Graphs of Uniformly Accelerated Motion: x(t), v(t), a(t)

Set the initial position x₀, the initial velocity v₀ and the acceleration a — the cart starts to move, and three graphs are drawn in sync below it: position, velocity and acceleration. The time cursor ties them together: at every moment you see where the cart is, how fast it goes and which way the acceleration points. The area under the velocity graph turns into the displacement on the x(t) graph right before your eyes, and the slope of the tangent to the x(t) parabola equals the velocity. Choose the “Turning point” scenario to see why distance and displacement are not the same thing. After the run, drag the mouse or a finger over the graphs to return to any moment of the motion.

x— position v— velocity component a— acceleration component area where v > 0 — distance forward area where v < 0 — distance back

The signed area under v(t) equals the displacement Δx (the arrow on the x(t) graph), and the area under a(t) equals the change in velocity Δv. The slope of the white tangent to x(t) equals the velocity. The green and red lanes under the cart are the distance forward and back; their lengths equal the areas of the same colour.

x₀ initial position 0 m

v₀ initial velocity (component) 0.0 m/s

a acceleration (component) +2.0 m/s²

T duration of the run 5.0 s

Brakes: at v = 0 the body stops (a → 0) instead of moving back

Playback speed

Areas under the v(t) and a(t) graphs

Slope: tangent to x(t) and triangle on v(t)

Strobe (marks every Δt)

Compare with uniform motion (v = v₀)

Predicted graphs (dashed)

The motion is computed from the exact formulas x = x₀ + v₀t + at²/2 and v = v₀ + at, so the numbers on screen match the answers to problems and the playback speed does not affect them. Moving a slider changes the conditions and resets the run.

t — time

x — position

v_x — velocity

a_x — acceleration

Δx — displacement

s — distance

Δx / t — average velocity

s / t — average speed

Solve the problem, enter the answer and press “Check” — the simulator will replay this case right away.

The idea that the distance travelled equals the area under the velocity graph is almost 300 years older than Descartes’ coordinate system. In the 1330s the scholars of Merton College, Oxford (William Heytesbury, Richard Swineshead) stated the “mean speed rule”: a body uniformly accelerating from rest covers the same distance as a body moving all that time at the mean speed. Around 1350 the French scholar Nicole Oresme proved the rule with a drawing: he plotted time horizontally and speed vertically and noticed that the distance is the area of the resulting triangle, which equals the area of a rectangle half as tall. It is one of the first graphs in the history of science. In 1638, in his “Two New Sciences”, Galileo repeated Oresme’s reasoning and derived a law from it: when starting from rest, the distances covered in successive equal time intervals are in the ratio 1 : 3 : 5 : 7. Choose “Speeding up from rest”, switch on the strobe — and check Galileo yourself.

KEY FORMULAS

Position equation

$$x(t) = x_0 + v_{0x}\,t + \frac{a_x t^2}{2}$$

x₀ is the initial position, v₀ₓ the initial velocity component, aₓ the (constant) acceleration component, t the time since the start. The x(t) graph is a parabola

▶ Detailed Explanation

The formula gives the position of the body at any moment if three numbers are known: where the body was at the start (x₀), how fast it started (v₀ₓ) and its acceleration (aₓ). All three carry a sign: a component is positive if the vector points along the x axis and negative if it points against it.

Example (the “Turning point” scenario): x₀ = −4 m, v₀ = 6 m/s, a = −2 m/s². After 3 s: x = −4 + 6·3 + (−2)·3²/2 = −4 + 18 − 9 = 5 m. After 7 s: x = −4 + 42 − 49 = −11 m — the cart has gone past the starting point in the opposite direction. With a = 0 the formula becomes the uniform-motion equation x = x₀ + vt and the graph becomes a straight line.

Velocity and acceleration

$$v_x(t) = v_{0x} + a_x t, \qquad a_x = \frac{v_x - v_{0x}}{t} = \mathrm{const}$$

The velocity changes linearly: the v(t) graph is a straight line whose slope equals the acceleration aₓ. The a(t) graph is a horizontal line

▶ Detailed Explanation

Acceleration shows how much the velocity changes every second. With a = 2 m/s² the velocity grows by 2 m/s each second, so the v(t) graph is a line with slope 2 (m/s)/s. In the simulator the yellow triangle shows it: over 1 s horizontally the graph rises or falls by exactly a.

If v and a have the same sign, the speed grows (speeding up); if their signs are opposite, it drops (slowing down). When the velocity reaches zero during braking but the acceleration does not vanish, the velocity component changes sign — the body turns back. The v(t) graph simply crosses the time axis, and the a(t) graph does not change.

Displacement is the area under the velocity graph

$$\Delta x = x - x_0 = \frac{v_{0x} + v_x}{2}\,t, \qquad \Delta x = S_+ - S_-, \qquad s = S_+ + S_-$$

S₊ is the area under the v(t) graph where v > 0, S₋ where v < 0. Displacement keeps the sign, the distance s does not

▶ Detailed Explanation

In uniform motion the displacement vt is the area of the rectangle under the velocity graph. The same holds for any motion: the area under v(t) equals the displacement. For uniformly accelerated motion the figure is a trapezium with bases v₀ and v and height t, hence Δx = (v₀ + v)/2 · t. So with constant a the average velocity is the mean of the initial and final velocities.

If the body turns back, part of the graph goes below the time axis. That area is displacement backwards and is taken with a minus sign. Example: v₀ = 6 m/s, a = −2 m/s², t = 7 s. Before the turn (3 s) S₊ = 6·3/2 = 9 m; after it (4 s, velocity from 0 to −8 m/s) S₋ = 8·4/2 = 16 m. The displacement is Δx = 9 − 16 = −7 m, the distance s = 25 m. Check with the formula: Δx = (6 + (−8))/2 · 7 = −7 m.

Time-independent equation

$$v_x^2 - v_{0x}^2 = 2a_x\,\Delta x$$

Links the velocities and the displacement when the time is unknown or not needed. Braking distance to a stop: s = v₀²/(2|a|)

▶ Detailed Explanation

The formula follows from v = v₀ + at and Δx = (v₀ + v)/2 · t by eliminating the time: t = (v − v₀)/a, so Δx = (v + v₀)(v − v₀)/(2a) = (v² − v₀²)/(2a).

Its main use is the braking distance. A car moving at 10 m/s brakes at 2.5 m/s²: before stopping (v = 0) it covers s = 10²/(2·2.5) = 20 m. You can see this in the “Car braking” scenario. Note that the braking distance grows as the square of the speed — at twice the speed it is four times longer. That is why speeding is so dangerous. In the “Measurements” panel the simulator checks this formula at every moment of the motion.

Turning moment and the vertex of the parabola

$$t^{*} = -\frac{v_{0x}}{a_x}, \qquad x^{*} = x_0 - \frac{v_{0x}^2}{2a_x}$$

If v₀ and a point opposite ways, at the moment t* the velocity is zero and the position x* is the largest or smallest. The tangent to x(t) there is horizontal

▶ Detailed Explanation

The turning moment follows from the condition v = 0: v₀ + at* = 0, so t* = −v₀/a. Substituting t* into the position equation gives the vertex of the parabola x* = x₀ − v₀²/(2a). For the “Turning point” scenario: t* = −6/(−2) = 3 s, x* = −4 − 36/(2·(−2)) = −4 + 9 = 5 m.

The most common mistake is to think that at the turning point the body “stands still” and the acceleration is zero. No: the velocity is zero for just an instant, while the acceleration stays the same all the time, so the body immediately starts moving back. A ball thrown upwards behaves this way: at the top v = 0 but a = g. A car behaves differently: after it stops, the brakes hold it with static friction, the acceleration becomes zero and the car does not roll back — tick “Brakes” to see the difference on the a(t) graph.

Galileo’s odd-number rule

$$s_1 : s_2 : s_3 : s_4 : \ldots = 1 : 3 : 5 : 7 : \ldots \;(v_0 = 0), \qquad \Delta x_{n+1} - \Delta x_n = a\,\Delta t^2$$

sₙ is the distance covered in the n-th equal time interval Δt when starting from rest. For any uniformly accelerated motion the displacements in neighbouring intervals differ by aΔt²

▶ Detailed Explanation

Starting from rest, the distance covered in time t is at²/2, i.e. it grows as the square of the time: after 1, 2, 3, 4 intervals it is 1, 4, 9, 16 “units”. So each next interval adds 1, 3, 5, 7 units — the odd numbers. Galileo discovered this rule by rolling balls down an inclined groove and regarded it as the key sign of uniformly accelerated motion.

A more general rule holds for any initial velocity: displacements in neighbouring equal intervals Δt differ by exactly aΔt². In the “Turning point” scenario (a = −2 m/s², Δt = 1 s) the strobe gives displacements of +5, +3, +1, −1, −3 m: each is 2 m less than the previous one. This is how acceleration is measured from strobe photographs in the lab: a = (Δx₂ − Δx₁)/Δt².

What uniformly accelerated motion is. Motion is uniformly accelerated when the velocity changes by the same amount in any equal time intervals. The acceleration a = (v − v₀)/t is then constant. A ball thrown up or falling without air resistance, a car accelerating smoothly, a puck sliding on ice until it stops — all move this way. In problems every quantity is taken as a component along the x axis: if a vector points against the axis, its component is negative. That is why vₓ and aₓ in the simulator can carry a minus sign.

Three graphs — one motion. The acceleration graph a(t) is a horizontal line. The velocity graph v(t) = v₀ + at is a sloped line whose slope equals the acceleration: every second the velocity changes by a (the yellow triangle on the graph). The position graph x(t) = x₀ + v₀t + at²/2 is a parabola opening upwards when a > 0 and downwards when a < 0. The slope of the tangent to the parabola at any point equals the velocity at that moment (the white line). Where v = 0 the tangent is horizontal — this is the vertex of the parabola and the turning moment t* = −v₀/a.

Area under the velocity graph is displacement. The displacement Δx = x − x₀ equals the area of the figure under the v(t) graph. For uniformly accelerated motion it is a trapezium: Δx = (v₀ + v)/2 · t. Area above the time axis (v > 0, green) counts as plus, area below it (v < 0, red) counts as minus. The distance travelled is the sum of the areas without signs. In the “Turning point” scenario (x₀ = −4 m, v₀ = 6 m/s, a = −2 m/s²) the cart moves forward for 3 s and covers S₊ = 9 m, then moves back for 4 s and covers S₋ = 16 m. The displacement is Δx = 9 − 16 = −7 m, while the distance is s = 9 + 16 = 25 m. In the same way the area under the a(t) graph equals the change in velocity Δv.

Try it yourself. 1) “Speeding up from rest” (a = 2 m/s²): the strobe marks the cart every second, and the segments are 1, 3, 5, 7, 9 m — the odd-number rule. 2) “Car braking”: with v₀ = 10 m/s and a = −2.5 m/s² the car stops after 4 s, having covered v₀²/(2|a|) = 20 m, and then stays put — the brakes hold it with static friction. Untick “Brakes” and the cart moves back: that is how a stone thrown upwards moves. 3) Turn on “Compare with uniform motion”: the violet cart moves at the constant velocity v₀, and the gap between the two grows exactly as at²/2.

Common mistakes. Thinking that the acceleration is also zero at the turning point: in fact v = 0 but a keeps its value, so the body immediately starts moving back. Confusing distance and displacement when the body turns back. Taking the x(t) graph for the path: the cart moves along a straight line, and the parabola is the dependence of position on time. Try the problems in the “Test yourself” tab — after each answer the simulator replays that case.

Graphs of Uniformly Accelerated Motion: x(t), v(t), a(t)

KEY FORMULAS

Position equation

Velocity and acceleration

Displacement is the area under the velocity graph

Time-independent equation

Turning moment and the vertex of the parabola

Galileo’s odd-number rule

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