Position, Displacement, and Motion States

Describing Motion in Three Dimensions

A moving object in space can be described by recording its location at each time. The record is a vector-valued function $\mathbf{r}(t)$, whose three components specify the object's coordinates relative to a chosen origin and coordinate system.

But a list of locations is not the same thing as a complete description of motion. We also need to distinguish how the location changes over an interval from how the object is moving at a particular instant, and from how its motion is changing. These are different quantities: position, displacement, velocity, and acceleration.

This distinction matters in applications as well as in calculations. Two objects can arrive at the same location while taking different paths. An object can have a large position vector but be momentarily at rest, or have zero displacement over an interval despite moving throughout it.

Definition: A position vector $\mathbf{r}(t)$ gives the location of an object at time $t$, measured from a specified origin. The displacement from time $t_1$ to time $t_2$ is the change in position: $$ \Delta \mathbf{r} = \mathbf{r}(t_2)-\mathbf{r}(t_1). $$

Position Is Location, Not Motion

In three dimensions, position is commonly written in component form as

\mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle.

At any specified time, the three component values identify one point relative to the chosen origin. If the components are measured in metres, then $\mathbf{r}(t)$ is a position vector measured in metres. Its components describe where the object is, not how quickly it is moving.

The origin is a reference point. Changing the origin changes the position coordinates: the same physical location is then represented by a different position vector. However, the difference between two position vectors for the same object is unaffected by a fixed change of origin. This is one reason displacement is a useful description of an interval of motion.

Displacement Is the Net Change in Position

To find displacement over an interval, evaluate the position at the final time and subtract the position at the initial time. The order is important: final position minus initial position. If $\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle$, then

\Delta\mathbf{r} = \langle x(t_2)-x(t_1),\, y(t_2)-y(t_1),\, z(t_2)-z(t_1)\rangle.

Displacement is a vector. Its components tell how much the object's coordinates changed in each direction. A positive component indicates a net change in the positive coordinate direction; a negative component indicates a net change in the negative coordinate direction. Displacement depends on the two endpoint positions, not on the details of the route between them.

Worked Example: Finding Displacement from Two Positions

An object's position, in metres, at time $t$ seconds is $\mathbf{r}(t)=\langle 1+2t,\,4-t,\,3t\rangle$. Find its displacement from $t=1$ to $t=4$.

Step 1: Evaluate the initial position.

\mathbf{r}(1) = \langle 1+2(1),\,4-1,\,3(1)\rangle = \langle 3,3,3\rangle\text{ m}.

Step 2: Evaluate the final position.

\mathbf{r}(4) = \langle 1+2(4),\,4-4,\,3(4)\rangle = \langle 9,0,12\rangle\text{ m}.

Step 3: Subtract initial position from final position.

\Delta\mathbf{r} = \langle 9,0,12\rangle-\langle 3,3,3\rangle = \langle 9-3,\,0-3,\,12-3\rangle = \langle 6,-3,9\rangle\text{ m}.

Check: Adding the displacement to the initial position recovers the final position: $\langle 3,3,3\rangle+\langle 6,-3,9\rangle=\langle 9,0,12\rangle$ m. The negative second component correctly records a net decrease of $3$ m in the $y$-coordinate.

Displacement Does Not Record the Whole Path

A position function may describe a curved route, while displacement records only the vector from the starting point to the ending point. In particular, an object can move a substantial distance and still have zero displacement if it returns to where it began.

Worked Example: Returning to the Starting Point

Consider the position function $\mathbf{r}(t)=\langle 2\cos t,\,2\sin t,\,0\rangle$ metres. Find the displacement from $t=0$ to $t=2\pi$, and check whether the positions at those times agree.

Step 1: Find the initial position. Since $\cos 0=1$ and $\sin 0=0$,

\mathbf{r}(0) = \langle 2(1),\,2(0),\,0\rangle = \langle 2,0,0\rangle\text{ m}.

Step 2: Find the final position. Since $\cos(2\pi)=1$ and $\sin(2\pi)=0$,

\mathbf{r}(2\pi) = \langle 2(1),\,2(0),\,0\rangle = \langle 2,0,0\rangle\text{ m}.

Step 3: Subtract the endpoint positions.

\Delta\mathbf{r} = \mathbf{r}(2\pi)-\mathbf{r}(0) = \langle 2,0,0\rangle-\langle 2,0,0\rangle = \langle 0,0,0\rangle\text{ m}.

Check that the object has not stayed at one position throughout. At $t=\pi$, its position is

\mathbf{r}(\pi) = \langle 2(-1),\,2(0),\,0\rangle = \langle -2,0,0\rangle\text{ m},

which differs from $\mathbf{r}(0)=\langle 2,0,0\rangle$ m. Thus the object moves away and returns, even though the net displacement is zero. The endpoint calculation is correct because displacement measures net change, not the full route.

Why Displacement Depends Only on the Endpoints

The endpoint rule follows directly from what change in position means. If the object goes from $\mathbf{r}(t_1)$ to $\mathbf{r}(t_2)$, the vector needed to move from the initial position to the final position is the final vector minus the initial vector. The following calculation also shows how displacements over consecutive intervals combine.

Write the displacements on two consecutive intervals.
For times $t_1<t_2<t_3$, the first displacement is $\mathbf{r}(t_2)-\mathbf{r}(t_1)$, and the second is $\mathbf{r}(t_3)-\mathbf{r}(t_2)$.

Add the vectors.
The shared middle position appears once with a positive sign and once with a negative sign: $$ [\mathbf{r}(t_2)-\mathbf{r}(t_1)] + [\mathbf{r}(t_3)-\mathbf{r}(t_2)] = \mathbf{r}(t_3)-\mathbf{r}(t_1). $$

Interpret the result.
The sum is exactly the displacement from the first time to the last time. The intermediate position cancels, so the net displacement is determined by the two outer endpoints.

This argument is sometimes called telescoping: the intermediate terms cancel when successive changes are added. It explains why displacement is additive across consecutive time intervals, while the path taken within each interval is not specified by the displacement vectors alone.

Four Different Motion Quantities

Position, displacement, velocity, and acceleration are related, but they answer different questions. Velocity describes the rate and direction of motion at an instant; acceleration describes the rate and direction of change of velocity. In the standard calculus description, for a differentiable position function, velocity is $\mathbf{v}(t)=\mathbf{r}'(t)$; when velocity is differentiable, acceleration is $\mathbf{a}(t)=\mathbf{v}'(t)$. Computing these derivatives is a separate task from finding displacement between two times.

Quantity	What it describes	Typical units
Position $\mathbf{r}(t)$	Location relative to the chosen origin at time $t$	metres (m)
Displacement $\Delta\mathbf{r}$	Net change in location between two specified times	metres (m)
Velocity $\mathbf{v}(t)$	Instantaneous rate and direction of motion	metres per second (m/s)
Acceleration $\mathbf{a}(t)$	Instantaneous rate and direction of change of velocity	metres per second squared (m/s²)

All four quantities may be represented by vectors in three dimensions, but their units and meanings prevent them from being interchangeable. For example, subtracting two positions gives a displacement, measured in metres. It does not give an instantaneous velocity, measured in metres per second.

Worked Example: Identifying Position, Displacement, Velocity, and Acceleration

Suppose an object's position in metres is $\mathbf{r}(t)=\langle t^2,\,2t,\,5-t\rangle$, with $t$ measured in seconds. Suppose its velocity and acceleration are also provided as $\mathbf{v}(t)=\langle 2t,\,2,\,-1\rangle$ m/s and $\mathbf{a}(t)=\langle 2,\,0,\,0\rangle$ m/s². Identify the three motion states at $t=2$, then find the displacement from $t=1$ to $t=3$.

Step 1: Evaluate position at $t=2$.

\mathbf{r}(2) = \langle 2^2,\,2(2),\,5-2\rangle = \langle 4,4,3\rangle\text{ m}.

This is a location: it specifies the object's coordinates relative to the origin.

Step 2: Evaluate the supplied velocity at $t=2$.

\mathbf{v}(2) = \langle 2(2),\,2,\,-1\rangle = \langle 4,2,-1\rangle\text{ m/s}.

This describes the instantaneous rate and direction of motion, not the object's location.

Step 3: Evaluate the supplied acceleration at $t=2$.

\mathbf{a}(2) = \langle 2,0,0\rangle\text{ m/s}^2.

This describes how velocity is changing at that instant. Its units differ from both position and velocity.

Step 4: Calculate the displacement from $t=1$ to $t=3$. First evaluate both endpoint positions:

\mathbf{r}(1) = \langle 1^2,\,2(1),\,5-1\rangle = \langle 1,2,4\rangle\text{ m}, \qquad \mathbf{r}(3) = \langle 3^2,\,2(3),\,5-3\rangle = \langle 9,6,2\rangle\text{ m}.

Now subtract in the required order:

\Delta\mathbf{r} = \langle 9,6,2\rangle-\langle 1,2,4\rangle = \langle 8,4,-2\rangle\text{ m}.

Check: Adding this displacement to the initial position gives $\langle 1,2,4\rangle+\langle 8,4,-2\rangle=\langle 9,6,2\rangle$ m, the final position. The displacement is measured in metres, unlike the velocity and acceleration evaluated at $t=2$.

A Common Error: Reversing the Subtraction

A frequent mistake is to subtract the final position from the initial position while still calling the result the displacement from the initial time to the final time. For example, in the first worked example, that incorrect order would give

\mathbf{r}(1)-\mathbf{r}(4) = \langle 3,3,3\rangle-\langle 9,0,12\rangle = \langle -6,3,-9\rangle\text{ m}.

This is the displacement in the reverse direction, from the position at $t=4$ back to the position at $t=1$. The correction is to follow the stated direction of time: final minus initial.

Check the order: For motion from $t_1$ to $t_2$, calculate $\mathbf{r}(t_2)-\mathbf{r}(t_1)$. Then verify that $\mathbf{r}(t_1)+\Delta\mathbf{r}=\mathbf{r}(t_2)$.

Reading Motion Models Carefully

Before calculating, identify what information the problem gives and what it asks for. A vector labelled $\mathbf{r}(t)$ gives position; subtracting its values at two times gives displacement. A vector labelled $\mathbf{v}(t)$ gives velocity, and one labelled $\mathbf{a}(t)$ gives acceleration. Their component notation may look alike, but their meanings, units, and roles are distinct.

When interpreting a result, retain its units and read each component. For instance, a displacement $\langle 6,-3,9\rangle$ m means that the object's net changes in the three coordinate directions are $6$ m, $-3$ m, and $9$ m. It does not claim that the object traveled along a straight path or that its instantaneous motion was in that direction at every moment.

Check Your Understanding

Answer each question using the definitions and distinctions in this tutorial.

If $\mathbf{r}(t_1)=\langle 2,-1,4\rangle$ m and $\mathbf{r}(t_2)=\langle 5,3,0\rangle$ m, what is the displacement from $t_1$ to $t_2$?
Can an object move during an interval and have zero displacement over that interval? Explain what must be true of its endpoint positions.
What does a position vector describe, and how does that differ from velocity?
Why are displacement and acceleration not interchangeable, even though both are vectors?
How can you check that a calculated displacement has the correct subtraction order?

Quantity	What it describes	Typical units
Position \(\mathbf{r}(t)\)	Location relative to the chosen origin at time \(t\)	metres (m)
Displacement \(\Delta\mathbf{r}\)	Net change in location between two specified times	metres (m)
Velocity \(\mathbf{v}(t)\)	Instantaneous rate and direction of motion	metres per second (m/s)
Acceleration \(\mathbf{a}(t)\)	Instantaneous rate and direction of change of velocity	metres per second squared (m/s²)