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Physics Lesson 3.4.3 - Displacement and Distance in Space (in three dimensions)
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Welcome to our Physics lesson on Displacement and Distance in Space (in three dimensions), this is the third lesson of our suite of physics lessons covering the topic of Displacement and Distance in 2 and 3 Dimensions, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Displacement and Distance in Space (in three dimensions)
When an object is moving in space, we call it "3D motion." In this case, an additional coordinate (usually denoted by z) is required to represent the third dimension.
However, the approach is the same as for 2D motion. We simply write the z-coordinate besides x and y coordinates we discussed before and all calculations rely on these three coordinates.
Examples of 3D motion include the flight of airplanes, birds, objects exploding in air etc. We will take an airplane's flight as an example to illustrate this kind of motion.
The vector r⃗i shows the initial position and the vector r⃗f the final position of the airplane. Both these vectors have three coordinates each: xi, yi and zi for r⃗i and xf, yf and zf for r⃗f. Therefore, the displacement vector
will have three coordinates as well. They are:
= xf - xi
= yf - yi
= zf - zi
The magnitude of the Displacement vector ∆r⃗ therefore is
Or
Conceptually, there is nothing new here; the only difference with the study of 2-D motion is the new coordinate z, which makes the equations longer, but the structure is the same.
The final figure containing all the necessary info is shown below.
As for the distance, if the motion is not linear, more info regarding the path's shape are needed. They require a more complicated mathematical apparatus that goes beyond the level of this course.
Example 3
An airplane takes off from the coordinates (120m, 200m, 0) and moves linearly for a few seconds until it goes at (400m, 300m, 100m). What is the displacement and the distance travelled by the airplane during this process?
Solution 3
From the clues, we have:
yi = 200m
zi = 0
xf = 400m
yf = 300m
zf = 100m
From the 3D displacement formula, we have
Substituting the known values, we obtain
= √(280 m)2 + (100 m)2 + (100 m)2
= √78400 m2 + 10000 m2 + 10000 m2
= √98400 m2
≈ 314m
You have reach the end of Physics lesson 3.4.3 Displacement and Distance in Space (in three dimensions). There are 3 lessons in this physics tutorial covering Displacement and Distance in 2 and 3 Dimensions, you can access all the lessons from this tutorial below.
More Displacement and Distance in 2 and 3 Dimensions Lessons and Learning Resources
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