Question

Explain clearly, with examples, the distinction between : 

1. magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; 
2. magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].

Answer 1

(a) Magnitude of displacement vs total path length

Magnitude of displacement is the absolute value of the change in position of a particle between its initial and final points during a given time interval.

Total path length is the actual distance travelled by the particle along its path during the same time interval.

Example:

Consider one-dimensional motion along a straight line.
A particle starts from x = 0 m, moves to x = 5 m, and then comes back to x = 2 m.

Magnitude of displacement = |final position − initial position|
= |2 − 0| = 2 m

Total path length = distance travelled forward + distance travelled backward
= 5 + 3 = 8 m

Thus,
Total path length (8 m) > magnitude of displacement (2 m)

In general, total path length is always greater than or equal to the magnitude of displacement.

Equality holds only when the particle moves in one direction without changing direction.

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(b) Magnitude of average velocity vs average speed

Average velocity over a time interval is defined as:

Average velocity = displacement / time interval

The magnitude of average velocity depends only on the net displacement.

Average speed over the same time interval is defined as:

Average speed = total path length / time interval

Example:

Using the same motion as above, suppose the total time taken is 4 s.

Magnitude of average velocity = displacement / time
= 2 / 4 = 0.5 m s⁻¹

Average speed = total path length / time
= 8 / 4 = 2 m s⁻¹

Thus,
Average speed (2 m s⁻¹) > magnitude of average velocity (0.5 m s⁻¹)

In general, average speed is always greater than or equal to the magnitude of average velocity.

Equality holds only when the particle moves in a straight line without reversing its direction.

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Conclusion:

In both cases (a) and (b), the second quantity (total path length or average speed) is always greater than or equal to the first (magnitude of displacement or magnitude of average velocity). Equality occurs only for motion along a straight line in one direction.