If an object begins at the origin, will its displacement from the origin be the same as its position?

2 Answers
Aug 15, 2016

Yes, if the displacement is measured as the shortest distance in a straight line between the point of origin and the point we are interested in.

Explanation:

The displacement is the most direct and shortest distance from the point of origin to the point in which we are interested.

Average velocity is the change in displacement divided by the time taken, average speed is the distance covered divided by the time taken.

If the motion is in a straight line and only in one direction (no back-tracking) then the displacement and the distance are the same, and the velocity and the speed are the same.

If the motion is in some kind of more-complex path, then the distance and displacement will be different. One simple example:

An Olympic sprinter runs a 100 m race, but her shoe falls off at the 20 m mark and she goes back to collect it. When standing by her shoe, she has a displacement of 20 m from her point of origin, but has covered a distance of 180 m.

Aug 16, 2016

Yes. But the correct notation is as below.
#vec(Δx)# (position) #=vecs# (displacement)

Explanation:

It is stated that

  1. Average speed is #x/t#
    Must remember that speed is a scalar quantity.
    Distance #x# traveled in time #t# .
    It is assumed that distance is also a scalar quantity here.

  2. Average velocity is #(Δx)/t# (correction applied).
    Without commenting on what have been seen on YouTube, it needs to be remembered here that velocity is a vector quantity. As such correct expression would be
    Average velocity is #vec(Δx)/t#.
    Therefore, displacement is also a vector quantity.

If we use #vecs# symbol for displacement, it needs to be written as

#vecs=(vecr-vecr_0)#
where #vecr# is position vector.

As we are at liberty to choose the coordinate axes, we can rewrite above expression as
#vecs=(vecx-vecx_0)#
where movement is along #x#-axis.

If initial position vector #vecx_0# coincides with origin then expression reduces to
#vecs=vecx#

Answer to your question, yes
#vec(Δx)# (position) #=vecs# (displacement)