Is our reasoning about constant motion accurate?

[jedimath]

Homework Statement

In a relay race the fourth rider of team A receives the baton to run the last 85m. At the same time the last athlete of team B is at a disadvantage of 2.6m.
The athlete of team A runs at a speed of 9.45ms, while that of team B is able to run at a speed of 9.80ms.

Does the last team B runner manage to win the race?

Relevant Equations

s = s_0 + vt

NOTE: Sorry for my english. I use Google Translate!

Comparing the performance with a friend of mine who is passionate about physics (and he is studying it by himself) we came to the same conclusion. In other words, we have calculated the time taken by both riders to reach the finish line. From the calculations it results that the runner B manages to win the race because, to cover 87.6m it takes 8.94s while the rider of the quadra takes 9.00s to cover 85m.

Is this reasoning right? Because the book makes a different one.
Practically we calculate the positions of the two athletes and then ask: when the two relay runners occupy the same position? And so talking ...

It is true that different reasoning can lead to the same result but ... in this case my reasoning and that of my friend is right?

Thanks.

 

[PeroK]

Homework Statement: In a relay race the fourth rider of team A receives the baton to run the last 85m. At the same time the last athlete of team B is at a disadvantage of 2.6m.
The athlete of team A runs at a speed of 9.45ms, while that of team B is able to run at a speed of 9.80ms.

Does the last team B runner manage to win the race?
Homework Equations: s = s_0 + vt

NOTE: Sorry for my english. I use Google Translate!

Comparing the performance with a friend of mine who is passionate about physics (and he is studying it by himself) we came to the same conclusion. In other words, we have calculated the time taken by both riders to reach the finish line. From the calculations it results that the runner B manages to win the race because, to cover 87.6m it takes 8.94s while the rider of the quadra takes 9.00s to cover 85m.

Is this reasoning right? Because the book makes a different one.
Practically we calculate the positions of the two athletes and then ask: when the two relay runners occupy the same position? And so talking ...

It is true that different reasoning can lead to the same result but ... in this case my reasoning and that of my friend is right?

Thanks.

You have solved the problem where the final runners receive the baton at the same time, but B is 2.6m behind A.

The problem might be that when B receives the baton, he has 85m to run, but A is already 2.6m ahead. Is that what the book does?

 

[jedimath]

Hi, when B receives the baton, he has 85m + 2.6m to run for a total of 87.6m to run.
We have this situation, at same instant:
- A receives the baton and have 85m to run
- B receives the baton and have 87.6m to run

The book, for solve the problem, calculate when A and B have same position.

At least that's what I understood :)

 

[PeroK]

Hi, when B receives the baton, he has 85m + 2.6m to run for a total of 87.6m to run.
We have this situation, at same instant:
- A receives the baton and have 85m to run
- B receives the baton and have 87.6m to run

At least that's what I understood :)

Looks right to me.

 

[jedimath]

Thanks. So, both solutions (my and that of book) are right? What I would like to understand is why the book uses that method.

 

[PeroK]

Thanks. So, both solutions (my and that of book) are right? What I would like to understand is why the book uses that method.

What's the book solution?

 

[jbriggs444]

The book, for solve the problem, calculate when A and B have same position.

One could solve the problem that way -- calculate the time to intercept. Then calculate the position of intercept. Then determine whether that position is on the far side of the finish line or the near.

It seems more straightforward to calculate, as you have, the time to finish for each runner and determine which occurs first.

 

[jedimath]

Book solution:

- We choose the point from which the team athlete $A$ starts as the origin of the positions. Thus the rider of the square B has an initial position $S_ {0} = -2.6m$.
- The law of the athlete's position $A$ is $s_ {A} = (9.45 m / s) t$.
- The law of the athlete's position $B$ is $s_ {B} = (9.80 m / s) t - 2.6m$
- Athlete B reaches (and passes) runner A when the two relay runners occupy the same position $s_ {B} = s_ {A}$.
- From the previous condition is obtained
$(9.80 m / s) t - 2.6m = (9.45 m / s) t$
$(9.80 m / s) t - (9.45 m / s) t = 2.6m$
$(0.35 m / s) t = 2.6m$
$t = \frac {2.6 m} {0.35 m / s} = 7.4s$

At the instant $t = 7.4s$ the runner $A$ is in the position:
$s_ {A} = (9.45 m / s) \times (7.4s) = 70m$

From the moment that $70m$ is less than the length (equal to $85 m$) from the last leg of the race, the rider of the quadra $B$ risks overcoming the athlete of the team A

 

[PeroK]

Book solution:

- We choose the point from which the team athlete $A$ starts as the origin of the positions. Thus the rider of the square B has an initial position $S_ {0} = -2.6m$.
- The law of the athlete's position $A$ is $s_ {A} = (9.45 m / s) t$.
- The law of the athlete's position $B$ is $s_ {B} = (9.80 m / s) t - 2.6m$
- Athlete B reaches (and passes) runner A when the two relay runners occupy the same position $s_ {B} = s_ {A}$.
- From the previous condition is obtained
$(9.80 m / s) t - 2.6m = (9.45 m / s) t$
$(9.80 m / s) t - (9.45 m / s) t = 2.6m$
$(0.35 m / s) t = 2.6m$
$t = \frac {2.6 m} {0.35 m / s} = 7.4s$

At the instant $t = 7.4s$ the runner $A$ is in the position:
$S_ {A} = (9.45 m / s) \times (7.4s) = 70m$

From the moment that $70m$ is less than the length (equal to $85 m$) from the last leg of the race, the rider of the quadra $B$ risks overcoming the athlete of the team A

Your method is much better!