If a man is moving on a horizontal belt (with constant velocity w.r.t belt)which is also moving in the same direction with some velocity , then time w.r.t belt and w.r.t ground to travel some distance on the belt is coming different, how is.this possible?

Time w.r.t belt = distance/velocity of man w.r.t belt

Time w.r.t ground = distance/velocity of man w.r.t belt+ velocity of belt

How.can the two times be different as time does not depend on reference frame

closed as unclear what you're asking by John Rennie, Bill N, ZeroTheHero, Kyle Kanos, Jon Custer Sep 18 at 13:05

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    Your wording is a little unclear. Do you want to ask how it is possible to move, say, 10 meters with respect to the ground in different time to 10 meters with respect to the belt? – user93146 Sep 14 at 13:45
  • the times do not add linearly, they add harmonically. That is, the rates add linearly. – JEB Sep 14 at 15:34

time w.r.t. belt = $\frac{distance..w.r.t..belt}{velocity..w.r.t...belt}$.

Not $\frac{distance..w.r.t..ground}{velocity..w.r.t..belt}$.

  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Jon Custer Sep 14 at 20:11
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    @JonCuster, this is the answer. Short and sound. Not critique nor clarification request. – npojo Sep 14 at 20:28

The two times are different because they are the times for different things to happen. In my example, I've got the man moving 1 m/s relative to the belt, and the belt moving 2 m/s relative to the ground. After 1 second, the belt has moved 2 m, and the man has moved 1 meter ahead of the belt. And so he has moved 1 m more along the ground than the belt did, or 3 m.

enter image description here

So to go 3 m w.r.t. the ground takes the man only 1 second. But to go 3 m w.r.t. to the belt means he has to get 3 m ahead of the belt. And that will take 3 seconds. By that time he will have moved 9 m w.r.t. the ground. In my picture he will have run out of belt.

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