# Time on moving sidewalk

1. Oct 23, 2009

### waldvocm

Lance takes 167 seconds to get down the gall on the moving sidewalk. While walking (not on the moving sidewalk) it took him 88 seconds. If he walks while on the sidewalk how long will it take him.

I came to the answer to easily so I am not sure if I did it right or if I am missing something

I just took the time it takes him standing on the moving sidewalk and subtracted the time it takes him to walk, not on the moving sidewalk.

My answer is 79 seconds while walking on the moving sidewalk. Is this correct? It seems too easy!

2. Oct 23, 2009

### Delphi51

Quite often your intuition works when the equations are linear, but not in this case! You must use the formula for motion at constant speed and figure it out. Don't worry about not knowing the distance - just leave the "d" in and it will cancel out in the end.

3. Oct 23, 2009

### waldvocm

hmmmm...........would this require me to know the velocity as well? I am struggling to figure out which equation you mean.

4. Oct 23, 2009

### Delphi51

For constant velocity motion, use d = v*t.
If you want an expression for the velocity, just solve the equation for v. The result will have a d in it, but that's okay - d cancels out in the end when you find the time for the guy on the moving sidewalk.

I forgot to suggest you try out your intuition on some cases. For instance, you could say "what if the walker also takes 167 seconds?" In that case your intuitive calculation would suggest a time of zero for the combined motion, which is impossible!

5. Oct 26, 2009

### waldvocm

I have the equations

d1=v(167) standing on the moving sidewalk
d2=v(88) walking off of the moving sidewalk

d1=d2

v(167)=v(88)

I don't know where to go here

d3=v*t3

6. Oct 26, 2009

### Delphi51

Better make that
d=v1(167) standing on the moving sidewalk
d=v2(88) walking off of the moving sidewalk
d = v3*t3 walking on the moving sidewalk
because the d's are the same and the v's are different.

You know something about that v3, too! Notice that and you'll have one less variable and it will be possible to solve for t3.

7. Oct 26, 2009

### waldvocm

v3=88v-167v

v3=79v

Ha that brought me back to my original thinking!

I don't know I am really confused

8. Oct 26, 2009

### Delphi51

says a velocity is equal to a distance - no good!

Say you have a slidewalk moving at 5 km/h and you run at 10 km/h on it. How fast will you be going?

9. Oct 27, 2009

### waldvocm

15km/h

So, do I add the velocities?

v1+v2=v3

d/167+d/88=d/t3

10. Oct 27, 2009

### Delphi51

Excellent! Cancel those d's and you can calculate your t3.