Calculating Relative Motion Using Time and Distance Equations

Click For Summary

Homework Help Overview

The problem involves calculating the ratio of a man's running speed to the speed of a moving sidewalk, based on the time taken to traverse the sidewalk in both directions. The subject area relates to relative motion and kinematics.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up equations for the velocities based on the time taken for each run but expresses uncertainty about how to proceed. Some participants question the lack of an initial attempt at a solution, while others suggest finding expressions for the distances traveled in both directions.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to relate the distances traveled during each run. There is a suggestion to equate the distances since they are the same, which may lead to further insights.

Contextual Notes

Participants note the challenge of not knowing the length of the sidewalk, which is a key piece of information for solving the problem.

uberbandgeek6
Messages
6
Reaction score
0
A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking 2.50 s. Then security agents appear and the man runs as fast as he can back along the sidewalk to his starting point, taking 8.00 s. What is the ratio of the man's running speed to the sidewalk's speed? (running speed / sidewalk speed)

Thanks
 
Physics news on Phys.org
Where is your attempt at the solution?
 
I'm not sure how to go about it. I set up the velocities as V(with)=length/2.5 and V(against)=length/8, but I'm not sure what to do from there.
 
How do I work around the fact that i don't know the length of the sidewalk?
 
You know that both times he travels the same distance. What you need to do is find an expression that would give you the distance for run 1 and run 2. You can then equate them because you know the distance should be the same and you should be able to find the answer from there.
 

Similar threads

Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
8
Views
6K
  • · Replies 18 ·
Replies
18
Views
5K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
4K