One-dimensional kinematics

A cart returns, but it has traveled

Picture a marked point on a cart moving along a straight track. Choose a fixed coordinate x that increases to the right. At the clock’s start, the point is at 2 m and moves left at 3 m/s. Throughout the next 3 s, assign it a constant acceleration of 2 m/s² toward the right. These are exact teaching assumptions, rather than reported observations of a cart.

The point first slows, turns, and returns to 2 m at 3 s. Its final coordinate equals its initial coordinate, yet it has moved along the track. We will calculate the turn and the distance, then use two time graphs to explain this difference. Kinematics describes that motion; identifying the forces that produce it is a separate question.

Useful preparation is substitution, signed arithmetic and keeping units in products and ratios. Units, standards, and physical constants and Algebra, functions, graphs, and logarithms provide that published preparation. The motion quantities and graph reasoning needed here are introduced locally. Start with the cart and its table; calculus is an optional return path for the rate and area arguments.

Declare the coordinate, clock and rates

Track the same marked point throughout the motion and name the fixed frame and axis. A negative coordinate means a position on the negative side of the chosen origin; a negative velocity means motion toward decreasing coordinates. Those are different statements. Our cart can have a positive position while moving left. The model describes this one coordinate, without describing the cart’s size, rotation or forces.

Let elapsed time be t, with t = 0 at the stated start. Initial position x₀ and initial velocity v₀ belong to that same event. For an interval t₁ < t₂, displacement is Δx = x(t₂) − x(t₁), and average velocity is Δx/(t₂ − t₁). The symbol Δ means a difference. Velocity has units m/s; instantaneous speed is the magnitude |v| and cannot be negative.

Instantaneous velocity is the local position-change rate: the limit of a position difference divided by its elapsed-time difference, when that limit exists. Acceleration is the local velocity-change rate, with units m/s². On a position–time graph, velocity is the local slope; on a velocity–time graph, acceleration is the local slope. A graph’s vertical height gives the quantity on its vertical axis, rather than that quantity’s rate. The smooth models here have these rates inside their domains and continuous one-sided rate values at the displayed endpoints.

Start with a constant-velocity check

For constant signed velocity u, equal elapsed times give equal displacements: x(t) = x₀ + ut. The product ut has units (m/s)s = m, so it can be added to a position. The velocity remains u and acceleration is zero. A constant speed alone would not specify a constant signed velocity if the direction changed.

As a separate example, let x₀ = −1 m and u = +0.5 m/s for 0 ≤ t ≤ 4 s. At 4 s, x = −1 m + (0.5 m/s)(4 s) = 1 m. Displacement is 2 m. Because this point moves only rightward on the interval, its distance is also 2 m, and its average velocity and average speed are both 0.5 m/s.

The position graph is a straight line of slope +0.5 m/s, and the velocity graph is a horizontal line at +0.5 m/s. This simple case checks the meanings of slope, height and units before we add acceleration. The equation is used only within its stated 0–4 s domain.

Use the constant-acceleration premise

Constant acceleration a means that the signed velocity changes by aΔt in every elapsed interval Δt. Starting at t = 0, this gives v(t) = v₀ + at. Velocity is therefore a straight-line function of time. Its mean over an interval is the arithmetic mean of its endpoint values: earlier and later values at equal distances from the midpoint pair to the same sum. This endpoint-mean rule needs the affine velocity supplied by constant acceleration.

Displacement equals mean velocity multiplied by duration. For t > 0, from 0 to t, Δx = [(v₀ + v(t))/2]t = v₀t + (1/2)at². Thus x(t) = x₀ + v₀t + (1/2)at²; at t = 0 this position formula gives x₀. The factor 1/2 is dimensionless; at² has units m. Constant acceleration is required throughout the interval, rather than only at the start and finish. If the acceleration varies, a different motion calculation is needed.

For our cart, x₀ = 2 m, v₀ = −3 m/s and a = +2 m/s² on 0 ≤ t ≤ 3 s. Substitution gives x(t) = 2 m − (3 m/s)t + (1 m/s²)t² and v(t) = −3 m/s + (2 m/s²)t. Positive acceleration increases the signed velocity. While v is negative, this makes its magnitude smaller; after v becomes positive, the magnitude grows. Positive acceleration does not always mean increasing speed.

Read position and velocity against the same clock

Predict which part of the position curve slopes downward before looking at the paired graphs. Both horizontal axes show the same elapsed time from 0 to 3 s. The upper vertical axis is position in metres; the lower is velocity in metres per second. Their heights and vertical scales have different meanings. The dashed upper guide joins the equal start and end positions; it is a coordinate reference, not the cart’s motion.

The position curve reaches its minimum when its slope is zero at 1.5 s. At the same time, the velocity line crosses zero. Open circles mark the start, diamonds the turn and squares the finish. Before the turn, velocity is negative; after it, velocity is positive. The cart turns because this model’s velocity changes sign, not merely because its velocity equals zero for an instant.

Signed area between the velocity line and the zero-velocity axis gives displacement. Here each part is a triangle, so ordinary base-times-height arithmetic is enough: −(1/2)(1.5 s)(3 m/s) = −2.25 m before the turn, and +(1/2)(1.5 s)(3 m/s) = +2.25 m after it. The contributions cancel. Their magnitudes add to the distance because each part of this motion has one direction. The triangle’s units are metres; it is not a physical area on the track.

Find a turn before adding distances

Solve v(t) = 0: −3 m/s + (2 m/s²)t = 0 gives t = 1.5 s. This time lies inside 0–3 s. Substituting into x(t) gives −0.25 m. Check the signs on either side: the point moves left before this time and right afterward. A zero found outside the supplied interval would not establish a turn inside it.

The first leg runs from 2 m to −0.25 m, a distance of 2.25 m. The second runs from −0.25 m back to 2 m, another 2.25 m. Total distance is 4.5 m; total displacement is 2 m − 2 m = 0 m. Over 3 s, average speed is 4.5 m/3 s = 1.5 m/s, while average velocity is 0 m/3 s = 0 m/s.

Eliminating time from the constant-acceleration equations also gives v² = v₀² + 2a(x − x₀). This can be useful when time is unknown, but the square does not retain the velocity’s sign. In our model, x = 0 m occurs at both 1 s and 2 s, with velocities −1 m/s and +1 m/s. Use time, direction and the domain to choose a signed velocity; taking a positive square root by habit can select the wrong leg.

Separate a model fit from its physical premise

Suppose we instead know three exact position–time pairs: (0 s, 2 m), (1 s, 0 m) and (3 s, 2 m). Within the chosen constant-acceleration family, write x(t) = 2 m + ut + (1/2)at², with unknown initial velocity u and constant acceleration a. The later two pairs supply two independent equations for these two unknowns.

At 1 s, u(1 s) + (1/2)a(1 s)² = −2 m. At 3 s, u(3 s) + (1/2)a(3 s)² = 0 m. Subtract three times the first equation from the second: a(3 s²) = 6 m, so a = 2 m/s². Substitution gives u = −3 m/s. These coefficients are uniquely determined within this family by these exact pairs.

The positions do not prove that acceleration really remained constant. For a concrete counterexample on the same 0–3 s domain, add (2 m/(9 s³))t(t − 1 s)(t − 3 s) to the cart’s position function. This extra term is zero at all three sampled times, but at 1.5 s it is −0.25 m, giving an alternative position of −0.5 m. The added cubic term has a changing acceleration. Matching a finite list of positions therefore does not identify an unrestricted trajectory or establish the physical assumptions behind the fitted family.

Check the event, premise and valid interval

List the known quantities, their signs and units, the requested event, and the valid time interval before choosing an equation. Keep the time origin consistent: if a clock reads T and motion starts when it reads 10 s, elapsed time is t = T − 10 s. Inserting T into a formula initialized at t = 0 would describe a different initial event.

The endpoint arithmetic mean of velocity belongs to constant acceleration, not to every motion. For example, on 0 ≤ t ≤ 1 s, the supplied pair y(t) = (1 m/s³)t³ and vᵧ(t) = (3 m/s³)t² has displacement 1 m and average velocity 1 m/s. Its endpoint velocities are 0 m/s and 3 m/s, whose arithmetic mean is 1.5 m/s. The velocity is curved in time, so that endpoint shortcut fails.

For measured motion, assess how the observations, uncertainty and untested times bear on the assumed family. A few matching values neither establish constant acceleration nor identify the forces. Derivatives, Taylor series, and local approximation and Integrals and multivariable calculus offer optional published return paths for general local-rate and signed-area calculations. The next map subject, Two- and three-dimensional motion, extends the coordinate description beyond a single axis.

Worked examples

Try it yourself

1. At 0.5 s in the main cart model, find position, velocity, speed and acceleration. Is the cart speeding up at that instant?

   Show the answer and reasoning

   x = 0.75 m, v = −2 m/s, speed = 2 m/s and a = +2 m/s²; it is slowing down.

   Substitute 0.5 s into the supplied functions. Velocity is negative while acceleration is positive, so the magnitude of velocity decreases locally before the turn. A positive position and positive acceleration do not imply rightward motion or increasing speed.

2. At which times in 0–3 s is the main cart at x = 0 m? What is its velocity at each?

   Show the answer and reasoning

   At 1 s, velocity is −1 m/s; at 2 s, velocity is +1 m/s.

   Let q = t/(1 s), the dimensionless number of seconds. At x = 0 m, 2 − 3q + q² = (q − 1)(q − 2) = 0. Thus q = 1 or 2, corresponding to 1 s and 2 s, both inside the domain. Substitution into v(t) distinguishes the two directions; the velocity-squared equation alone gives the same squared value at either event.

3. The main cart’s start is relabeled as clock reading T = 10 s, without changing its physical initial state. Find its position and velocity when that clock reads 11 s.

   Show the answer and reasoning

   Elapsed time is 1 s, so position is 0 m and velocity is −1 m/s.

   Use t = T − 10 s = 1 s in the formulas initialized at t = 0. The same motion now occupies clock readings 10–13 s. Using 11 s directly as elapsed time would use the wrong start and lie outside the supplied 0–3 s elapsed-time domain.

4. Keep the origin and clock fixed but reverse the coordinate axis, defining z = −x for the main cart. What are its initial position, initial velocity, acceleration, total displacement and distance?

   Show the answer and reasoning

   z₀ = −2 m, initial velocity +3 m/s, acceleration −2 m/s², total displacement 0 m and distance 4.5 m.

   A fixed axis reversal changes the signs of the position and both time rates. Signed displacement changes sign too, so zero remains zero. Physical leg lengths and elapsed times remain unchanged; distance is still 4.5 m.

5. For y(t) = (1 m/s³)t³ and vᵧ(t) = (3 m/s³)t² on 0–1 s, compare average velocity from displacement with the arithmetic mean of endpoint velocities. Can the constant-acceleration shortcut be used?

   Show the answer and reasoning

   Average velocity is 1 m/s, while the endpoint arithmetic mean is 1.5 m/s. The shortcut cannot be used.

   The point moves from 0 m to 1 m in 1 s. Its endpoint velocities are 0 m/s and 3 m/s, but the velocity function is quadratic, with changing acceleration. The constant-acceleration premise needed for the endpoint-mean rule is absent.

6. Within the constant-acceleration family, a separate exact record gives x = 1 m at 0 s, 2 m at 1 s and 5 m at 2 s. Find its initial velocity and acceleration. Does the finite record alone prove the family?

   Show the answer and reasoning

   Initial velocity is 0 m/s and acceleration is 2 m/s² within the chosen family; the record alone does not prove the family.

   With x₀ = 1 m, the later pairs give u(1 s) + a(0.5 s²) = 1 m and u(2 s) + a(2 s²) = 4 m. Subtract twice the first equation from the second to obtain a(1 s²) = 2 m. Back-substitution gives u = 0 m/s. As in the worked counterexample, other smooth position functions can match a finite set of samples.

What to carry forward

• A signed one-axis description needs a consistent fixed coordinate, tracked point, time origin and domain.
• Constant velocity gives a linear position function; constant acceleration throughout an interval gives an affine velocity and a quadratic position function.
• Graph height, local slope and signed velocity–time area represent different quantities with different units.
• A reversal requires a velocity sign change; split a path there before adding leg lengths.
• Displacement and average velocity can vanish even when distance and average speed are positive.
• Exact samples can identify coefficients inside a declared family without establishing that family’s physical validity.