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Welcome to our Physics lesson on Displacement and Distance in two dimensions, this is the first 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.
Let's consider a visual example before dealing with numbers. Look at the figure below.
Path 1 represents the Distance travelled by the object if it moves according the curved line given. It is not linear, so if there is no measuring tool available but we are dealing with coordinates only, we must divide the Path 1 into smaller sections that are almost linear to find the path's length (Distance). Look at the figure:
Thus, we can consider the segments AC, CD, DE, EF, FG and GB as linear and then calculating the length of each segment by splitting them into components.
After doing this, we can use the Pythagorean Theorem to calculate the length of each segment (it is the hypotenuse of each right triangle formed) and then adding them mathematically. Thus, we find the Distance with a good approximation (with a small error). Remember, we usually label the horizontal direction by x and the vertical direction with y.
Hence, we can write for the distance s:
As for the displacement, it is much easier to calculate it, as we have to deal only with two components: ABx and ABy. Therefore, the Pythagorean Theorem is used only once, as the hypotenuse of the right triangle represents the magnitude of the displacement.
Thus, if we denote the displacement by ∆r⃗ we can write:
If the path resembles a known geometric figure, we must simply calculate the perimeter of this figure in order to find the Distance. For example, if the object moves from A to B according the path shown in the figure,
where R is the radius of circle, it is obvious that the Distance s is the length of the arc shown, which it equal to 3/4 of the circumference.
On the other hand, since Displacement represents the shortest path from A to B, here it is equal to the length of the chord AB. Look at the figure:
Combining the knowledge obtained in Geometry with the Kinematics concepts, we can write:
and
For example, if the Radius is 14m, and giving that π = 3.14 = 22/7, we obtain for Distance and the magnitude of Displacement
and
You may have noticed in all examples discussed so far that displacement is always shorter or equal to the distance. This is obvious since displacement vector AB⃗ represents the shortest possible distance from A to B.
Calculate the Displacement and the Distance travelled by the object shown in the figure if the object has moved according the trajectory ABCDE.
Displacement ∆r⃗ is the shortest path from A to E, i.e it is found by moving due right from A to E as shown below.
The Displacement vector Δr⃗ is coloured in indigo. Since the figure now is a rectangle, and given that the opposite sides of a rectangle are parallel and equal, it is obvious that
Hence, we can write for the Displacement:
As for the distance s, we have to take the sum of all segments in black, i.e. the perimeter of the rectangle BCDE minus the length of the vector AE⃗. Thus, we have:
You have reach the end of Physics lesson 3.4.1 Displacement and Distance in two 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.
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