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].

Answer 1

(a)

Displacement is a vector quantity and distance is a scalar quantity. If a particle is moving in a circular path and returns to its starting point then the displacement of the particle is zero, but the distance covered by the particle is equal to the circumference of the circular path.

Therefore, the total distance is greater than or equal to the total displacement.

(b)

The average velocity is,

v= total displacement time taken In part (a), the displacement is zero therefore, the average velocity is zero.

The average speed is,

s= total distance time taken The total distance is non zero; therefore, the average speed is also not zero and is greater than the average velocity.

If the displacement is non zero then the total distance and total displacement is equal due to which both average velocity and average speed is equal.

When the particle moves in a straight line then the total distance and total displacement is equal.

Similarly, the average velocity and average speed is equal.