38.
Explain clearly with examples, the distinction
between
(i) magnitude of displacement (sometimes
called distance) over an interval of time and
the total length of the path covered by a
particle over the same interval.
(ii) 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 first. When is the equality sign true?
[For simplicity, consider one-dimensional
motion only)
INCERT)
Answers
The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle.
The total path length of a particle is the actual path length covered by the particle in a given interval of time.
For example, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle = AC.
Whereas, total path length = AB + BC
It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.
(b) Magnitude of average velocity = Magnitude of displacement / Time interval
For the given particle,
Average velocity = AC / t
Average speed = Total path length / Time interval
= (AB + BC) / t
Since (AB + BC) > AC, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.
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