Two bicyclists finish a race in 2 hours and 15 minutes. Prove that at some time during the race, the bicyclists are traveling at the same velocity.

Question

I need help ASAP please!!

Answer 1

Let f(t) be the distance traveled by the first cyclist in t hours and let g(t) be the distance traveled by the second cyclist in t hours.  Then their velocities are f'(t) and g'(t).   

 

Let D(t) = f(t) - g(t)

 

Then, D(0) = 0 and D(2.25) = 0 (since f(0)=g(0) and f(2.25)=g(2.25))

 

If we assume that D(t) is continuous on [0, 2.25] and is differentiable on (0, 2.25), then we can apply the Mean Value Theorem to conclude that there is at least one number, c, in the interval (0, 2.25) so that 

 

D'(c) = [D(2.25)-D(0)]/[2.25-0] = 0

 

But, D'(c) = f'(c)-g'(c) = 0

 

               So, f'(c) = g'(c).  That is, their velocities are the same when

                                       t = c.

 

 

Answer 2

Hi, Kathryn!

 

I'm not sure what level class this is for, but we can prove that the bicyclists traveled at the same velocity sometime during the race by simple logic.  To begin the proof, we need to address all three possible ways for the race to start:  The bicyclists start at 1) the same speed 2) Bicyclist A started faster than Bicyclist B and 3) Bicyclist B started faster than Bicyclist A.  Then we prove that Bicyclists A and B traveled at the same speed sometime in each situation.

 

Situation 1: Bicyclists A & B start the race at the same speed.

Proof: Regardless of their speed during the rest of the race, they traveled at the same velocity when they started the race.

 

Situation 2: Bicyclist A started faster than Bicyclist B.

Proof: In order for Bicyclist A & B to finish the race in the same amount of time, their average velocities would have to be the same (v = total distance/total time), so Bicyclist A's speed would have to decrease at some point during the race to compensate for him starting faster than Bicyclist B.  As his speed decreases, it will have to cross the speed of Bicyclist B.  When this occurs, they will be traveling at the same velocity.

 

Situation 3: Bicyclist B started faster than Bicyclist A.

Proof: (Same as Situation 2, just switch the bicyclists.)  In order for Bicyclist A & B to finish the race in the same amount of time, their average velocities would have to be the same (v = total distance/total time), so Bicyclist B's speed would have to decrease at some point during the race to compensate for him starting faster than Bicyclist A. As his speed decreases, it will have to cross the speed of Bicyclist A. When this occurs, they will be traveling at the same velocity.

 

The key to proving this is that the average velocity of Bicyclist A is the same as the average velocity of Bicyclist B.  Both are calculated using the formula velocity = distance/time.  The race is the same distance for both bicyclists and they completed it in the same time.  Therefore they have the same average velocity.  The idea of an average velocity is that, at some point during the race, both Bicyclists will travel at the average velocity - the thing is, that doesn't necessarily happen at the same time.

 

Hope this is helpful!

 

Patty