Relative velocity during a race?

[Sagrebella]

Hello,

could someone please check my answer for this problem. All of my work and equations are clearly presented in the picture attached below. (If my answer is wrong, please give me pointers on how I could obtain the correct solution).

Thanks!

Problem 5:

On their way to play soccer in the World Cup, Mia and Brandi get stranded at O'Hare Airport in Chicago because of bad weather. Late at night, with nobody else around, they decide to have a race to see who is faster. They run down to a particular point, then turn around and return to their starting point. That race turns out to be a tie. They race again, this time with Mia running on a moving sidewalk. They start at the same time, run east at top speed a distance L through the airport terminal, and then turn around and run west back to the starting point. Brandi runs at a constant speed v, while Mia runs on a moving sidewalk that travels at a constant speed of v/3. Mia runs at a constant speed v relative to the moving sidewalk. Neglect the time it takes the women to turn around at the halfway point. Let's say that v = 8.00 m/s and L = 48.0 m.

(a) Mia reaches the turnaround point first, because she starts off running on the moving sidewalk in the direction the sidewalk is moving. What is the distance between the two women when Mia is at the turnaround point?

(b) After Mia turns around, she is running opposite to the direction the moving sidewalk is moving. What is the distance between the two women when Brandi is at the turnaround point?

(c) What is the distance between the two women when the winner of the race reaches the finish line?

 

[phinds]

Just as a suggestion, if you are going to ask people for help, it's a really bad idea to make it hard for them to GIVE you help. Why would you post your work sideways so it's necessary to get a neck cramp to look at it?

 

[Student100]

Hello,

could someone please check my answer for this problem. All of my work and equations are clearly presented in the picture attached below. (If my answer is wrong, please give me pointers on how I could obtain the correct solution).

Thanks!

Problem 5:

On their way to play soccer in the World Cup, Mia and Brandi get stranded at O'Hare Airport in Chicago because of bad weather. Late at night, with nobody else around, they decide to have a race to see who is faster. They run down to a particular point, then turn around and return to their starting point. That race turns out to be a tie. They race again, this time with Mia running on a moving sidewalk. They start at the same time, run east at top speed a distance L through the airport terminal, and then turn around and run west back to the starting point. Brandi runs at a constant speed v, while Mia runs on a moving sidewalk that travels at a constant speed of v/3. Mia runs at a constant speed v relative to the moving sidewalk. Neglect the time it takes the women to turn around at the halfway point. Let's say that v = 8.00 m/s and L = 48.0 m.

(a) Mia reaches the turnaround point first, because she starts off running on the moving sidewalk in the direction the sidewalk is moving. What is the distance between the two women when Mia is at the turnaround point?

(b) After Mia turns around, she is running opposite to the direction the moving sidewalk is moving. What is the distance between the two women when Brandi is at the turnaround point?

(c) What is the distance between the two women when the winner of the race reaches the finish line?View attachment 113093 View attachment 113094 View attachment 113095

Part A looks fine. Part B and C are wrong numerically, unless I made a calculation error (Brandi does indeed win).

Also, your work is hard to follow. Not because it's sideways, but because you're complicating it by plugging numbers in too soon. Did the problem specify the velocity and distance by the way? You should be able to solve this algebraically pretty easily.

For part B, how much further does Mia run in the time it takes Brandi to reach the turn around point? Check your work.

 

[Sagrebella]

Just as a suggestion, if you are going to ask people for help, it's a really bad idea to make it hard for them to GIVE you help. Why would you post your work sideways so it's necessary to get a neck cramp to look at it?

 

[Sagrebella]

Just as a suggestion, if you are going to ask people for help, it's a really bad idea to make it hard for them to GIVE you help. Why would you post your work sideways so it's necessary to get a neck cramp to look at it?

sorry about that, I posted my pictures again.

 

[Sagrebella]

Part A looks fine. Part B and C are wrong numerically, unless I made a calculation error (Brandi does indeed win).

Also, your work is hard to follow. Not because it's sideways, but because you're complicating it by plugging numbers in too soon. Did the problem specify the velocity and distance by the way? You should be able to solve this algebraically pretty easily.

For part B, how much further does Mia run in the time it takes Brandi to reach the turn around point? Check your work.

Hello; thanks for trying to help me even though my work was sideways! I uploaded the pictures again. Would you mind taking a look at it just one more time.
Yes, the problem does specify the velocity and distance. The length of one part of the race is 48 m, and the velocity v is 8 m/s. The moving sidewalk travels at v/3.

 

[Student100]

B and C are still wrong, did you recheck?

 

[Sagrebella]

B and C are still wrong, did you recheck?

Yes, now I rechecked.
For B) I got 8 meters
but for C) I still got 16.02 meters

 

[Student100]

Yes, now I rechecked.
For B) I got 8 meters
but for C) I still got 16.02 meters

B sounds good now. I can't remember what i got for C now, but that ain't it.

So what you have so far: Mia is at 40 meters from the finish line when Brandi is 48. How long does it take Brandi to cross the finish line? Then find the distance Mia can travel in that time.

 

[Sagrebella]

B sounds good now. I can't remember what i got for C now, but that ain't it.

So what you have so far: Mia is at 40 meters from the finish line when Brandi is 48. How long does it take Brandi to cross the finish line? Then find the distance Mia can travel in that time.

Ok, so for part C I did this.

I found that it takes Brandi 6 seconds to reach the finish line.
In 6 seconds, Mia ends up at the position 8.02 meters.

x_i=40
v_i=-5.33 m/s
t= 6 seconds

x_f = 40 - 5.33(6) = 8.02

so, the final distance between the winner and loser is 8.02 meters, right?

Hopefully that's right. This problem seems very simple but I keep struggling to get the right answer for some reason :P

 

[Student100]

Ok, so for part C I did this.

I found that it takes Brandi 6 seconds to reach the finish line.
In 6 seconds, Mia ends up at the position 8.02 meters.

x_i=40
v_i=-5.33 m/s
t= 6 seconds

x_f = 40 - 5.33(6) = 8.02

so, the final distance between the winner and loser is 8.02 meters, right?

Hopefully that's right. This problem seems very simple but I keep struggling to get the right answer for some reason :P

Thats right (basically, don't round velocity and you should get 8 meters. )

Now can you reason how that would be, how does Mia gain 8 meters but lose 16 meters on the way back?

 

[Student100]

While you're chewing on that, let me show you a way to minimize mistakes and make your work more general/easier to read.

Here's kind of how I would approach the problem:

First, let's figure out Brandi's times... We're assuming that both runners are starting initially from their maximum velocity, so that acceleration is a constant 0. We can also assume we begin our race from a point we'll define as zero. So to figure out here time to run to the turn around point is:

$$x = x_0 + v_0(t) + \frac{1}{2}at^2$$

Or after making substitutions for what we know/assume:

$$L = v_0(t)$$ $$t= \frac{L}{v_0}=T$$

Using symmetry, we know that the total time it takes her to run is just twice T.

For Brandi's motion we're going to do the same thing..

On her trip there her velocity is $v_0+\frac{1}{3}v_0$ or, $\frac{4}{3}v_0$, that way we can conclude..

$$L = \frac{4}{3}v_0t$$ $$ t= \frac{3L}{4v_0}$$ or that $\frac{3T}{4}$, so she's faster than Brandi at this point.

Next we'll look at her return trip, $-v_0 +\frac{1}{3}v_0$ or $-\frac{2}{3}v_0$

So we get,

$$0 =L - \frac{2}{3}v_0t $$ $$t= \frac{3L}{2v_0}$$ $\frac{3T}{2}$ Summing her two times we find that it takes her $\frac{9T}{4}$ to complete the race. So we already know at this point she loses.

So part A we would find Mia's first time using $t= \frac{3L}{4v_0}$, since we know she's quicker, then find the how far Brandi has ran in the same time using $L = v_0(t)$

For part B we would find Brandi's time $t= \frac{L}{v_0}$, subtract the time found in part A, and then use $0 =L - \frac{2}{3}v_0t$ to find how far Mia has gone back on her return trip and then find the distance between them.

For part C we would use Brandi's time from part B, since it takes her an equal amount of time to run back and then use $0 =L - \frac{2}{3}v_0t$ setting L to the length we found in part B.

Do you maybe see how it's more useful to do it this way, given any velocity or length we can solve this problem now, and it's easier to write/less error prone than plugging in numbers, in my opinion.

 

[Sagrebella]

While you're chewing on that, let me show you a way to minimize mistakes and make your work more general/easier to read.

Here's kind of how I would approach the problem:

First, let's figure out Brandi's times... We're assuming that both runners are starting initially from their maximum velocity, so that acceleration is a constant 0. We can also assume we begin our race from a point we'll define as zero. So to figure out here time to run to the turn around point is:

$$x = x_0 + v_0(t) + \frac{1}{2}at^2$$

Or after making substitutions for what we know/assume:

$$L = v_0(t)$$ $$t= \frac{L}{v_0}=T$$

Using symmetry, we know that the total time it takes her to run is just twice T.

For Brandi's motion we're going to do the same thing..

On her trip there her velocity is $v_0+\frac{1}{3}v_0$ or, $\frac{4}{3}v_0$, that way we can conclude..

$$L = \frac{4}{3}v_0t$$ $$ t= \frac{3L}{4v_0}$$ or that $\frac{3T}{4}$, so she's faster than Brandi at this point.

Next we'll look at her return trip, $-v_0 +\frac{1}{3}v_0$ or $-\frac{2}{3}v_0$

So we get,

$$0 =L - \frac{2}{3}v_0t $$ $$t= \frac{3L}{2v_0}$$ $\frac{3T}{2}$ Summing her two times we find that it takes her $\frac{9T}{4}$ to complete the race. So we already know at this point she loses.

So part A we would find Mia's first time using $t= \frac{3L}{4v_0}$, since we know she's quicker, then find the how far Brandi has ran in the same time using $L = v_0(t)$

For part B we would find Brandi's time $t= \frac{L}{v_0}$, subtract the time found in part A, and then use $0 =L - \frac{2}{3}v_0t$ to find how far Mia has gone back on her return trip and then find the distance between them.

For part C we would use Brandi's time from part B, since it takes her an equal amount of time to run back and then use $0 =L - \frac{2}{3}v_0t$ setting L to the length we found in part B.

Do you maybe see how it's more useful to do it this way, given any velocity or length we can solve this problem now, and it's easier to write/less error prone than plugging in numbers, in my opinion.

Yes, thank you for your thorough explanation. Using variables instead of actual numbers is much cleaner and easier to follow. Also, your method gives a concrete explanation as to why Mia lost the race. It all had to do with the moving sidewalk either adding to her velocity, or slowing it down.

Do you mind taking a look at my other questions perhaps and seeing if my answers are wrong or correct? I appreciate the explanations and feel that your pointers could help me a lot on future homework problems and exams. Thanks again