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# Displacement

Module by: Sunil Kumar Singh. E-mail the author

Summary: Motion involves two types of measurements : one which depends on the end points (displacement) and the other which depends on all points(distance) of motion.

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Displacement is a measurement of change of position. Its magnitude and direction are measured by the length and direction of the straight line joining initial and final positions of the particle under motion.

Definition 1: Displacement
Displacement is the vector joining the initial position of the particle to its final position during an interval of time.

Displacement conveys the meaning of shortest distance and direction of motion between two time instants (i.e two positions). Initial and final positions of the point object are the only important consideration for measuring displacement. Actual path between two positions has no consequence. The quantum of displacement is measured by the length of the straight line joining two ends. If there is no change in the position at the end of a motion, the displacement is zero.

Consider the motion of a particle from A to B to C. The displacement vector is represented by the vector AC and its magnitude by the length of AC.

Once motion has begun, displacement may increase or decrease (at a slow, fast or constant rate) or may even be zero, if the object return to its initial position.

Consider another example of the motion of a particle along the rectangular path from A to B to C to D to A. Magnitude of displacement, shown by dotted vectors, is increasing during motion from A to B to C. Magnitude of displacement, however, is decreasing as the particle moves from C to D to A, eventually being equal to zero, when the particle returns to A.

However, displacement is essentially a measurement of length combined with direction. As direction has no dimension, its dimension formula is also [L] like that of distance and likewise, its SI measurement unit is ‘meter’.

## Displacement and Position vector

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 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 law of triangle (moving from O to A to B to O), we have :

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

The symbol “T” pronounced as “del” signifies the change in the quantity before which it appears. The change in quantity is equal to the difference between final and initial values.

Thus, displacement is equal to the difference between final initial position vectors. The displacement can also be expressed in terms of coordinates by expanding position vectors in terms of components :

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

Position vectors and displacement vectors are different vectors or 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 (r1 and r2).

In the above figure, 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). 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 simply 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.

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 )

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. Let us consider following points to elucidate the subtleties involved :

1 : Magnitude of displacement is a scalar quantity representing shortest distance between initial and final positions.

2 : Distance is the length of the actual path.

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 along a straight line; otherwise, distance is greater than the magnitude of displacement.

s | Δ r | s | Δ r |

This inequality is important. The important aspect here is 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 |AB| etc. and not by “s”.

### Example 2: Displacement

QuestionPosition (in meter) of a moving particle as a function of time (in second) is given by :

r = ( 3 t 2 - 3 ) i + ( 4 - 7 t ) j + ( - t 3 ) k r = ( 3 t 2 - 3 ) i + ( 4 - 7 t ) j + ( - t 3 ) k

Find the displacement in first 2 seconds.

Solution : The position vector at t = 0 and 2 seconds are calculated to identify initial and final positions.

When t = 0 (start of the motion)

r 1 = ( 3 x 0 - 3 ) i + ( 4 - 7 x 0 ) j + ( - 0 ) k = - 3 i + 4 j r 1 = ( 3 x 0 - 3 ) i + ( 4 - 7 x 0 ) j + ( - 0 ) k = - 3 i + 4 j

When t = 2 s,

r 2 = ( 3 x 2 2 - 3 ) i + ( 4 - 7 x 2 ) j + ( - 2 3 ) k = 9 i - 10 j - 8 k r 2 = ( 3 x 2 2 - 3 ) i + ( 4 - 7 x 2 ) j + ( - 2 3 ) k = 9 i - 10 j - 8 k

The displacement, Δr, is given by :

Δ r = r 2 - r 1 = ( 9 i - 10 j - 8 k ) - ( - 3 i + 4 j ) Δ r = r 2 - r 1 = ( 9 i - 10 j - 8 k ) - ( - 3 i + 4 j )

Δ r = r 2 - r 1 = ( 9 i - 10 j - 8 k + 3 i - 4 j = 12 i - 14 j - 8 k Δ r = r 2 - r 1 = ( 9 i - 10 j - 8 k + 3 i - 4 j = 12 i - 14 j - 8 k

Magnitude of displacement is given by :

| r | = ( 12 2 + ( - 14 ) 2 + ( - 8 ) 2 ) = 404 = 20.1 m | r | = ( 12 2 + ( - 14 ) 2 + ( - 8 ) 2 ) = 404 = 20.1 m

## 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 reduces to :

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 Δ r = Δ x i + Δ y j ; Δ z

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

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

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

"r = rx i ; iz = =y = 0 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 :

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

## Displacement – time plot

Plotting displacement vector requires three axes. Displacement – time plot would, therefore, need a fourth axis. Displacement – time plot, therefore, 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.

These two attributes of rectilinear (linear or one dimensional) motion allow 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.

Following construct is used for this purpose :

1: Assign an axis along the motion, which is treated as positive.

2: Assign the origin with the start of motion. It is, however, a matter of convenience and is not a requirement of the construct.

3: Use all quantities describing motion in the direction of axis as positive.

4: Use all quantities describing motion in the opposite direction of axis as negative.

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.

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.

These data for displacement reveals that assigning 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.

In order to plot these values of the magnitude of displacement, let us assume that time instants to the points O, A, B, A(again on return), O(again on return) and C are 0, 1, 2, 3, 4 and 5 seconds respectively. Note that origin here coincides with initial position. Hence, x – coordinate values represent the magnitude of displacement.

In this case, it is to be noted that displacement – time and position time plots for one dimensional motion are equivalent.

### Example 3: Displacement – time plot

Question : A ball falling from an height ‘h’ strikes a hard horizontal surface. On each rebound, the height reached by the ball is half of the height it fell from. Draw displacement – time plot for the motion covering two consecutive strikes, emphasizing the nature of curve (ignore actual calculation).

Solution : Let us consider that the origin of coordinate system coincides with the initial position of the ball. Let the upward direction is considered positive. Now, the ball falls along the vertical axis of the coordinate system under the influence of gravity. Thus, when ball reaches the surface, displacement is "-h" (A on plot). On first rebound, ball rises to reach half of the original height i.e. "h/2"(B on the plot), which is "-h/2" from the original position. The ball again travels down to the surface (C on the plot), which is "-h" away from the original position.

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