1D Displacement Explained: The Cleanest Way to Understand Motion in One Dimension
- Feb 3
- 4 min read
Updated: 20 hours ago
If you are starting kinematics, the very first idea you need to get crystal clear is displacement. Students often mix it up with distance, and that single mix-up causes a bunch of later problems with velocity, acceleration, and graphs.
This post will make displacement feel straightforward, then we’ll level up into multi-step “multiple movements” questions (the kind that actually show up on tests).
What “1D” means
1D motion means motion along a straight line. Examples:
moving east/west along a road
moving up/down in an elevator
walking forward/backward in a hallway
Because it’s one-dimensional, we can pick a sign convention:
Choose one direction as positive (commonly right or up)
The opposite direction is negative
Once we decide that, we can represent position with a single number.
Key vocabulary (and the #1 confusion)
Position
Your position is your location relative to a chosen origin (a “zero point”). Example: if the origin is your house and you are 200 m east, your position could be +200 m (if east is positive).
Displacement
Displacement is the change in position:
Δx = xf − xi
xi = initial position
xf = final position
Δx = displacement
Displacement is a vector in 1D, meaning it has a direction (sign matters).
Distance (different from displacement)
Distance is how much ground you covered total. It does not care about direction. Distance is always positive (or zero).
A sign convention you can use every time
In most 1D kinematics problems, you can safely do this:
Right (or east) = positive
Left (or west) = negative
Or if it’s vertical:
Up = positive
Down = negative
The key is consistency. You can choose the opposite if you want, but don’t switch halfway through.
Example 1 (simple): displacement from positions
A student starts at xi = +3 m and ends at xf = +11 m. Find the displacement.
Δx = xf − xi = 11 − 3 = +8 m
Answer: Δx = +8 m (8 m in the positive direction)
Example 2 (simple): negative displacement
A cart starts at xi = +5 m and ends at xf = −2 m. Find the displacement.
Δx = −2 − 5 = −7 m
Answer: Δx = −7 m (7 m in the negative direction)
Example 3 (distance vs displacement)
You walk 6 m east, then 6 m west, ending where you started. Find your displacement and the distance you walked.
Displacement: you ended at the starting point, so Δx = 0 m
Distance: you walked 6 + 6 = 12 m
Answer: displacement = 0 m, distance = 12 m
This is the classic proof that distance and displacement are not the same thing.
Multiple-movement examples
For these, we’ll use: right = positive, left = negative.
A very reliable method is:
Convert each movement into a signed value
Add them to get displacement
Add absolute values to get distance (only if asked)
Example 4 (moderate): three movements
A robot moves:
10 m right
4 m left
3 m right
Find the displacement.
Signed movements:
+10
−4
+3
Δx = 10 − 4 + 3 = +9 m
Answer: Δx = +9 m
Example 5 (moderate): finding final position
A cyclist starts at xi = −12 m. She moves:
25 m right
9 m left
Find her final position xf and displacement Δx.
Signed movements: +25, −9 Net change:
Δx = 25 − 9 = +16 m
Final position:
xf = xi + Δx = −12 + 16 = +4 m
Answer: xf = +4 m and Δx = +16 m
Example 6 (harder): “back and forth” with a trap
A person starts at x = 0 m. They walk:
30 m east
18 m west
12 m west
7 m east
a) Find displacement
b) Find distance traveled
Displacement
Signed movements:
+30
−18
−12
+7
Δx = 30 − 18 − 12 + 7 = +7 m
Displacement: Δx = +7 m
Distance
Add magnitudes: d = 30 + 18 + 12 + 7 = 67 m
Answer: Δx = +7 m, distance = 67 m
Example 7 (harder): unknown segment
A toy car starts at xi = +2 m. It ends at xf = −11 m. Along the way it moves:
6 m right
then some unknown distance left, call it (L)
Find L.
First find displacement using positions:
Δx = xf - xi = −11 − 2 = −13 m
Now write displacement as the sum of signed movements:
6 m right is +6
L left is −L
So: +6 − L = −13
Solve:
Subtract 6 from both sides: −L = −19
Multiply both sides by −1: (L = 19)
Answer: L = 19 m left
Example 8 (challenge): multiple movements + final position + comparison
A runner starts at x = −40 m. Over a drill, she runs:
55 m right
20 m left
10 m left
8 m right
a) Find net displacement
b) Find final position
c) Did she end up to the left or right of the origin?
1) Net displacement Δx = +55 − 20 − 10 + 8 = +33 m
So Δx = +33 m
2) Final position xf = xi + Δx = −40 + 33 = −7 m
3) Side of the origin −7 m is negative, so she ends left of the origin.
Answer: Δx = +33 m, xf = −7 m, left of origin
Common mistakes
Using distance when the question asks for displacement Displacement is net change. Distance is total path length.
Forgetting the formula Δx = xf − xi Not “initial minus final.”
Dropping the sign A negative displacement is not “wrong,” it just means direction.
Changing your sign convention halfway through Decide once: right/up is positive is a great default.
Quick practice
Assume right is positive.
Start at xi = 4 m, end at xf = −9 m. Find Δx.
Start at x = 0 m: move 12 m right, 5 m left, 9 m left. Find displacement.
Start at x = -6 m. You move 14 m right, then L m left, and end at x = 1 m. Find L.
Answer key
Δx = −9 − 4 = −13 m
Δx = 12 − 5 − 9 = −2 m
First Δx = 1 − (−6) = +7. Then 14 − L = 7 ⇒ L = 7 m
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