The Bicycle Race

1. The Bicycle Race Dr. Spackman October 21, 2013

2. Goal: Find the “ideal” value of k (Carol’s speed) so that the support car can catch her and, at the instant the support car is alongside, their speeds are the same. Support car position Carol’s position

3. We want to find k so that In other words: and

4. We want to find k so that In other words: and Substituting the value of k given by the second equation into the first, rearranging and factoring, yields: and thus Since t=0 corresponds to k=0 (in which case Carol is not racing at all), the only meaningful solution is t=5, in which case k=25/3.

5. We want to find k so that In other words: and Substituting the value of k given by the second equation into the first, rearranging and factoring, yields: and thus Since t=0 corresponds to k=0 (in which case Carol is not racing at all), the only meaningful solution is t=5, in which case k=25/3. Therefore, in order for the support car to catch Carol at the instant they have the same speed, Carol must be traveling at 25/3 meters per second, and the drink hand-off occurs exactly 5 seconds after Carol passes the refreshment station.

6. We want to find k so that In other words: and Substituting the value of k given by the second equation into the first, rearranging and factoring, yields: and thus Since t=0 corresponds to k=0 (in which case Carol is not racing at all), the only meaningful solution is t=5, in which case k=25/3. Therefore, in order for the support car to catch Carol at the instant they have the same speed, Carol must be traveling at 25/3 meters per second, and the drink hand-off occurs exactly 5 seconds after Carol passes the refreshment station. NOTE: The cubic polynomial has a double root at t=0. We’ll address that later.

7. Just for fun, let’s find out what would happen if Carol were travelling faster or slower than the ideal speed of 25/3 m/sec. by solving the equation C(t)=S(t) for t. Rearranging produces

8. Just for fun, let’s find out what would happen if Carol were travelling faster or slower than the ideal speed of 25/3 m/sec. by solving the equation C(t)=S(t) for t. Rearranging produces Applying the quadratic formula yields solutions

9. Just for fun, let’s find out what would happen if Carol were travelling faster or slower than the ideal speed of 25/3 m/sec. by solving the equation C(t)=S(t) for t. Rearranging produces Applying the quadratic formula yields solutions Therefore, if k>25/3, there are no times (after t=0) that the support car and Carol are in the same position—Carol is travelling too fast for the support car to catch up! And if 0<k<25/3, there are two times when Carol and the support car are in the same position; in other words if Carol is going too slowly the support car passes her (going too fast to safely pass the drink), and then Carol passes the support car (which is now travelling too slowly to safely pass the drink).

10. Just for fun, let’s find out what would happen if Carol were travelling faster or slower than the ideal speed of 25/3 m/sec. by solving the equation C(t)=S(t) for t. Rearranging produces Applying the quadratic formula yields solutions Therefore, if k>25/3, there are no times (after t=0) that the support car and Carol are in the same position—Carol is travelling too fast for the support car to catch up! And if 0<k<25/3, there are two times when Carol and the support car are in the same position; in other words if Carol is going too slowly the support car passes her (going too fast to safely pass the drink), and then Carol passes the support car (which is now travelling too slowly to safely pass the drink). NOTE: When k=25/3, the equation has a double root of t=5. An explanation will follow.

11. Let’s see what these findings look like graphically. First, here is the support car’s position function graphed with Carol’s position function at her ideal speed of 25/3 m/sec.

12. Next, here is what the two position graphs look like when Carol’s speed is 10 meters per second.

13. And here is the support car’s position function graphed with Carol’s position function, assuming her speed is lower than 25/3—here her speed is 7 meters per second.

14. Finally, there were two instances where we encountered a cubic polynomial with a double root: the support car’s position function and the cubic polynomial equation expressing the times that Carol and the support car meet when Carol is travelling at ideal speed From the first, we can deduce that both S(0)=0 and S’(0)=0. From the second we deduce that both C(5)=S(5) and C’(5)=S’(5).

15. Finally, there were two instances where we encountered a cubic polynomial with a double root: the support car’s position function and the cubic polynomial equation expressing the times that Carol and the support car meet when Carol is travelling at ideal speed From the first, we can deduce that both S(0)=0 and S’(0)=0. From the second we deduce that both C(5)=S(5) and C’(5)=S’(5). In general, if P(x) is a cubic polynomial with a double root at x=a, then both P(a)=0 and P’(a)=0. Proof:

16. Finally, there were two instances where we encountered a cubic polynomial with a double root: the support car’s position function and the cubic polynomial equation expressing the times that Carol and the support car meet when Carol is travelling at ideal speed From the first, we can deduce that both S(0)=0 and S’(0)=0. From the second we deduce that both C(5)=S(5) and C’(5)=S’(5). In general, if P(x) is a cubic polynomial with a double root at x=a, then both P(a)=0 and P’(a)=0. Proof: A cubic polynomial with a double root at x=a must have the form

17. Finally, there were two instances where we encountered a cubic polynomial with a double root: the support car’s position function and the cubic polynomial equation expressing the times that Carol and the support car meet when Carol is travelling at ideal speed From the first, we can deduce that both S(0)=0 and S’(0)=0. From the second we deduce that both C(5)=S(5) and C’(5)=S’(5). In general, if P(x) is a cubic polynomial with a double root at x=a, then both P(a)=0 and P’(a)=0. Proof: A cubic polynomial with a double root at x=a must have the form Then by the Product Rule,

18. Finally, there were two instances where we encountered a cubic polynomial with a double root: the support car’s position function and the cubic polynomial equation expressing the times that Carol and the support car meet when Carol is travelling at ideal speed From the first, we can deduce that both S(0)=0 and S’(0)=0. From the second we deduce that both C(5)=S(5) and C’(5)=S’(5). In general, if P(x) is a cubic polynomial with a double root at x=a, then both P(a)=0 and P’(a)=0. Proof: A cubic polynomial with a double root at x=a must have the form Then by the Product Rule, Since both P(x) and P’(x) have (x-a) as a factor, they both have x=a as a root.

19. Got it!