Motion in a Straight Line

Question : 13 of 27

Marks: +1, -0

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].

Solution:

(a) Magnitude of displacement of a particle inmotion for a given time is the shortest distance betweenthe initial and final positions of the particle in that time,whereas the total length of the path covered by particleis the actual path traversed by the particle in the giventime. If a particle goes from A to B and B to C in time tas shown in figure, then

Magnitude of displacement = distance AC.

Total path length = distance AB + distance BC.

From above we note that total path length (AB + BC) is greater thanmagnitude of displacement (AC).

If there is a motion of the particle in one dimension i.e. along a straightline, then the magnitude of displacement becomes equal to total pathlength traversed by the particle in the given time.

(b) Magnitude of average velocity

= \frac{\text{ magnitude of displacement }}{\text{ time interval }}

= \frac{\text{A C}}{t}

and Average speed

= \frac{\text{ total path length }}{\text{ time interval }}

= \frac{(AB+BC)}{t}

As, (AB + BC) > AC, so average speed is greater than the magnitude ofaverage velocity. If the particle is moving along a straight line, then ina given time the magnitude of displacement is equal to total path lengthtraversed by particle in that time, so average speed is equal to magnitudeof average velocity.

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