Explain clearly, with ezamples, the distaintion between :
(a) magnitude of displacemnt (sometimes called distance ) overand interval of time, and the total length of the path coverd by a particle over the same interval.
(b) magnitude of average velocity over an intercal of time , and the average speed ocer the same interval. [ Average speed of a particle over an interval of time is defined as the toal path length divided by the time intrval]. Show in both (a) and (b) that the second quantity is either greater than or equal to first. When is the equality sing true ? [ For simplocity, consider one- dimensional motion only]

Explanation

(a) Magnitude of displacement of a particle in motion for a given time is the shortest distance between the initial and final position of the particle in that time, whereas the total length of the path covered by particle is the actual path traversed by the particle in the given time. If a particle goes from ` A to B` to C` in time (t) as shown in Fig. 2 (NCT) . 4, then Magnitude of displacrmt= distance ` Ab+ distance BC`
From above we note that total path length (AB + BC) is greater than magnitude of dispacement (AC).
If there is a motion of the particle in one dimension i.e. along a straight line, then the magnitude of displacement beconmes equal to total path length traversed by the particle in the given time.
(b) Magnitude of average velocity = ( magnitude of displacment ) /(time interval) = (AC)/t`
and ` Acerage speed ` =(total pathe length )/( time interval ) = (AB+BC) /t`
As, (AB+ BC) gt AC, so average speed is greater than the magnitude of average velocity. If the particle is moving along a straight line, then in a given time the magniude of displacemnt is equal to total path length traversed by particle in that time, so average speed is equal to magnitude of average velocity.