We have the liberty to describe displacement vector as an independent vector ( AB) or in terms of position vectors ( r 1 r 1 and r 2 r 2 ). The choice depends on the problem in hand. The description, however, is equivalent.

Let us consider that a point object moves from point A (represented by position vector r 1 r 1 ) to point B (represented by position vector r 2 r 2 ) as shown in the figure. Now, using triangle law (moving from O to A to B to O), we have :

Figure 3: Displacement is equal to the difference between final and initial position vectors

Displacement in terms of position vectors

OA = AB + OB r 1 + AB = r 2 ⇒ AB = r 2 - r 1 = ∆ r OA = AB + OB r 1 + AB = r 2 ⇒ AB = r 2 - r 1 = ∆ r

Thus, displacement is equal to the difference between final initial position vectors. It is important to note that we obtain the difference between initial and final position vectors by drawing a third vector starting from the tip of the initial position vector and ending at the tip of the final position vector. This approach helps us to quickly draw the vector representing the difference of vectors and is a helpful procedural technique that can be used without any ambiguity.

r 1 = x 1 i + y 1 j + z 1 k r 2 = x 2 i + y 2 j + z 2 k ⇒ AB = ∆ r = ( x 2 - x 1 ) i + ( y 2 - y 1 ) j + ( z 2 - z 1 ) k ⇒ AB = ∆ r = ∆ x i + ∆ y j + ∆ x k r 1 = x 1 i + y 1 j + z 1 k r 2 = x 2 i + y 2 j + z 2 k ⇒ AB = ∆ r = ( x 2 - x 1 ) i + ( y 2 - y 1 ) j + ( z 2 - z 1 ) k ⇒ AB = ∆ r = ∆ x i + ∆ y j + ∆ x k

We must emphasize here that position vectors and displacement vectors are different vectors quantities. We need to investigate the relation between position vectors and displacement vector - a bit more closely. It is very important to mentally note that the difference of position vectors i.e. displacement,Δr, has different directional property to that of the position vectors themselves ( r 1 r 1 and r 2 r 2 ).

In the figure above, the position vectors r 1 r 1 and r 2 r 2 are directed along OA and OB respectively, while displacement vector, Δr, is directed along AB. This means that the direction of displacement vector need not be same as that of either of the position vectors. Now, what would be the situation, when the motion begins from origin O instead of A? In that case, initial position vector is zero (null vector). Now, let the final position vector be denoted as r. Then

r 1 = 0 r 2 = r r 1 = 0 r 2 = r

and displacement is :

⇒ AB = Δ r = r 2 - r 2 = r - 0 = r ⇒ AB = Δ r = r 2 - r 2 = r - 0 = r

This is a special case, when final position vector itself is equal to the displacement. For this reason, when motion is studied from the origin of reference or origin of reference is chosen to coincide with initial position, then displacement and final position vectors are equivalent and denoted by the symbol r.

In general, however, it is the difference between final and initial position vectors, which is equal to the displacement and we refer displacement in terms of change in position vector and use the symbol Δr to represent displacement.

Magnitude of displacement is equal to the absolute value of the displacement vector. In physical sense, the magnitude of displacement is equal to the distance between original and final positions along the straight line joining two positions i.e. the shortest distance between initial and final positions. This value may or may not be equal to the distance along the actual path of motion. In other words, magnitude of displacement represents the minimum value of distance between any two positions.

Example 1: Displacement

Question : Consider a person walking from point A to B to C as shown in the figure. Find distance, displacement and magnitude of displacement.

Figure 4:

Solution : The distance covered, s, during the motion from A to C is to the sum of the lengths AB and BC.

s = 4 + 3 = 7 m s = 4 + 3 = 7 m

Displacement, AC, is given as :

AC = Δ r = r 2 - r 1 AC = Δ r = ( 6 - 2 ) i + ( 5 - 2 ) j = 4 i + 3 j AC = Δ r = r 2 - r 1 AC = Δ r = ( 6 - 2 ) i + ( 5 - 2 ) j = 4 i + 3 j

The displacement vector makes an angle with the x – axis given by :

θ = tan - 1 ( 3 / 4 ) θ = tan - 1 ( 3 / 4 )

and the magnitude of the displacement is :

| AB | = | r 2 r 1 | ⇒ | AB | = | Δ r | = | 4 i + 3 j | = 4 2 + 3 2 = 5 m | AB | = | r 2 r 1 | ⇒ | AB | = | Δ r | = | 4 i + 3 j | = 4 2 + 3 2 = 5 m

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 other than the 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.

s ≥ | Δ r | s ≥ | Δ r |

This inequality is important. It implies that displacement is not distance plus direction as may loosely be considered. 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, |Δr| is equal to distance, “s”. Matter of fact, 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 |Δr| or in terms of displacement vector like |AB| and not by “s”.

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 :

Δ r = Δ x i + Δ y j ; Δ z = 0 Δ r = Δ x i + Δ y j ; Δ z = 0

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

Δ r = r = x i + y j ; z = 0 Δ r = r = x i + y j ; z = 0

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

Δ r = Δ x i ; Δ y = Δ z = 0 Δ r = Δ x i ; Δ y = Δ z = 0

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

Δ r = r = x i ; y = z = 0 Δ r = r = x i ; y = z = 0

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.

One dimensional motion, having only two directions – along or opposite to the positive direction of axis, allows plotting displacement – time graph. One dimensional motion involves only one way of changing direction i.e. the particle under motion can reverse its direction of motion. Any other change of direction is not possible; otherwise the motion would not remain one dimensional motion.

The simplication, in the case of one directional motion, allows us to do away with the need to use vector notation and vector algebra for quantities with directional attributes. Instead, the vectors are treated simply as scalars with one qualification that vectors in the direction of chosen reference is considered positive and vectors in the opposite direction to chosen reference is considered negative.

Representation of a displacement vector as a scalar quantity uses following construct :

To illustrate the construct, let us consider a motion of a ball which transverses from O to A to B to C to O along x-axis as shown in the figure.

Figure 5:

The displacements (in meters) at various points of motion are :

x 1 = OA = 5 i x 2 = OB = 10 i x 3 = OC = - 5 i x 1 = OA = 5 i x 2 = OB = 10 i x 3 = OC = - 5 i

It is important to note from above data that when origin is chosen to coincide with initial position of the particle, then displacement and position vectors are equal.

The data for displacement as obtained above also reveals that by assigning proper sign to a scalar value, we can represent directional attribute of a vector quantity. In other words, plotting magnitude of displacement in one dimension with appropriate sign would completely represent the displacement vector in both magnitude and direction.