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By the end of this section, you will be able to:
  • Calculate position vectors in a multidimensional displacement problem.
  • Solve for the displacement in two or three dimensions.
  • Calculate the velocity vector given the position vector as a function of time.
  • Calculate the average velocity in multiple dimensions.

Displacement and velocity in two or three dimensions are straightforward extensions of the one-dimensional definitions. However, now they are vector quantities, so calculations with them have to follow the rules of vector algebra, not scalar algebra.

Displacement vector

To describe motion in two and three dimensions, we must first establish a coordinate system and a convention for the axes. We generally use the coordinates x , y , and z to locate a particle at point P ( x , y , z ) in three dimensions. If the particle is moving, the variables x , y , and z are functions of time ( t ):

x = x ( t ) y = y ( t ) z = z ( t ) .

The position vector    from the origin of the coordinate system to point P is r ( t ) . In unit vector notation, introduced in Coordinate Systems and Components of a Vector , r ( t ) is

r ( t ) = x ( t ) i ^ + y ( t ) j ^ + z ( t ) k ^ .

[link] shows the coordinate system and the vector to point P , where a particle could be located at a particular time t . Note the orientation of the x , y , and z axes. This orientation is called a right-handed coordinate system ( Coordinate Systems and Components of a Vector ) and it is used throughout the chapter.

An x y z coordinate system is shown, with positive x out of the page, positive y to the right, and positive z up. A point P, with coordinates x of t, y of t, and z of t is shown. All of P’s coordinates are positive. The vector r of t from the origin to P is also shown as a purple arrow. The coordinates x of t, y of t and z of t are shown as dashed lines. X of t is a segment in the x y plane, parallel to the x axis, y of t is a segment in the x y plane, parallel to the y axis, and z of t is a segment parallel to the z axis.
A three-dimensional coordinate system with a particle at position P ( x ( t ), y ( t ), z ( t )).

With our definition of the position of a particle in three-dimensional space, we can formulate the three-dimensional displacement. [link] shows a particle at time t 1 located at P 1 with position vector r ( t 1 ) . At a later time t 2 , the particle is located at P 2 with position vector r ( t 2 ) . The displacement vector     Δ r is found by subtracting r ( t 1 ) from r ( t 2 ) :

Δ r = r ( t 2 ) r ( t 1 ) .

Vector addition is discussed in Vectors . Note that this is the same operation we did in one dimension, but now the vectors are in three-dimensional space.

An x y z coordinate system is shown, with positive x out of the page, positive y to the right, and positive z up. Two points, P 1 and P 2 are shown. The vector r of t 1 from the origin to P 1 and the vector r of t 2 from the origin to P 2 are shown as purple arrows. The vector delta r is shown as a purple arrow whose tail is at P 1 and head at P 2.
The displacement Δ r = r ( t 2 ) r ( t 1 ) is the vector from P 1 to P 2 .

The following examples illustrate the concept of displacement in multiple dimensions.

Polar orbiting satellite

A satellite is in a circular polar orbit around Earth at an altitude of 400 km—meaning, it passes directly overhead at the North and South Poles. What is the magnitude and direction of the displacement vector from when it is directly over the North Pole to when it is at −45 ° latitude?

Strategy

We make a picture of the problem to visualize the solution graphically. This will aid in our understanding of the displacement. We then use unit vectors to solve for the displacement.

Solution

[link] shows the surface of Earth and a circle that represents the orbit of the satellite. Although satellites are moving in three-dimensional space, they follow trajectories of ellipses, which can be graphed in two dimensions. The position vectors are drawn from the center of Earth, which we take to be the origin of the coordinate system, with the y -axis as north and the x -axis as east. The vector between them is the displacement of the satellite. We take the radius of Earth as 6370 km, so the length of each position vector is 6770 km.

An x y coordinate system, centered on the earth, is shown. Positive x is to the east and positive y to the north. A blue circle larger than and concentric with the earth is shown. Vector r of t 1 is an orange arrow from the origin to the location where the blue circle crosses the y axis (90 degrees counter clockwise from the positive x axis.) Vector r of t 2 is an orange arrow from the origin to the location on the blue circle at minus 45 degrees. Delta r vector is shown as a purple arrow pointing down and to the right, starting at the head of vector r of t 1 and ending at the head of vector r of t 2.
Two position vectors are drawn from the center of Earth, which is the origin of the coordinate system, with the y -axis as north and the x -axis as east. The vector between them is the displacement of the satellite.
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Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Practice Key Terms 3

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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