Design a fair race in which two animals have an equal chance of winning if they race at their top speed. The animals are a coyote with speed of 43mph, and rabbit with speed of 35mph. Write a system of two linear equations showing the distance each animal can travel to model the fair race. Be sure to define all variables. (tips: the coyote and the rabbit must finish the race at the same time, you can give slower animal shorter distance, and the animals can run at a constant speed.
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One way to resolve this is to work out the distance each must run to take the same time to complete the race. Distance=speed times time, so time=distance÷speed. So if the distances are D1 and D2 we can write D1/43=D2/35 (time for coyote=time for rabbit) and that means D1/D2=43/35 (about 1.23). If the ratio of the distances is the inverse of the speeds then the race will be fair. (D1=43T (coyote) and D2=35T (rabbit); divide one equation by the other: D1/D2=43T/35T so D1/D2=43/35.)

TRACK ARRANGEMENTS FOR FAIR RACE

We could have two parallel linear tracks but the rabbit starts at a different point along its track. If the longer track is 43n miles long then the shorter track is 35n miles long where n is a number, probably a fraction, for example 1/10. The rabbit's start-line is 43n-35n=8n miles beyond the coyote's. The rabbit runs only 35n miles while the coyote runs 43n miles. The finishing-lines are in line. The race will take n hours if both animals run at top speed on average throughout the race. So if, for example, n=1/10 the race will last for 1/10 hr = 6 minutes=4.3/43=3.5/35. The rabbit runs 3.5 miles while the coyote runs 4.3 miles.

Another way is to use circular tracks. Since the circumference of a circle is a fixed multiple (2π) of the radius, we have two concentric circular tracks where the radius of the inner track is R1 (for the rabbit) and radius of the outer track is R2 so that R1/R2=35/43=2πR1/2πR2, where 2πR1 and 2πR2 are the lengths of the inner and outer tracks. The advantage of circular tracks is that the race can be run over a number of laps rather than just one lap. The race is more exciting because spectators can surround the track and observe the progress of the race better.

A third solution is to have semicircular tracks placed at the two ends of a rectangular circuit. The curvature of the circles compensates for the difference in speeds of the animals. The width of the rectangle is 2R1, where R1 is the radius of the inner track. If L=length of the rectangle then the length of the tracks are (2L+2πR1) and (2L+2πR2), and (2L+2πR1)/(2L+2πR2)=35/43=(L+πR1)/(L+πR2), so 43L+43πR1=35L+35πR2; 8L=π(35R2-43R1) (35R2>43R1, so R1/R2<35/43). This equation relates the length of the rectangular part of the track to the inner and outer radii of the semicircular parts of the tracks. If this equation is satisfied the start and finish lines will be in line. If the equation is not satisfied then, the start and finish lines will be out of line by the value of the expression 8L-π(35R2-43R1).

(EXAMPLE: If R2=2R1 for the above type of track, 8L=π(70R1-43R1)=27πR1, and L=27πR1/8=10.6R1 approx. or 5.3W where W=2R1 is the width of the rectangle.)

This answer probably goes into more detail than you require, but I hope you get the general idea.

 

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