What Displacement Tells You
A particle can move along a line, change direction, and eventually finish somewhere other than where it started. To describe its net change in position over a time interval, compare where it ends with where it began. This quantity is called displacement.
Displacement records the change from an initial position to a final position. It does not describe every position the particle occupied along the way. The order of subtraction matters: the final position comes first, and the initial position is subtracted from it.
Here, \(x(t)\) is the particle’s coordinate on a line. Assume the positions are defined at both endpoints and that \(t_f>t_i\). The sign of the displacement depends on the chosen positive coordinate direction. A positive result means the final coordinate is greater than the initial coordinate; a negative result means it is smaller. Neither sign, by itself, says how the particle moved between the endpoints.
How to Calculate and Interpret It
For a calculation, identify the two endpoint positions before subtracting. Use the position at the final time as the first value and the position at the initial time as the second. Then include the position unit in the result.
Record the initial time \(t_i\) and final time \(t_f\).
Find \(x(t_i)\) and \(x(t_f)\), including their units.
Calculate final position minus initial position.
Compare the result with the chosen positive coordinate direction.
A positive displacement indicates a net change toward the positive direction, and a negative displacement indicates a net change toward the negative direction. If the displacement is zero, the final and initial positions are equal. It does not follow that the particle stayed still: it may have moved and later returned to its starting position.
Displacement is not the same as distance traveled. Displacement depends only on the two endpoint positions. Distance traveled describes how much ground the particle covered along its motion, so it can depend on what happened between those endpoints. Endpoint data alone do not generally tell you that full path.
Worked Example: Displacement in the Positive Direction
A particle is at \(x(2)=-3\) meters and later at \(x(7)=5\) meters. Find its displacement from \(t=2\) to \(t=7\), and interpret the sign.
The initial time is \(t_i=2\) seconds, so the initial position is \(x(t_i)=-3\) meters. The final time is \(t_f=7\) seconds, so the final position is \(x(t_f)=5\) meters.
Subtract initial position from final position:
The displacement is \(8\) meters. It is positive, so the final coordinate is 8 meters greater than the initial coordinate: the net change is in the positive coordinate direction.
Check: Add the displacement to the initial position. \(-3\text{ m}+8\text{ m}=5\text{ m}\), which matches the given final position.
Why the Endpoints Are Enough for Displacement
Suppose we record several positions in sequence, starting at \(x_0\), passing through \(x_1\) and \(x_2\), and ending at \(x_3\). The signed changes in position across those stages are \(x_1-x_0\), \(x_2-x_1\), and \(x_3-x_2\). Adding them gives
The intermediate positions cancel: \(+x_1\) cancels with \(-x_1\), and \(+x_2\) cancels with \(-x_2\). The remaining change is the final position minus the initial position. The same cancellation occurs for any finite sequence of intermediate positions.
This reasoning also explains why a return to the starting position gives zero displacement. The signed changes along the way cancel overall. Zero describes the net change in position, not necessarily the absence of motion.
Worked Examples with Position Data
Worked Example: Negative Displacement from a Table
A particle’s position is \(12\) meters at \(t=0\) seconds and \(3\) meters at \(t=4\) seconds. Find its displacement and the average velocity over this interval.
First calculate displacement using the endpoint positions:
The negative sign means the final coordinate is 9 meters less than the initial coordinate. The net change is toward the negative coordinate direction.
Average velocity is displacement divided by elapsed time. The elapsed time is \(4\text{ s}-0\text{ s}=4\text{ s}\), so
Check: The position change implied by this average velocity over 4 seconds is \(\left(-\frac{9}{4}\text{ m/s}\right)(4\text{ s})=-9\text{ m}\), matching the displacement. Also, \(12\text{ m}+(-9\text{ m})=3\text{ m}\), the final position.
The average-velocity calculation uses displacement, not a description of every part of the particle’s motion. A negative average velocity means the net displacement is negative over the interval; it does not by itself say that the particle moved in the negative direction at every instant.
Worked Example: Equal Endpoints Do Not Describe the Path
A particle has position \(x(t)=t^2-6t+4\) meters, where \(t\) is measured in seconds. Find its displacement from \(t=1\) to \(t=5\), and use an intermediate position to explain what the result does and does not show.
Evaluate the initial position:
Evaluate the final position:
Now subtract the initial position from the final position:
The particle has zero displacement because its endpoint positions are equal. To see why this does not mean it stayed at \(-1\) meter, evaluate its position at \(t=3\) seconds:
At the intermediate time, the particle’s position is \(-5\) meters, different from its position at either endpoint. Thus the zero displacement states only that the final position equals the initial position; it does not specify what occurred between those times.
Check: The computed displacement gives \(x(1)+\Delta x=-1\text{ m}+0\text{ m}=-1\text{ m}\), which agrees with \(x(5)\).
A Common Subtraction Error
A frequent mistake is to subtract in the reverse order. For the positions in the first example, the incorrect calculation would be
This is not the displacement from \(t=2\) to \(t=7\). It gives the change in the opposite order, from the later position back to the earlier one. The correction is to use final minus initial:
A useful check is to add your proposed displacement to the initial position. The result should be the final position. The incorrect value fails this check because \(-3\text{ m}+(-8\text{ m})=-11\text{ m}\), not \(5\text{ m}\). The correct value passes: \(-3\text{ m}+8\text{ m}=5\text{ m}\).
Interpreting the Sign and Units
The coordinate system determines what positive and negative mean. If the positive direction is east, a positive displacement is a net change eastward and a negative displacement is a net change westward. If the positive direction is chosen to be west instead, those interpretations reverse. The numerical positions and the chosen direction must be understood together.
| Displacement | What it tells you | What it does not tell you |
|---|---|---|
| Positive | The final coordinate exceeds the initial coordinate. | That the particle moved positively throughout the interval. |
| Negative | The final coordinate is less than the initial coordinate. | That the particle moved negatively throughout the interval. |
| Zero | The final and initial coordinates are equal. | That the particle remained at rest. |
Because displacement is a difference of positions, its units are position units: for example, meters, feet, or kilometers. Do not report displacement in units of time or speed. When positions are given with units, carry those units through the subtraction.
For AP Calculus motion questions, keep the endpoint calculation separate from any question about the route taken. First find the signed displacement from the initial and final positions. Then interpret the sign using the stated coordinate direction. If a problem asks about the path or the total amount of ground covered, endpoint displacement alone is not enough information to answer that different question.
Check Your Understanding
For each question, use final position minus initial position and state what the sign means when appropriate.
- A particle moves from \(x=4\) meters to \(x=-2\) meters. What is its displacement?
- At \(t=3\) seconds a particle is at \(7\) feet, and at \(t=8\) seconds it is at \(10\) feet. Find the displacement and its sign.
- If a particle’s initial and final positions are equal, what can you conclude about its displacement? What can you not conclude about its motion?
- A proposed displacement is \(-6\) meters from an initial position of \(2\) meters. What final position does this imply?
- Why can two particles with the same initial and final positions have the same displacement even if their motion between those times differs?
