Question

Explain clearly, with examples, the distinction between:
(a) 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;
(b) 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].

Accepted Answer

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 $= A C$ .

Whereas, total path length $= A B + B C$ 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.

$The magnitude of average velocity = Magnitude of displacement / Time Interval$
For the given particle,
Average velocity $= \frac{A C}{t}$
Average Speed $= \frac{Total path length}{Time interval}$
$= \frac{AB+BC}{t}$
Since $(A B + B C) > A C$ , average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a unidirectional straight line .

Important Questions on Motion in a Straight Line

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