How fast must the women run?

In summary: I'm clueless.Try it...if it doesn't work, I'm clueless.In summary, a woman is on a train leaving the station at 3.5 m/s, with a friend running alongside the car at 7.0 m/s in the same direction. To keep up with her friend, the woman would need to walk at a speed of 3.5 m/s. Once the train reaches a speed of 10 m/s, the woman would need to walk at a speed of 3.0 m/s to keep up with her friend.
  • #1
emily081715
208
4

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


vxw=vxt+vtw

The Attempt at a Solution


v= 10-3.5
 
Physics news on Phys.org
  • #2
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
phinds said:
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
 
  • #4
emily081715 said:
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.
 
  • #5
phinds said:
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
 
  • #6
emily081715 said:
i still need help finding the answer
Well, good luck with that.
 
  • #7
phinds said:
Well, good luck with that.
do you have any suggestions ?
 
  • #8
emily081715 said:
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.
 
Last edited:
  • Like
Likes billy_joule
  • #9
I like Serena said:
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
 
  • #10
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
Please, I want people to check this...
 
  • #12
obanz2nd said:
Please, I want people to check this...
that anwser was incorrect too
 
  • Like
Likes obanz2nd
  • #13
emily081715 said:

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


vxw=vxt+vtw

The Attempt at a Solution


v= 10-3.5
Is that the question as given to you, word for word ?
 
  • #14
SammyS said:
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:?
 
  • #15
emily081715 said:
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) .
 
  • #16
SammyS said:
Now let's see your attempt(s) .
10-3.5= 6.5 m/s
 
  • #17
emily081715 said:
10-3.5= 6.5 m/s
In part B), What is it that moves at 3.5 m/s ?
 
  • #18
SammyS said:
In part B), What is it that moves at 3.5 m/s ?
SammyS said:
In part B), What is it that moves at 3.5 m/s ?
should it be 10-7?
 
  • #19
emily081715 said:
should it be 10-7?
since the bus is moving at 10 and the friend is running at 7
 
  • #20
emily081715 said:
since the bus train is moving at 10 and the friend is running at 7
Try it .
 
  • #21
SammyS said:
Try it .
it was 7-10= -3 m/s
 
  • #22
emily081715 said:
it was 7-10= -3 m/s
What is the significance of the negative sign?
 
  • #23
SammyS said:
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
 
  • #24
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
phinds said:
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
 
  • #26
emily081715 said:
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.
 
  • #27
emily081715 said:
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.
 

1. What is the average speed of a woman?

The average speed of a woman can vary greatly depending on age, physical fitness, and other factors. However, according to studies, the average running speed of women ranges from 6 to 8 miles per hour (mph).

2. How does a woman's running speed compare to a man's?

Generally, men have a higher average running speed compared to women due to biological and physiological differences. However, there are always exceptions and women can also achieve high running speeds through proper training and conditioning.

3. Is there a specific speed that women must run at?

No, there is no specific speed that women must run at. It depends on individual goals and abilities. However, for competitive races, there may be specific time requirements that women must meet in order to qualify or win.

4. Does a woman's running speed decrease with age?

As with any physical activity, a woman's running speed may decrease with age due to changes in muscle mass, flexibility, and stamina. However, with proper training and exercise, women can maintain and even improve their running speed as they age.

5. What factors affect a woman's running speed?

Several factors can affect a woman's running speed, including but not limited to: genetics, physical fitness and training, nutrition, age, and health conditions. Additionally, external factors such as weather, terrain, and elevation can also impact a woman's running speed.

Similar threads

Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
3
Views
5K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Replies
19
Views
1K
Back
Top