Is it Better to Use a Diagram or Math to Calculate Displacement?

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Calculating displacement can be effectively done using the formula delta d = df - di, where positions can be assigned positive or negative values based on direction. For example, if the initial position is 23m West and the final position is 34m East, treating West as negative and East as positive allows for the calculation of displacement as 34 - (-23). Visualizing the scenario with a diagram can enhance understanding, particularly when both points lie on the same horizontal line, making the calculation straightforward. However, if the points are not aligned, such as when one point has a vertical component, using the Pythagorean Theorem becomes necessary to find the correct displacement. Overall, both mathematical calculations and diagrams serve important roles in accurately determining displacement.
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If you're calculating displacement using delta d=df-di, do you make one position positive and one negative and then add?
For example, if your first position was 23m[W] and the final position was 34 m [E], would you just do, 34- (-23), making the position that is west negative and the east position positive?
 
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Balsam said:
If you're calculating displacement using delta d=df-di, do you make one position positive and one negative and then add?
For example, if your first position was 23m[W] and the final position was 34 m [E], would you just do, 34- (-23), making the position that is west negative and the east position positive?
You can do this for this example. With how you described it, it like taking the x-axis. There, you would be setting the 0 point at the origin, such that any distance westward (in your case, the 23m) is negative and anything eastward positive, which is legitimate. The only time this may get a bit trickier thinking of it this way is if you aren't given two points along the x-axis. Say, for example you were given the first point 23m West and the second point 34m East and 10m North. In this case, you would not just add 23m+34m+10m to get 67m because the 2 points do not lie on the same line. However, in your case, this works.
 
Yosty22 said:
You can do this for this example. With how you described it, it like taking the x-axis. There, you would be setting the 0 point at the origin, such that any distance westward (in your case, the 23m) is negative and anything eastward positive, which is legitimate. The only time this may get a bit trickier thinking of it this way is if you aren't given two points along the x-axis. Say, for example you were given the first point 23m West and the second point 34m East and 10m North. In this case, you would not just add 23m+34m+10m to get 67m because the 2 points do not lie on the same line. However, in your case, this works.

Would it make more sense just to draw a diagram and see the distance and direction of the final point from the starting point?
 
Balsam said:
Would it make more sense just to draw a diagram and see the distance and direction of the final point from the starting point?
In general, I believe pictures always make this more intuitive and easy to understand. If you drew a picture of the case you presented in the original post, it would be evident that the two points lie on the same horizontal line. In this case, your method works.

However, in the case I gave as an example, if you plotted these two points and connected them with a straight line, it is obvious that the line connecting the two points has a non-zero slope. In this case, one can simply just use the Pythagorean Theorem to solve for the hypotenuse of the right triangle to get the displacement.
 
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