1.12 Displacement  (Page 3/5)

The example above brings out nuances associated with terms used in describing motion. In particular, we see that distance and magnitude of displacement are not equal. This inequality arises due to the path of motion, which may be other than the shortest linear path between initial and final positions.

This means that distance and magnitude of displacement may not be equal. They are equal as a limiting case when particle moves in one direction without reversing direction; otherwise, distance is greater than the magnitude of displacement in most of the real time situation.

This inequality is important. It implies that displacement is not distance plus direction as may loosely be considered. As a matter of fact, displacement is shortest distance plus direction . For this reason, we need to avoid representing displacement by the symbol “s” as a vector counterpart of scalar distance, represented by “s”. In vector algebra, modulus of a vector, A , is represented by its non bold type face letter “A”. Going by this convention, if “ s ” and “s” represent displacement and distance respectively, then s = | s |, which is incorrect.

When a body moves in a straight line maintaining its direction (unidirectional linear motion), then magnitude of displacement, |Δ r | is equal to distance, “s”. Often, this situational equality gives the impression that two quantities are always equal, which is not so. For this reason, we would be careful to write magnitude of displacement by the modulus |Δ r | or in terms of displacement vector like | AB | and not by "| s |".

Displacement and dimension of motion

We have so far discussed displacement as a general case in three dimensions. The treatment of displacement in one or two dimensions is relatively simplified. The expression for displacement in component form for these cases are given here :

1. Motion in two dimension : Let the motion takes place in the plane determined by x and y axes, then :

$$\Delta \mathbf{r}=\Delta x\mathbf{i}+\Delta y\mathbf{j};\Delta z=0$$

If the initial position of the particle coincides with the origin of reference system, then :

$$\Delta \mathbf{r}=\mathbf{r}=x\mathbf{i}+y\mathbf{j};z=0$$

2. Motion in one dimension : Let the motion takes place along the straight line parallel to x - axis, then :

$$\Delta \mathbf{r}=\Delta x\mathbf{i};\Delta y=\Delta z=0$$

If the initial position of the particle coincides with the origin of reference system, then :

$$\Delta \mathbf{r}=\mathbf{r}=x\mathbf{i};y=z=0$$

Displacement – time plot

Plotting displacement vector requires three axes. Displacement – time plot will, therefore, need a fourth axis for representing time. As such, displacement – time plot can not be represented on a three dimensional Cartesian coordinate system. Even plotting two dimensional displacement with time is complicated.