The Central Idea
Kinematics describes motion without first asking what forces caused it. Before studying speed, velocity, or acceleration, we need a precise way to describe where an object is and how its location changes.
Four basic ideas organize that description:
- position tells where an object is;
- path tells which route the object follows;
- displacement compares final position with initial position;
- distance measures the total length of the route traveled.
The distinction between displacement and distance is especially important. An object can travel a large distance and still have zero displacement.
What Is Position?
Position is the location of an object relative to a chosen reference system.
On a number line, position can be represented by a single coordinate such as
\[ x=5. \]In a plane, position may require two coordinates:
\[ (x,y)=(3,-2). \]In three-dimensional space, three coordinates are commonly used:
\[ (x,y,z)=(3,-2,7). \]Position therefore depends on both the object's location and the coordinate system used to describe it.
Position Requires a Reference
A statement such as “the object is at \(5\)” is incomplete unless the reference system is understood.
For example, \(x=5\) might mean:
- 5 meters east of an origin;
- 5 kilometers along a road;
- coordinate 5 on a horizontal axis;
- 5 feet above a selected reference level.
The same physical location can receive different coordinates when a different origin or orientation is chosen.
The Origin
The origin is the point assigned coordinate zero in a one-dimensional coordinate system.
\[ x=0. \]Positions to one side may be assigned positive coordinates, while positions on the opposite side are assigned negative coordinates.
A common convention is
\[ \text{right or east} \rightarrow \text{positive} \]and
\[ \text{left or west} \rightarrow \text{negative}. \]Other conventions are possible. The important requirement is to choose a convention and use it consistently.
Position on a Number Line
Suppose a cart moves along a straight track. Let the origin be the center of the track, with positive \(x\) pointing to the right.
If the cart is 4 meters to the right of the origin, then
\[ x=4\text{ m}. \]If it is 3 meters to the left, then
\[ x=-3\text{ m}. \]The sign is part of the position coordinate. It indicates location relative to the chosen origin and direction.
Initial and Final Position
Motion problems often distinguish between the position at the beginning of an interval and the position at the end.
We can write the initial position as
\[ x_i \]and the final position as
\[ x_f. \]For example,
\[ x_i=-2\text{ m}, \qquad x_f=7\text{ m}. \]These two values are enough to determine displacement, but not necessarily the total distance traveled.
What Is a Path?
A path is the route traced by a moving object through space.
A path may be:
- a straight line;
- a curved road;
- a circle;
- an arc;
- a parabola;
- a complicated curve through two- or three-dimensional space.
The path contains information that the initial and final positions alone do not provide.
Same Endpoints, Different Paths
Imagine traveling from point \(A\) to point \(B\).
One traveler walks directly from \(A\) to \(B\). Another follows a winding path but reaches the same endpoint.
Both travelers have the same initial and final positions, so they have the same displacement.
Their distances traveled can be very different because their paths have different lengths.
What Is Displacement?
Displacement is the change in position.
In one dimension,
\[ \boxed{\Delta x=x_f-x_i}. \]The symbol \(\Delta\), read “delta,” commonly means change in a quantity.
Therefore,
\[ \Delta x = \text{final position} - \text{initial position}. \]Example: Positive Displacement
Suppose
\[ x_i=2\text{ m} \]and
\[ x_f=9\text{ m}. \]Then
\[ \Delta x=x_f-x_i \] \[ \Delta x=9-2 \] \[ \boxed{\Delta x=7\text{ m}}. \]The positive sign indicates displacement in the positive \(x\)-direction.
Example: Negative Displacement
Suppose
\[ x_i=6\text{ m}, \qquad x_f=-1\text{ m}. \]Then
\[ \Delta x=-1-6 \] \[ \boxed{\Delta x=-7\text{ m}}. \]The negative sign indicates that the final position lies 7 meters in the negative direction from the initial position.
Displacement Is Not Final Position
Final position and displacement are different quantities.
Suppose
\[ x_i=10\text{ m}, \qquad x_f=13\text{ m}. \]The final position is
\[ 13\text{ m}, \]but the displacement is
\[ 13-10=3\text{ m}. \]Position tells where the object is. Displacement tells how much its position changed.
Displacement Depends Only on the Endpoints
Displacement depends only on the initial and final positions.
If an object begins at \(x=1\text{ m}\) and ends at \(x=8\text{ m}\), then
\[ \Delta x=8-1=7\text{ m}. \]The object might have moved directly from 1 to 8, or it might have wandered back and forth before reaching 8.
As long as the initial and final positions remain the same, the displacement remains
\[ 7\text{ m}. \]What Is Distance?
Distance traveled is the total length of the path followed by an object.
Distance does not include a directional sign in the same way displacement does.
If a person walks 4 meters east and then 3 meters west, the total distance traveled is
\[ 4+3=7\text{ m}. \]Distance adds the lengths of the pieces of the trip.
Distance Is Nonnegative
Ordinary distance traveled cannot be negative.
\[ \boxed{d\ge 0}. \]A negative sign can indicate direction in a coordinate or displacement, but distance measures path length.
Thus a statement such as
\[ d=-5\text{ m} \]is not a valid ordinary distance traveled.
Distance and Displacement Are Not the Same
Suppose an object begins at
\[ x=0, \]moves to
\[ x=10, \]and then returns to
\[ x=4. \]Its displacement is
\[ \Delta x=4-0=4. \]Its distance traveled is
\[ 10+6=16. \]Therefore,
\[ \boxed{\text{displacement}=4,\qquad \text{distance}=16}. \]Round Trips
A round trip returns to its starting position.
Therefore,
\[ x_f=x_i. \]The displacement is
\[ \Delta x=x_f-x_i=0. \]But the distance traveled is usually positive.
A runner who completes one 400-meter lap has
\[ \boxed{\text{distance}=400\text{ m}} \]but
\[ \boxed{\text{displacement}=0}. \]When Distance Equals the Magnitude of Displacement
In one-dimensional motion, distance equals the magnitude of displacement when the object moves from its starting point to its ending point without reversing direction.
For example, moving directly from \(x=2\) to \(x=9\) gives
\[ \Delta x=7 \]and
\[ d=7. \]In this case,
\[ \boxed{d=|\Delta x|}. \]Distance Is at Least the Magnitude of Displacement
For motion between two positions,
\[ \boxed{d\ge |\Delta x|}. \]The shortest possible route between two points in one-dimensional motion is the direct route without reversal.
Any backtracking increases the distance while leaving the endpoint displacement unchanged.
Example: Motion with One Reversal
An object starts at
\[ x=-2\text{ m}, \]moves to
\[ x=5\text{ m}, \]and then moves back to
\[ x=1\text{ m}. \]The displacement is
\[ \Delta x=1-(-2)=3\text{ m}. \]The first segment has length
\[ |5-(-2)|=7\text{ m}. \]The second segment has length
\[ |1-5|=4\text{ m}. \]Therefore, the total distance is
\[ \boxed{d=7+4=11\text{ m}}. \]A Reliable One-Dimensional Distance Method
If a motion is described by a sequence of positions
\[ x_0,x_1,x_2,\ldots,x_n, \]then the total distance over those successive straight-line segments is
\[ \boxed{ d= |x_1-x_0| + |x_2-x_1| +\cdots+ |x_n-x_{n-1}| }. \]Absolute value is used because each segment contributes a nonnegative length.
Example: Several Reversals
Suppose the successive positions are
\[ -3,\quad 4,\quad 1,\quad 6,\quad 2. \]The displacement is determined only by the first and last positions:
\[ \Delta x=2-(-3)=5. \]The distance is
\[ |4-(-3)|+|1-4|+|6-1|+|2-6|. \]Therefore,
\[ d=7+3+5+4 \] \[ \boxed{d=19}. \]Position as a Function of Time
A moving object's position generally changes with time.
We can represent one-dimensional position by a function
\[ x=x(t). \]This notation means that the position \(x\) depends on time \(t\).
For example,
\[ x(t)=2t+1 \]assigns a position to each time in the function's domain.
Reading a Position Function
Suppose
\[ x(t)=3t-4, \]where \(x\) is measured in meters and \(t\) in seconds.
At \(t=0\),
\[ x(0)=3(0)-4=-4\text{ m}. \]At \(t=5\),
\[ x(5)=3(5)-4=11\text{ m}. \]The displacement over this interval is
\[ \Delta x=11-(-4)=15\text{ m}. \]Displacement over a Time Interval
If position is given by \(x(t)\), then the displacement from time \(t_1\) to time \(t_2\) is
\[ \boxed{\Delta x=x(t_2)-x(t_1)}. \]This is the same final-minus-initial rule, written using a position function.
Example: Displacement from a Nonlinear Position Function
Let
\[ x(t)=t^2-4t+3. \]Find the displacement from \(t=1\) to \(t=5\).
First compute the initial position:
\[ x(1)=1-4+3=0. \]Then compute the final position:
\[ x(5)=25-20+3=8. \]Therefore,
\[ \boxed{\Delta x=8-0=8}. \]This calculation gives displacement. It does not by itself determine the total distance if the object reverses direction between \(t=1\) and \(t=5\).
Why Distance Can Require More Information
Knowing only
\[ x(t_1) \]and
\[ x(t_2) \]determines displacement, but not necessarily distance traveled.
To determine total distance, we need to know whether the object changes direction and where those reversals occur.
Later lessons will connect those direction changes to velocity.
Position in the Coordinate Plane
In two dimensions, position can be written as an ordered pair
\[ (x,y). \]For example,
\[ P=(2,3) \]and
\[ Q=(8,7). \]A moving object may travel from \(P\) to \(Q\) along many possible paths.
Displacement in Two Dimensions
In two dimensions, displacement is naturally represented by a vector.
If the initial position is
\[ (x_i,y_i) \]and the final position is
\[ (x_f,y_f), \]then the displacement vector is
\[ \boxed{ \Delta\mathbf{r} = \langle x_f-x_i,\;y_f-y_i\rangle }. \]Example: Two-Dimensional Displacement
Suppose an object moves from
\[ P=(2,-1) \]to
\[ Q=(7,3). \]Then
\[ \Delta\mathbf{r} = \langle 7-2,\;3-(-1)\rangle \] \[ \boxed{ \Delta\mathbf{r} = \langle 5,4\rangle }. \]The object ends 5 coordinate units in the positive \(x\)-direction and 4 coordinate units in the positive \(y\)-direction relative to where it began.
Magnitude of a Displacement Vector
The magnitude of a two-dimensional displacement vector
\[ \Delta\mathbf{r} = \langle \Delta x,\Delta y\rangle \]is
\[ \boxed{ |\Delta\mathbf{r}| = \sqrt{(\Delta x)^2+(\Delta y)^2} }. \]This magnitude gives the straight-line distance between the initial and final positions.
Example: Magnitude of Displacement
For
\[ \Delta\mathbf{r}=\langle 5,4\rangle, \]the magnitude is
\[ |\Delta\mathbf{r}| = \sqrt{5^2+4^2} \] \[ = \sqrt{41}. \]Therefore,
\[ \boxed{ |\Delta\mathbf{r}|=\sqrt{41} }. \]If the object followed a curved or indirect route, its actual distance traveled would be greater than \(\sqrt{41}\).
Distance Along a Piecewise Path
Suppose an object moves from
\[ A=(0,0) \]to
\[ B=(3,0) \]and then to
\[ C=(3,4). \]The distance traveled is
\[ AB+BC=3+4=7. \]The displacement from \(A\) to \(C\) is
\[ \langle 3,4\rangle \]with magnitude
\[ \sqrt{3^2+4^2}=5. \]Thus,
\[ \boxed{\text{distance}=7,\qquad |\text{displacement}|=5}. \]Straight-Line Motion Between Two Points
If an object moves directly from
\[ A=(x_1,y_1) \]to
\[ B=(x_2,y_2), \]without deviating from the straight segment joining the points, then the distance traveled equals the magnitude of the displacement:
\[ \boxed{ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }. \]Position Vectors
A position in the plane or in space can be represented by a position vector.
In two dimensions,
\[ \mathbf{r} = \langle x,y\rangle. \]In three dimensions,
\[ \mathbf{r} = \langle x,y,z\rangle. \]The vector points from the chosen origin to the object's position.
Vector Form of Displacement
If the initial position vector is
\[ \mathbf{r}_i \]and the final position vector is
\[ \mathbf{r}_f, \]then displacement is
\[ \boxed{ \Delta\mathbf{r} = \mathbf{r}_f-\mathbf{r}_i }. \]This is the vector version of final position minus initial position.
Three-Dimensional Example
Suppose
\[ \mathbf{r}_i=\langle 1,-2,4\rangle \]and
\[ \mathbf{r}_f=\langle 5,3,1\rangle. \]Then
\[ \Delta\mathbf{r} = \langle 5-1,\;3-(-2),\;1-4\rangle \] \[ \boxed{ \Delta\mathbf{r} = \langle 4,5,-3\rangle }. \]Its magnitude is
\[ |\Delta\mathbf{r}| = \sqrt{4^2+5^2+(-3)^2} \] \[ \boxed{ |\Delta\mathbf{r}|=\sqrt{50}=5\sqrt{2} }. \]A Path Can Be Parameterized
A path through the plane can be described by coordinates that depend on a parameter, often time:
\[ x=x(t), \qquad y=y(t). \]Equivalently, we may write a position vector
\[ \mathbf{r}(t) = \langle x(t),y(t)\rangle. \]This representation will become central in later lessons on parametric motion and vector-valued position functions.
The Path Is Not the Same as the Position Function
The path is the geometric route traced in space.
A position function describes where the object is at each time.
Two objects can trace the same path while moving along it at different rates or in opposite directions.
Thus the path describes geometry, while the time-dependent position also describes the timing of the motion.
Example: Same Path, Different Motion
Consider
\[ \mathbf{r}_1(t)=\langle t,t^2\rangle \]and
\[ \mathbf{r}_2(t)=\langle 2t,4t^2\rangle. \]For the first motion,
\[ y=x^2. \]For the second,
\[ x=2t \]so
\[ t=\frac{x}{2}. \]Then
\[ y=4\left(\frac{x}{2}\right)^2=x^2. \]Both motions trace the same parabolic path, even though the parameterization differs.
Curved Paths and Distance
For a curved path, the total distance traveled is the length of the curve actually followed.
That length is generally greater than the straight-line distance between the endpoints.
In later mathematics, curve length can be computed using arc-length methods. In kinematics, the same idea appears through the accumulation of speed over time.
Position-Time Graphs
A graph of position versus time places time on the horizontal axis and position on the vertical axis.
A point such as
\[ (3,8) \]on an \(x\)-versus-\(t\) graph means
\[ x(3)=8. \]In words: at time \(t=3\), the object is at position \(x=8\).
Displacement from a Position-Time Graph
To find displacement over a time interval from a position-time graph, subtract the initial vertical coordinate from the final vertical coordinate.
If
\[ x(2)=5 \]and
\[ x(7)=-1, \]then
\[ \Delta x=x(7)-x(2) \] \[ =-1-5 \] \[ \boxed{\Delta x=-6}. \]A Position-Time Graph Is Not a Picture of the Path
This distinction is essential.
A curved line on a position-time graph does not mean that the object is physically traveling along a curved road.
The horizontal axis represents time, not a second spatial coordinate.
The graph records how position changes with time.
Example: Reading Motion from a Position Table
Suppose an object's recorded positions are:
| Time \(t\) | Position \(x\) |
|---|---|
| 0 | 2 |
| 1 | 6 |
| 2 | 3 |
| 3 | 9 |
The total displacement is
\[ 9-2=7. \]Assuming the object moves directly between each listed position, the distance is
\[ |6-2|+|3-6|+|9-3| \] \[ =4+3+6 \] \[ \boxed{13}. \]Coordinate Units Matter
Position, displacement, and distance carry units of length.
Common units include:
- meters;
- kilometers;
- centimeters;
- feet;
- miles.
If the position is measured in meters, then displacement and distance over the same problem should also be expressed in meters unless a unit conversion is performed.
Changing the Origin Changes Position Coordinates
Suppose an object's coordinate is
\[ x=12 \]in one coordinate system.
If a new origin is placed 5 units to the right of the old origin, then the same physical location has new coordinate
\[ x'=12-5=7. \]The physical object has not moved. Only the coordinate description has changed.
Displacement Survives a Simple Translation of the Origin
Suppose every position coordinate is shifted by the same constant \(c\):
\[ x'=x-c. \]Then
\[ \Delta x' = x_f'-x_i'. \]Substituting,
\[ \Delta x' = (x_f-c)-(x_i-c) \] \[ = x_f-x_i. \]Therefore,
\[ \boxed{\Delta x'=\Delta x}. \]A simple shift of the origin changes individual position coordinates but not the physical displacement.
Direction and Sign Conventions
In one dimension, the sign of displacement depends on the direction chosen as positive.
If east is positive, a 5-meter westward displacement is
\[ -5\text{ m}. \]If west were chosen as positive instead, the same physical displacement would be written
\[ +5\text{ m}. \]The coordinate sign changes, but the physical motion does not.
Common Error: Adding Positions to Find Displacement
Suppose
\[ x_i=3, \qquad x_f=10. \]Displacement is not
\[ 3+10=13. \]It is final minus initial:
\[ \boxed{\Delta x=10-3=7}. \]Common Error: Reversing Final and Initial Position
If
\[ x_i=8, \qquad x_f=2, \]then
\[ \Delta x=2-8=-6. \]Writing \(8-2=6\) gives the magnitude of the change but loses the direction encoded by the sign.
Common Error: Treating Distance as Signed
If an object travels 4 meters in the negative \(x\)-direction, its displacement may be
\[ -4\text{ m}, \]but its distance traveled is
\[ \boxed{4\text{ m}}. \]Direction belongs to displacement. Ordinary path length remains nonnegative.
Common Error: Assuming Distance Is Always \(|x_f-x_i|\)
The expression
\[ |x_f-x_i| \]gives the magnitude of displacement.
It equals total distance only when the path contains no extra motion or reversal that increases the route length.
For motion from \(0\) to \(10\) and back to \(4\),
\[ |x_f-x_i|=|4-0|=4, \]while the distance is
\[ 10+6=16. \]Common Error: Confusing a Path with a Position-Time Graph
A path exists in physical or geometric space.
A position-time graph exists in a graph whose horizontal coordinate is time.
These are different objects even if both are drawn as curves.
A Reliable Procedure for One-Dimensional Motion
- Choose or identify the positive direction.
- Record the initial position \(x_i\).
- Record the final position \(x_f\).
- Compute displacement using \(\Delta x=x_f-x_i\).
- To find distance, identify every segment of the route.
- Take the absolute value of each segment change.
- Add the segment lengths.
- Attach the correct units.
- Check that \(d\ge |\Delta x|\).
A Reliable Procedure for Coordinate Motion
- Write the initial position vector.
- Write the final position vector.
- Subtract component by component.
- Write the displacement vector.
- Find its magnitude if straight-line separation is needed.
- Use the actual path length, not merely the endpoint separation, for total distance traveled.
Example: Full One-Dimensional Workflow
A cart starts at
\[ x=-4\text{ m}, \]moves to
\[ x=6\text{ m}, \]then to
\[ x=1\text{ m}, \]and finally to
\[ x=8\text{ m}. \]The displacement is
\[ \Delta x=8-(-4)=12\text{ m}. \]The distance is
\[ |6-(-4)|+|1-6|+|8-1| \] \[ =10+5+7 \] \[ \boxed{22\text{ m}}. \]Check:
\[ 22\ge |12|. \]The result is consistent.
Example: Full Two-Dimensional Workflow
An object moves from
\[ A=(-2,1) \]to
\[ B=(4,5). \]The displacement vector is
\[ \Delta\mathbf{r} = \langle 4-(-2),5-1\rangle \] \[ = \langle 6,4\rangle. \]Its magnitude is
\[ |\Delta\mathbf{r}| = \sqrt{6^2+4^2} \] \[ = \sqrt{52} \] \[ \boxed{ |\Delta\mathbf{r}|=2\sqrt{13} }. \]If the object traveled directly along the straight segment from \(A\) to \(B\), its distance traveled is also \(2\sqrt{13}\). If it followed an indirect route, its distance is larger.
Practice 1: Basic Displacement
An object moves from \(x=3\text{ m}\) to \(x=14\text{ m}\). Find its displacement.
Practice 2: Negative Displacement
An object moves from \(x=7\text{ m}\) to \(x=-5\text{ m}\). Find its displacement.
Practice 3: Distance with a Reversal
A cart starts at \(x=1\text{ m}\), moves to \(x=9\text{ m}\), and then returns to \(x=4\text{ m}\). Find both displacement and distance.
Practice 4: Round Trip
A runner begins at a marker, runs 250 meters along a straight trail, and returns to the same marker. Find the runner's displacement and distance.
Practice 5: Several Position Changes
An object occupies the successive positions
\[ -4,\quad 2,\quad -1,\quad 5. \]Find total displacement and total distance.
Practice 6: Position Function
Let
\[ x(t)=2t^2-3t+1. \]Find the displacement from \(t=1\) to \(t=4\).
Practice 7: Two-Dimensional Displacement
Find the displacement vector from
\[ A=(1,-3) \]to
\[ B=(6,5). \]Practice 8: Magnitude of Displacement
For the displacement vector
\[ \langle 5,12\rangle, \]find the magnitude.
Practice 9: Piecewise Path
An object moves from
\[ A=(0,0) \]to
\[ B=(6,0) \]and then to
\[ C=(6,8). \]Find the distance traveled and the magnitude of the total displacement.
Practice 10: Identify the Error
A student says:
“An object moved from \(x=2\) to \(x=10\), then back to \(x=6\). Its distance traveled is \(|6-2|=4\).”
Explain the error and find the correct distance.
Practice Answers
Practice 1
\[ \Delta x=14-3 \] \[ \boxed{\Delta x=11\text{ m}}. \]Practice 2
\[ \Delta x=-5-7 \] \[ \boxed{\Delta x=-12\text{ m}}. \]Practice 3
Displacement:
\[ \Delta x=4-1=3\text{ m}. \]Distance:
\[ |9-1|+|4-9| = 8+5 \] \[ \boxed{d=13\text{ m}}. \]Practice 4
The runner returns to the starting point, so
\[ \boxed{\Delta x=0}. \]The total distance is
\[ 250+250 \] \[ \boxed{d=500\text{ m}}. \]Practice 5
Displacement:
\[ 5-(-4)=9. \]Distance:
\[ |2-(-4)|+|-1-2|+|5-(-1)| \] \[ =6+3+6 \] \[ \boxed{d=15}. \]Therefore,
\[ \boxed{\Delta x=9,\qquad d=15}. \]Practice 6
Initial position:
\[ x(1)=2(1)^2-3(1)+1=0. \]Final position:
\[ x(4)=2(16)-12+1=21. \]Therefore,
\[ \boxed{\Delta x=21}. \]Practice 7
\[ \Delta\mathbf{r} = \langle 6-1,\;5-(-3)\rangle \] \[ \boxed{ \Delta\mathbf{r} = \langle 5,8\rangle }. \]Practice 8
\[ |\Delta\mathbf{r}| = \sqrt{5^2+12^2} \] \[ = \sqrt{25+144} \] \[ \boxed{|\Delta\mathbf{r}|=13}. \]Practice 9
Distance:
\[ 6+8=14. \]Total displacement is
\[ \langle 6,8\rangle, \]so its magnitude is
\[ \sqrt{6^2+8^2} = 10. \]Therefore,
\[ \boxed{\text{distance}=14,\qquad |\text{displacement}|=10}. \]Practice 10
The student computed only the magnitude of the total displacement from \(x=2\) to \(x=6\).
The actual path has two segments:
\[ |10-2|=8 \]and
\[ |6-10|=4. \]Therefore,
\[ \boxed{d=8+4=12}. \]What to Remember
Position identifies location relative to a chosen reference system.
Displacement compares endpoints:
\[ \boxed{\Delta x=x_f-x_i}. \]In vector form,
\[ \boxed{ \Delta\mathbf{r} = \mathbf{r}_f-\mathbf{r}_i }. \]Distance measures the total length of the path actually traveled.
The core distinction is:
- position tells where;
- path tells which route;
- displacement tells the net change in position;
- distance tells the total amount of travel.
For any trip,
\[ \boxed{\text{distance}\ge |\text{displacement}|}. \]Where We Go Next
Position describes where an object is and displacement describes how that position changes. The next step is to include time explicitly and measure how quickly motion occurs.
The next lesson, Speed and Velocity, distinguishes the rate at which distance accumulates from the rate at which position changes.