How Fast Must the Woman Walk to Match Her Friend's Speed?

[emily081715]

Homework Statement

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?

Homework Equations

v_xw=v_xt+v_tw

The Attempt at a Solution

v= 10-3.5

 

[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.

 

[emily081715]

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.

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

 

[phinds]

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

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.

 

[emily081715]

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.

i still need help finding the answer

 

[phinds]

i still need help finding the answer

Well, good luck with that.

 

[emily081715]

Well, good luck with that.

do you have any suggestions ?

 

[I like Serena]

do you have any suggestions ?

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.

 

[emily081715]

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.

Yea I tried negative and still wrong

 

[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...

 

[obanz2nd]

Please, I want people to check this...

 

[emily081715]

Please, I want people to check this...

that anwser was incorrect too

 

[SammyS]

Homework Statement

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?

Homework Equations

v_xw=v_xt+v_tw

The Attempt at a Solution

v= 10-3.5

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

 

[emily081715]

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

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?
Answer:?

 

[SammyS]

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?
Answer:?

Now let's see your attempt(s) .

 

[emily081715]

Now let's see your attempt(s) .

10-3.5= 6.5 m/s

 

[SammyS]

10-3.5= 6.5 m/s

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

 

[emily081715]

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

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

should it be 10-7?

 

[emily081715]

should it be 10-7?

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

 

[SammyS]

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

Try it .

 

[emily081715]

Try it .

it was 7-10= -3 m/s

 

[SammyS]

it was 7-10= -3 m/s

What is the significance of the negative sign?

 

[emily081715]

What is the significance of the negative sign?

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

 

[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.

 

[emily081715]

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.

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

 

[SammyS]

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

Yes.

So apparently the first answer being positive implied that the direction was toward the front of the train, and as you stated, negative implied motion toward the rear.

 

[phinds]

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

Then your instructor is being VERY sloppy. That doesn't make it right.