Does Joe Still Win even though he has to run 10 meters more?

Yessir!  When Sam is at 90 meters so is Joe.  They are tied. Remember that Joe (faster) runs 100 meters in the same time that Sam runs 90.  Because they both have 10 meters to go, Joe being faster, wins.

That works, if you happen to “see” it.  How about using a more systematic approach?  Use the formula D = R * T, or R = D/T or T = D/R, where D = distance, R = rate, or speed. and T = time (the * means multiply, and the / means divide)

Here’s a way to think about it.  What do we want to know?  Is Joe’s time less than Sam’s.  So we have to figure out how long each takes in the second race — but we don’t know the times or rates of the runners.  We do know (better: can figure out) the relative rates.

You get that from the first race. (we’re assume the same speeds in the second race).  Then using that information you can calculate the relative times, T, for each — for Joe to run 110 meters, and Sam to run 100 meters.  Notation: I’ll use (j) for Joes’ variables and (s) for Sam’s.

From the first race:

Joe: R(j) = 100/T,  Sam: R(s) = 90/T  (T = same for both)

Solve both for T and then equate them.

T = 100/R(j) = 90/R(s) so,

R(s)/R(j) = 90/100.   (Sam runs 90% as fast as Joe).

Now for the second race: Here the times may be different. Solve for the time for each, and see if we can get that ratio of rates.  If we can, then we just plug in the 90/100.

T(j) = 110/R(j) and T(s) = 100/R(s)

Now, divide the entire two equations:

T(j)/T(s) = [110/R(j)] / [100/R(s)]

T(j)/T(s) = 110/100 * R(s)/R(j).

Substituting: = 110/100 * 90/100 = 9900/10000 = 0.99.  Joes time is 99% of Sam’s.

I heard you.  “What a pain!  Confusing! Why work so hard to do something that you can just figure out in your head!?”

You are right — sort of.  What happens is that there are many problems, analyses, and opportunities that you can’t just “see” the answer.  You need to go through some logic. Each step builds on it’s predecessor.  You also have to get very good at manipulating symbols — and develop habits that help keep you from making mistakes — especially typos.

Exercise for the reader:  How far back does Joe have to start for them to tie?

This may be overkill, but here’s how I’d solve the problem in a more general way.  (Parens hold current problem’s data)

Let d = race distance (100 meters)

Let x = “loss distance” (10 meters)

let y = “setback distance” for the second race (10 Meters)

From the first race we know that R(j)/R(s) = d/(d-x).  Makes sense.  Ratio of speeds inverse to ratio of distances.

For second race Joe runs d+y, while Sam runs d.  If you go through the same steps we did before we get

T(j)/T(s) = (d+y)*(d-x) / (d*d),  and if y = x

T(j)/T(s) = (d*d – x*x) / (d*d), clearly less than 1 (could have written d*d as d^2 or figured out how to write exponents  🙂

If you want to find out what y is for them to tie, you just have to set the above expression = 1 and solve for y.

(d+y)*(d-x) = d*d,  divide (d-x) both sides

d+y = d*d / (d-x), subtract d both sides

y = d*d / (d-x) – d.  Plugging in d = 100 and x = 10 we get

y = 11.1111 meters.  You could also expand the (d+y)*(d-x) and collect terms. Then y = x*d / (d-x), plug in the numbers and you get the same answer.


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