Explain clearly, with ezamples, the difference 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
over 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) Displacement is a vector quantity. When a body changes its position with respect to time, then a straight line joining from initial to final position of a body is called the displacement independent of the nature of the actual path taken by the body. Distance is the total path length covered by a body in a certain interval of time.
Examples
(i) Consider a particle is moving from A to B in 2 s and then from B to A in 2s.

Here distances covered `=AB+BA=4+4=8`m
But displacement will be zero as the initial and final position of the particle is same.
In this case, distance is greater than the displacement.
Distance will be equal to displacement when the body is moving in a straight line in same direction.
(ii) Average speed `=("distance covered")/("time taken")`
`=(AB+BA)/(t)`
`(4+4)/(2+2)=2m//s`
Average velocity`=("Displacement")/("time")`
=0
Hence in this case, average speed is greater than the average velocity.
Average speed is equal to average velocity when the body is moving in a straight line in same direction only.