300m/s \times t_a + \frac{1m/s^2 \times t_a^2}{2}=4761.9$$ $$v_{max}=300m/s + t_a$$ $$d_d=v_{max} \times t_d + \frac{-20m/s^2 \times t_d^2}{2}=238.1$$

Adding these up give $5,000$, and the time it takes to travel this entire distance is $t_a + t_d ~= 16.25s$.

Last turn

For the leg he must come to a complete stop before entering the water.