Position, displacement, velocity, and acceleration

Locate one point on a chosen axis

Choose a marked point on a cart and a fixed straight track. Put the origin at a mark on the track and choose right as positive. If the point starts at x = −1 m and later reaches x = +2 m, its coordinate has increased by 3 m. The minus sign says where the starting point lies relative to the origin; it does not say which way the cart is moving.

Position describes location at a stated time. Write x(t) for the point's signed coordinate in metres and t for elapsed time in seconds. Keep the origin and positive direction fixed throughout a calculation, and state when the clock reads zero. This lesson follows one point along one axis; describing an extended object's rotation requires more information.

Use the published Units, standards, and physical constants lesson if quantity values and m/s units need review. Algebra, functions, graphs, and logarithms prepares substitution and graph reading. Derivatives, Taylor series, and local approximation prepares the limiting-rate sections. On a first reading, follow the finite changes and supplied graph/table; then return to the derivative calculations. No Taylor expansion is needed here.

Separate displacement from distance

For initial time t₁ and later time t₂, displacement is the change in position. Its signed component along our axis is Δx = x(t₂) − x(t₁). The symbol Δ means final minus initial. Displacement is a vector; Δx is its signed one-dimensional component. A positive component points along the chosen positive direction, and a negative component points the other way.

Distance travelled is the length of the path followed and is nonnegative. On a monotone straight-line leg, with no reversal, it equals |Δx|. If the point reverses, add the lengths of its separate monotone legs. For example, moving from −2 m to +3 m and then back to 0 m gives displacement +2 m but distance 5 m + 3 m = 8 m.

Both quantities use metres, but they answer different questions. Two endpoints determine displacement. They do not determine distance unless the path between them is also specified. Returning to the starting point gives zero displacement even when the travelled distance is positive.

Compare average velocity and average speed

Let Δt = t₂ − t₁ > 0 be elapsed time. The signed component of average velocity is vₓ,avg = Δx/Δt. Average speed is distance travelled divided by the same elapsed time. Both have m/s units. Average velocity retains a direction through its sign; average speed is nonnegative.

Suppose the two monotone legs from −2 m to +3 m to 0 m take 2 s and 3 s. The whole trip has Δx = +2 m and Δt = 5 s, so average velocity is +0.4 m/s. Its distance is 8 m, so average speed is 1.6 m/s. Always use the total elapsed time, rather than an unweighted average of the two leg rates.

An average covers a finite interval. It does not specify the velocity at every instant in that interval. On a position–time graph, the slope of the straight line joining the two endpoints is average velocity: a change in metres divided by a change in seconds. A graph's time axis is elapsed time, not another spatial direction.

Take an instantaneous rate

At an interior time where x(t) is differentiable, instantaneous velocity is vₓ(t) = dx/dt. It is the limit of [x(t + h) − x(t)]/h as the nonzero time increment h approaches zero with t + h inside the domain. This derivative is the tangent slope of the position–time curve, with m/s units. At a domain endpoint, use the appropriate one-sided derivative when it exists.

Supply the exact teaching model x(t) = 1 m + (4 m/s)t − (1 m/s²)t² for 0 ≤ t ≤ 3 s. Every term has metre units. Its values at 0 s, 1 s, 2 s and 3 s are 1 m, 4 m, 5 m and 4 m. The smooth function is supplied between those times too; the four values alone would not determine it.

Substitution and subtraction give [x(t+h)−x(t)]/h = 4 m/s − (2 m/s²)t − (1 m/s²)h for an allowed nonzero h. Taking the limit yields vₓ(t) = 4 m/s − (2 m/s²)t. The first second has average velocity (4 m − 1 m)/(1 s) = 3 m/s, while the instantaneous velocity at 1 s is 2 m/s. These are different rates of the same motion.

In one dimension, instantaneous speed is |vₓ(t)|. Positive velocity means increasing coordinate and negative velocity means decreasing coordinate. A positive coordinate can therefore accompany negative velocity. At 2.5 s this model has x = 4.75 m and vₓ = −1 m/s: the point lies to the right of the origin while moving left.

Read acceleration as a change of velocity

The average acceleration component is [vₓ(t₂) − vₓ(t₁)]/(t₂ − t₁) for t₂ > t₁. Here the endpoint values are instantaneous velocities. Where vₓ is differentiable, instantaneous acceleration is aₓ(t) = dvₓ/dt; if x is twice differentiable, this is also its second time derivative. Acceleration has m/s² units because a velocity change in m/s is divided by time in seconds.

For the supplied quadratic, differentiating vₓ gives aₓ = −2 m/s² throughout the permitted interval, with one-sided endpoint derivatives. Its velocity decreases by 2 m/s in every additional second. At 2 s the velocity is zero but the acceleration is still −2 m/s². Being stationary for one instant does not mean the point will remain at rest.

A negative acceleration component means that the velocity component is becoming more negative. Before 2 s, positive velocity decreases toward zero, so speed decreases. After 2 s, velocity is negative and its magnitude grows, so speed increases. The signs of both velocity and acceleration matter; negative acceleration alone does not mean slowing down.

Compare three graphs of one motion

Compare all three graphs at the same time. The position curve rises to 5 m at 2 s and then falls; its tangent is horizontal at that maximum. The velocity line reaches zero at the same time and changes from positive to negative. The acceleration line stays at −2 m/s². The dashed vertical guides select 2 s in each panel; they are time comparisons, not a spatial path.

The graphs share a 0–3 s time scale, but their vertical axes have different quantities and units. Position is in metres, velocity in m/s and acceleration in m/s². Their numerical heights must be interpreted through those labels. The table contains every marked value; the curves also show the supplied model between the marks.

From 0 s to 2 s, displacement is 5 m − 1 m = +4 m. From 2 s to 3 s it is 4 m − 5 m = −1 m. In the velocity panel these are the signed areas of two triangles: (2 s)(4 m/s)/2 = +4 m and −(1 s)(2 m/s)/2 = −1 m. Net displacement is +3 m, while distance is 4 m + 1 m = 5 m. An area below the zero-velocity line contributes negatively to displacement; its geometric area is positive.

This shaded-area calculation works directly because the supplied velocity is linear on both legs. The broader derivative–integral relationship is developed in Integrals and multivariable calculus. Here the monotone position legs and triangle arithmetic already establish the displayed displacement and distance. The exact teaching function has not been fitted to measurements.

Check whether a stationary instant is a reversal

A stationary instant has vₓ = 0. To establish a reversal, also check the direction of motion immediately before and after it. The main model reverses at 2 s because its velocity changes from positive to negative there. A zero velocity by itself is insufficient.

For a separate exact model, supply q(t) = 2 m + (1 m/s³)(t − 1 s)³ on 0 ≤ t ≤ 2 s, using the same fixed positive direction. Here q names the coordinate in this separate case on the same fixed axis. Differentiation gives v_q(t) = (3 m/s³)(t − 1 s)² and a_q(t) = (6 m/s³)(t − 1 s).

At 1 s both derivatives are zero. Yet v_q is positive on both sides of 1 s: at 0.5 s and 1.5 s it is 0.75 m/s. The coordinate continues increasing through 1 m, 2 m and 3 m at the three integer times. There is a stationary instant without reversal. A flat tangent need not be a maximum or minimum.

Keep the coordinate choice and model conditions

Moving a fixed origin by a constant length c in metres gives the new coordinate x_new = x − c, so differences and time derivatives stay the same. Reversing the fixed positive direction gives x_new = −x; signed displacement, velocity and acceleration change signs, while speed and travelled distance remain the same. These are fixed coordinate changes. A moving or rotating frame needs further analysis.

Before interpreting a motion statement, check its point, axis, clock, units, time interval and supplied path or function. Two position samples give a finite displacement and average velocity. An instantaneous rate needs an appropriate differentiable function or further justified information; it does not follow from those two samples alone. Keep every calculation inside its model's domain.

These quantities describe the supplied motion. The next lesson, One-dimensional kinematics, develops solutions and checks under a stated constant-acceleration assumption. Later, Newton’s laws and free-body diagrams connect acceleration with interactions. The present graph does not determine which forces could produce it.

Worked examples

Try it yourself

1. On the initial fixed track, a point moves monotonically from −1 m to +2 m in 1 s. Find displacement, distance, average velocity and average speed. Does its negative starting coordinate establish a negative velocity?

   Show the answer and reasoning

   Displacement is +3 m, distance is 3 m, and both average rates are 3 m/s. A negative starting coordinate does not establish negative velocity.

   Final minus initial gives (+2 m) − (−1 m) = +3 m. The specified monotone straight leg has length 3 m. Divide each by 1 s. Position describes location; motion direction requires how that position changes.

2. For the main model, find average velocity from 1 s to 2 s and instantaneous velocity at 2 s. Explain why the two values need not match.

   Show the answer and reasoning

   Average velocity is +1 m/s, while instantaneous velocity at 2 s is 0 m/s.

   The finite change is 5 m − 4 m = +1 m in 1 s. The derivative at the endpoint of that interval gives 4 m/s − (2 m/s²)(2 s) = 0 m/s. One rate covers the whole interval; the other describes the tangent at one instant.

3. At 2.5 s in the main model, find position, velocity, acceleration and speed. Is speed increasing or decreasing just then?

   Show the answer and reasoning

   x = 4.75 m, vₓ = −1 m/s, aₓ = −2 m/s² and speed is 1 m/s. Speed is increasing.

   Substitute 2.5 s into the supplied functions. The point is to the right of the origin but moves left. Its negative velocity is becoming more negative, so its magnitude grows. Negative acceleration does not imply decreasing speed when velocity is also negative.

4. At 1 s in the main model, shift the fixed origin using x_new = x − 10 m. Find the new position and velocity. Separately reverse the original axis using x_reversed = −x: what happens to the 0–3 s displacement, distance and speed at 2.5 s?

   Show the answer and reasoning

   The shifted position is −6 m and velocity stays +2 m/s. Reversing the original axis gives displacement −3 m, distance 5 m and speed 1 m/s at 2.5 s.

   Subtracting the constant 10 m changes x(1 s) = 4 m to −6 m but its derivative is zero, so velocity is unchanged. Reversal changes every signed coordinate component's sign. Absolute lengths and speed magnitudes stay the same. Neither coordinate change moves the point.

5. Only two position samples are supplied: x(0) = 1 m and x(3 s) = 4 m on the same fixed axis. Which of displacement, average velocity, travelled distance and instantaneous acceleration are determined without a further motion rule?

   Show the answer and reasoning

   Displacement +3 m and average velocity +1 m/s are determined. Travelled distance and instantaneous acceleration are not.

   The two samples fix their difference and the 3 s duration. They do not state whether the point moved monotonically, reversed or paused between them, nor do they supply a twice-differentiable position function. The main quadratic is one compatible supplied model, not a consequence of those two samples.

6. Compare the main model's stationary instant at 2 s with the separate cubic's at 1 s. Which reverses direction, and what extra sign check establishes the distinction?

   Show the answer and reasoning

   The main quadratic reverses; the cubic does not. Check the velocity sign on both sides of each stationary instant.

   For the main model, 4 m/s − (2 m/s²)t changes from positive to negative across 2 s. For the cubic, (3 m/s³)(t − 1 s)² is positive on both sides of 1 s. Zero velocity establishes a stationary instant; the surrounding motion decides whether it is a reversal.

What to carry forward

• Position is a location relative to a stated origin and direction; its sign does not establish motion direction.
• Displacement uses endpoint difference, while distance requires the travelled path.
• Average velocity divides signed displacement by elapsed time; average speed divides distance by that time.
• Instantaneous velocity and acceleration are derivatives when their stated limits exist, with m/s and m/s² units.
• Negative acceleration can decrease or increase speed, depending on the velocity sign.
• A stationary instant is a reversal only when the direction of motion changes across it.