I was thinking on a basic high school problem about motion on one axis and came across one of those problem where displacement is zero because final and initial positions are same. My question is that why do we have concept of displacement where we take the shortest path an object takes? it is cause of many problems like if the "displacement" is zero then work done will also be zero (even if there is some distance covered) and this will be irrespective of energy consumed .Why don't we just ditch the idea of a hypothetical shortest path and only consider distances. One of reason i came across was that distance is scalar while velocity is vector I think we can devise a system where we can define if distance is being taken as scalar or vector....(as in if distance is only in straight line then distance will be considered vector else scalar) I know there is something stupid i am missing but motive of this post is to rediscover what that thing is....


(a) We are often interested in studying a body's motion throughout a period of time. Suppose that it followed a curved path. Then its displacements over successive time intervals would keep changing direction. So the body's velocity would keep changing direction, giving the body accelerations. That's one context in which we need the concept of displacement.

(b) You seem particularly interested in cases where a body performs a 'round trip' so its overall displacement is zero. You say that the work done on the body is zero. That's true for the work done on the body by a gravitational field, or, if the body has a charge, by an electric field due to static charges. I don't see why that's a problem, and I don't understand what you mean by "this will be irrespective of the energy consumed".

Note that not all forces on a body have this "conservative" property. If you push a book across a table, and then push it back to its starting point, you will have done a finite amount of work against friction – which always opposes the relative motion between surfaces.

  • $\begingroup$ "work done will be zero because displacement is zero" It depends: please see my third paragraph. What do you have in mind when you talk about energy being "consumed" to shift the object? $\endgroup$ Jul 12 at 14:19
  • $\begingroup$ Your comment, to which I have just replied, seems to have disappeared. I assume that you've deleted it. $\endgroup$ Jul 12 at 14:26
  • $\begingroup$ but in that book case result will be very much different we break the actions into components of a the book going from a to b (calculate the work done) and then from b to a $\endgroup$ Jul 12 at 14:27
  • $\begingroup$ I am actually having problem in conveying my idea so i'll very soon link some explanation of my doubt here in the comment section.. $\endgroup$ Jul 12 at 14:28
  • $\begingroup$ Yes actually i have deleted that comment because there was more to add to it and i accidentally posted that $\endgroup$ Jul 12 at 14:29

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