Displacement (vector)

A displacement is the shortest distance from the initial and final positions of a point P. Thus, it is the length of an imaginary straight path, typically distinct from the path actually travelled by P. A displacement vector represents the length and direction of that imaginary straight path.

Displacement versus distance traveled along a path.

A position vector expresses the position at a point in space in terms of displacement from an arbitrary reference point (typically the origin of a coordinate system). Namely, it indicates both the distance and direction of an imaginary motion along a straight line from the reference position to the position of the point. Thus, a displacement may be also described as a relative position: the final position of a point relative to its initial position, and a displacement vector can be mathematically defined as the difference between the final and initial position vectors.

In considering motions of objects over time the instantaneous velocity of the object is the rate of change of the displacement as a function of time. The velocity then is distinct from the instantaneous speed which is the time rate of change of the distance traveled along a specific path.

If a fixed origin is defined we may then equivalently define the velocity as the time rate of change of the position vector. However if one considers a time dependent choice of origin as in a moving coordinate system the rate of change of the position vector only defines a relative velocity.

For motion over a given interval of time, the displacement divided by the length of the time interval defines the average velocity.

In dealing with the motion of a rigid body, the term displacement may also include the rotations of the body. In this case, the displacement of a particle of the body is called linear displacement (displacement along a line), while the rotation is called angular displacement.

Contents

Distance Travelled

If the position of an object is described by a vector function

\mathbf{r}(t):\mathbf{R} \to \mathbf{R}^n,

then the distance traveled as a function of t is described by the integral of one with respect to arc length.

d(t)=\int_{0}^{t}1\,\mathrm{d}s

where

ds is the arc length differential

The arc length differential is described by the following equation:

\mathrm{d}s=\left|\mathbf{r}'(t)\right|\,\mathrm{d}t=\left|\mathbf{v}(t)\right|\,\mathrm{d}t=v(t)\,\mathrm{d}t

where

\mathbf{v}(t) is velocity
v(t)\, is speed.

Displacement, position, and velocity

A position vector can be viewed as a hypothetical displacement of a point or particle from the origin of a coordinate system to the location of a point at a given time.

On a graph representing the position of a particle with respect to time (position vs. time graph), the slope of the straight line joining two points on the graph is the average velocity of the particle during the corresponding time interval, while the slope of the tangent to the graph at a given point is the instantaneous velocity at the corresponding time (first derivative of the particle position).

Displacement and the equations of motion

Main article: SUVAT equations
Further information: Equations of motion

To calculate displacement all vectors and scalars must be taken into consideration.[1][2][3] The following formulas can be used to calculate displacement for, s, for an object undergoing constant acceleration.[1][2].

\mathbf{s} = {\mathbf{u}t+{1\over 2}\mathbf{a}t^2}[1][2]
\mathbf{v} = \mathbf{u}+ \mathbf{a}t
\mathbf {v^2} = \mathbf{u^2}+2\mathbf{as}

Where:[3]

\mathbf{u} Initial velocity
\mathbf{v} Final velocity
\mathbf{a} Acceleration
\mathbf{t} Time
\mathbf{s=}Δ\mathbf{x} Displacement
\mathbf{x} Position (in one dimension)
\mathbf{r} Position (in three dimensions)
  • It should also be emphasized that vector directions, negative and positive signs, are important when calculating displacement [1][2][3]

Height displacement

Height displacement is the distance an object peaks in height vertically.[1][2] If, for example, a ball was thrown up in the air and fell back into the owner's hand, the displacement would be zero, since displacement over a period of time is defined as the distance between an object's starting and finishing points.[3]

However one may use the general equation s = {ut+{1\over 2}at^2}[2][3] to calculate overall vertical height. This is modified to h = {ut-{1\over2} gt^2}[2][3] for the case of a ball in the presence of gravity. The height h is dependent upon the time t at which it is being measured. g is the acceleration caused by Earth's gravity; it stays constant at approximately 9.81\ \text{m}/\text{s}^2. The {1\over2} gt^2 term is preceded by a minus sign because gravity acts in the opposite direction of h and u, which signify a distance and speed, respectively, away from the Earth's center of mass.[3]

See also

References

  1. ^ a b c d e BBC - Education Scotland - Higher Bitesize Revision - Physics - Analysing Motion - Analysing Motion: Revision 2
  2. ^ a b c d e f g New Higher Physics, Author: Adrian Watt Editor: Jim Page, Chapter 1, section 1.12: pg.26-27 ISBN 978-0-340-84776-3 [Amazon-US | Amazon-UK]
  3. ^ a b c d e f g Revision Notes for Higher Grade Physics by Lyn Robinson - head teacher of Williamwood High School Clarkston, also by Campbell White and Editor: Jim Jardine pg.7-22 Explanation of Vector directions, height, trajectory and mathematical formulae, ISBN 1-870570-55-3 [Amazon-US | Amazon-UK] (1994, reprint 1996)

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