Kinematics of three runners in a race

[bkraabel]

Homework Statement

Runners A, B, and C run a 100-m race, each at a constant speed.
Runner A takes first place, beating runner B by 10 m. Runner
B takes second place, beating runner C by 10 m. By what time
interval does runner A beat runner C?

Homework Equations

d = 100 m
Δd = 10 m
Δt[itex]_{i}[/itex] = time for runner i to travel 100 m
v[itex]_{i}[/itex] = speed of runner i

The Attempt at a Solution

Can this problem be solved without knowing the speed of one of the runners? Here's a few equalities for this problem:

v[itex]_{a}[/itex] = [itex]\frac{d}{Δt_{a}}[/itex]
v[itex]_{b}[/itex] = [itex]\frac{d-Δd}{Δt_{a}}[/itex]
v[itex]_{c}[/itex] = [itex]\frac{v_{b}Δt_{b}-Δd}{Δt_{b}}[/itex]

 

[barryj]

Try assuming a couple of different velocities for runner A and then work out the time difference. Does the time difference vary with the velocity of runner A? This should answer your question.

 

[bkraabel]

Note that runner A beats runner C by more than 20 m.

 

[rude man]

Note that runner A beats runner C by more than 20 m.

Yikes, that's right!

I am deleting my post accordingly to think more about this.

 

[bkraabel]

I tried assuming a few speeds for runner A and found that the interval between runners A and C depends on the speed of runner A. In other words, there is no unique solution with the given information.

 

[voko]

The problem has six unknowns (3 speeds and 3 times) and five equations. It is fairly trivial to express any five unknowns in terms of the other one.

 

[rude man]

Right all around. At least I got the 'no unique solution' part right ...

 

[nil1996]

time for C

A beats C by a time period of 20/V_C

V_C=velocity of C

 

[voko]

A beats C by a time period of 20/V_C

V_C=velocity of C

This assumes that A beats C by 20 meters. This is not given. What is given is that B beats C by 10 meters, which happens when B is at the finish line. When A is at the finish line, B is not there, so you can't assume that he is 10 meters ahead of C yet.