# Homework Help: How fast must the women run?

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1. Oct 9, 2016

### emily081715

1. The problem statement, all variables and given/known data
A woman is on a train leaving the station at3.5 m/s , while a friend waving goodbye runs alongside the car she's in. Call the train's direction of motion the +x direction.
Once the train has reached a speed of 10 m/s, how fast must the woman walk, and in which direction, to keep up with her friend?

2. Relevant equations
vxw=vxt+vtw

3. The attempt at a solution
v= 10-3.5

2. Oct 9, 2016

### phinds

This seems to me like a VERY poorly worded question. I see various interpretations, but yes, your answer is correct using one of them, although even with that, I don't think the "keep up with her friend" is anything like standard English using since it is the friend who is "keeping up". It would have been more correct to say "stay even with her friend"

Also, you did not specify the direction of the passenger's motion relative to the train. You have v, which normally stands for "velocity", a vector, shown as a scalar.

3. Oct 9, 2016

### emily081715

its not the correct answer though. mastering physics says it wrong

4. Oct 9, 2016

### phinds

Well, then perhaps we are both misinterpreting the problem. As I said, I think it is poorly worded. Not clear, to me at least, exactly what they are asking for.

5. Oct 9, 2016

### emily081715

i still need help finding the answer

6. Oct 9, 2016

### phinds

Well, good luck with that.

7. Oct 9, 2016

### emily081715

do you have any suggestions ?

8. Oct 9, 2016

### I like Serena

Did you try -6.5?
Negative since she moves to the back of the train, while the train is moving in the positive x direction.
To be honest, it doesn't really make sense, since the problem says she's "walking", while 6.5 m/s is "sprinting".
Then again, seeing how poorly worded the problem is, perhaps it's still the correct answer.

Last edited: Oct 9, 2016
9. Oct 9, 2016

### emily081715

Yea I tried negative and still wrong

10. Oct 10, 2016

### obanz2nd

It is assumed that both activities occur at the same time interval...they move in same direction since one must approach the other from assumed reference start or origin.however,the question lacks good construct....The v=8.25m/s.... one way is to linearise it with equations of motion....using v^2= u^2 + 2ax, for the train ,v= 10, u= 6.5, both in m/s...so,we obtain ,ax=28.875m/s ..(1),also considering the time interval for covering x,we use v=u+ at, so,at= 3.5m/s ...(2),therefore, using both equations,we can divide 1 by 2 ,velocity with which the woman must move to catch up with the friend or thingy is given by v= 28.875/3.5 m/s. = 8.25m/s....

11. Oct 10, 2016

### obanz2nd

Please, I want people to check this.....

12. Oct 10, 2016

### emily081715

that anwser was incorrect too

13. Oct 10, 2016

### SammyS

Staff Emeritus
Is that the question as given to you, word for word ?

14. Oct 10, 2016

### emily081715

exact whole question:
A woman is on a train leaving the station at3.5 m/s , while a friend waving goodbye runs alongside the car she's in. Call the train's direction of motion the +x direction.
Part A)If the friend is running at 7.0 m/s and moving in the same direction as the train, how fast must the woman walk, and in which direction, to keep up with him? answer: 3.5 m/s
Part B) Once the train has reached a speed of 10 m/s, how fast must the woman walk, and in which direction, to keep up with her friend?

15. Oct 10, 2016

### SammyS

Staff Emeritus
Now let's see your attempt(s) .

16. Oct 10, 2016

### emily081715

10-3.5= 6.5 m/s

17. Oct 10, 2016

### SammyS

Staff Emeritus
In part B), What is it that moves at 3.5 m/s ?

18. Oct 10, 2016

### emily081715

should it be 10-7?

19. Oct 10, 2016

### emily081715

since the bus is moving at 10 and the friend is running at 7

20. Oct 10, 2016

### SammyS

Staff Emeritus
Try it .

21. Oct 10, 2016

### emily081715

it was 7-10= -3 m/s

22. Oct 10, 2016

### SammyS

Staff Emeritus
What is the significance of the negative sign?

23. Oct 10, 2016

### emily081715

the friend on the train is moving in the opposite direction of the train, so she is basically running to the back of the train.making it negative

24. Oct 10, 2016

### phinds

No, the answer in part a is NOT "3.5 m/s". As I have pointed out to you previously you keep expressing v as a scalar which is not correct.

25. Oct 10, 2016

### emily081715

both answers for the parts have already been marked correct so clearly i expressed them properly for this particular question