Graphs of Motion — padho.wiki

In short

A position–time graph encodes velocity as its slope and acceleration as its curvature. A velocity–time graph encodes acceleration as its slope and displacement as the area under the curve. An acceleration–time graph encodes the change in velocity as its area. Given any one graph, you can construct the other two — slope takes you from x–t to v–t to a–t, and area takes you back.

A Delhi Metro train pulls out of Rajiv Chowk station. For the first 20 seconds it accelerates — you feel yourself pressed back into the seat. Then the ride smooths out: constant speed for a couple of minutes. As the train approaches Patel Chowk, the brakes engage and you lurch forward slightly until the train stops.

You lived that journey in your body. But a physicist watching from the platform cannot feel the forces. All they have is data — position readings at each second, recorded by sensors along the track. From those numbers, they need to reconstruct everything you felt: when you accelerated, how fast you were going, when the brakes kicked in.

The tool that makes this possible is a graph. Not one graph — three. The position–time graph, the velocity–time graph, and the acceleration–time graph are three different windows into the same motion. Each one reveals something the others hide. And the connections between them — slope and area — are the most powerful reading tools in all of kinematics.

The position–time graph — where are you at each instant?

The simplest motion graph plots position x on the vertical axis against time t on the horizontal axis. Every point on the curve answers one question: "where was the object at this moment?"

But the shape of the curve tells you far more than position. It tells you how the object was moving.

Slope is velocity

Pick any two points on the x–t curve. The slope of the line joining them is:

\text{slope} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}

Why: this is the definition of average velocity — displacement divided by time. The slope of the x–t graph between two instants is the average velocity over that interval.

As you shrink the interval to zero, the slope of the secant becomes the slope of the tangent — and that gives you the instantaneous velocity:

v = \frac{\mathrm{d}x}{\mathrm{d}t} = \text{slope of the } x\text{–}t \text{ graph at that instant}

Why: the derivative \mathrm{d}x/\mathrm{d}t is the limit of \Delta x/\Delta t as \Delta t \to 0. Geometrically, that limit is the slope of the tangent line.

This single rule — slope of x–t = velocity — is the master key to reading position–time graphs. A steep upward slope means fast motion in the positive direction. A gentle slope means slow motion. A horizontal line means the object is not moving. A downward slope means motion in the negative direction.

Curvature tells you about acceleration

If the x–t curve is a straight line, the slope (velocity) is constant — there is no acceleration. If the curve bends, the slope is changing, which means the velocity is changing, which means there is acceleration.

A curve that bends upward (concave up) has an increasing slope — the velocity is growing — so the acceleration is positive.
A curve that bends downward (concave down) has a decreasing slope — the velocity is shrinking — so the acceleration is negative.

Why: acceleration is the rate of change of velocity. If the slope (velocity) is increasing, acceleration is positive. If the slope is decreasing, acceleration is negative. The curvature of the x–t graph is a visual fingerprint of acceleration.

The six canonical shapes

Every motion you will encounter in kinematics produces one of six basic shapes on the x–t graph. Recognising them on sight is like recognising letters — once you know the alphabet, you can read any word.

1. At rest

The object does not move. Position is constant at x = 5 m (or whatever value).

At rest. A horizontal line on the $x$–$t$ graph. Slope = 0, so velocity = 0. The object sits at $x = 5$ m and does not move.

2. Constant velocity (positive)

The object moves at a steady rate in the positive direction. x = 2 + 3t — position increases by 3 m every second.

Constant velocity. A straight line with positive slope. The slope is $3$ m/s everywhere — the velocity is constant at $+3$ m/s. No curvature means no acceleration.

3. Constant positive acceleration (starting from rest)

The object starts from rest and speeds up uniformly. x = \tfrac{1}{2}(2)t^2 = t^2 — a parabola opening upward.

Constant acceleration from rest. A parabola (concave up). The slope starts at zero (the object is initially at rest) and increases steadily — the object speeds up at a constant rate. The acceleration here is $2$ m/s$^2$.

4. Constant deceleration (slowing down)

The object starts fast and slows to a stop. x = 10t - t^2 — a parabola opening downward.

Constant deceleration. A parabola (concave down). The slope starts large and positive, decreases, and reaches zero at $t = 5$ s — the object slows to a stop. The acceleration is $-2$ m/s$^2$.

5. Increasing acceleration

The object speeds up, and the rate of speeding up itself increases. x = \tfrac{1}{6}t^3 — the curve bends upward more and more steeply.

Increasing acceleration. The curve gets steeper faster than a parabola. The slope (velocity) is growing, and the rate of growth (acceleration) is itself increasing.

6. Decreasing acceleration

The object speeds up, but the rate of speeding up diminishes. x = 10(1 - e^{-t/3}) — the curve flattens out, approaching a horizontal asymptote.

Decreasing acceleration. The curve rises quickly at first, then flattens. The slope (velocity) is still positive, but it increases more and more slowly. Think of an autorickshaw reaching its top speed — it accelerates hard at first, then the engine can barely push it any faster.

The velocity–time graph — how fast, and in which direction?

The v–t graph plots velocity on the vertical axis against time on the horizontal. It is the most information-rich of the three graphs, because it encodes both acceleration (as slope) and displacement (as area).

Slope is acceleration

The same argument that connected the x–t slope to velocity now connects the v–t slope to acceleration:

a = \frac{\mathrm{d}v}{\mathrm{d}t} = \text{slope of the } v\text{–}t \text{ graph}

Why: acceleration is the rate of change of velocity. On the v–t graph, \Delta v / \Delta t is literally the rise-over-run of the curve. In the limit, the slope of the tangent gives the instantaneous acceleration.

A straight line with a positive slope means constant positive acceleration.
A horizontal line means zero acceleration (constant velocity).
A straight line with a negative slope means constant deceleration.

Area under the curve is displacement

This is the powerful half of the v–t graph. The area enclosed between the velocity curve and the time axis, from t_1 to t_2, equals the displacement over that interval:

\Delta x = \int_{t_1}^{t_2} v \, \mathrm{d}t = \text{area under the } v\text{–}t \text{ curve}

Why: displacement is velocity times time. For a tiny interval \mathrm{d}t, the displacement is v \, \mathrm{d}t, which is the area of a thin vertical strip under the curve. Adding up all such strips from t_1 to t_2 gives the total displacement — that sum is the integral, which is the total area.

Area above the time axis counts as positive displacement. Area below the time axis (when velocity is negative) counts as negative displacement — the object is moving backward.

The acceleration–time graph — how is the push changing?

The a–t graph is the simplest of the three. For constant-acceleration problems (which dominate introductory kinematics), it is just a horizontal line. But its key property still matters:

\Delta v = \int_{t_1}^{t_2} a \, \mathrm{d}t = \text{area under the } a\text{–}t \text{ curve}

Why: the same logic as above. The change in velocity over a tiny interval is a \, \mathrm{d}t. Adding up all such changes gives the total change in velocity.

Feature	x–t graph	v–t graph	a–t graph
What the height tells you	Position at that instant	Velocity at that instant	Acceleration at that instant
What the slope tells you	Velocity	Acceleration	Rate of change of acceleration (jerk)
What the area tells you	(not commonly used)	Displacement	Change in velocity

The three graphs for one journey — the Delhi Metro example

Take the Delhi Metro ride from the opening. The train accelerates uniformly from rest for 20 s, then cruises at constant speed for 60 s, then decelerates uniformly to a stop over 20 s. Suppose the acceleration is 1 m/s^2 during the first phase and -1 m/s^2 during the last.

The v–t graph

During Phase 1 (0 to 20 s): velocity increases linearly from 0 to 20 m/s.

During Phase 2 (20 to 80 s): velocity stays constant at 20 m/s.

During Phase 3 (80 to 100 s): velocity decreases linearly from 20 m/s to 0.

Velocity–time graph for the Delhi Metro. Phase 1 (accelerate): a rising line. Phase 2 (cruise): a flat line at 20 m/s. Phase 3 (brake): a falling line. The total area under this trapezoid gives the displacement for the entire journey.

The displacement is the area of this trapezoid. Phase 1 is a triangle: \tfrac{1}{2} \times 20 \times 20 = 200 m. Phase 2 is a rectangle: 60 \times 20 = 1200 m. Phase 3 is a triangle: \tfrac{1}{2} \times 20 \times 20 = 200 m. Total: 1600 m.

Why calculate area as triangles and rectangles? Because the v–t graph here is made of straight segments — the area under each segment is a basic geometric shape. For curved v–t graphs, you would need calculus (integration), but for constant-acceleration phases, geometry is enough.

The a–t graph

The acceleration is constant within each phase, so the a–t graph is a step function.

Acceleration–time graph for the Delhi Metro. Phase 1: constant $+1$ m/s$^2$. Phase 2: zero. Phase 3: constant $-1$ m/s$^2$. The area from $t = 0$ to $t = 20$ is $20$ m/s — exactly the velocity gained. The area from $t = 80$ to $t = 100$ is $-20$ m/s — the velocity lost.

The x–t graph

Now build the position–time graph. During Phase 1 (a = 1 m/s^2, starting from rest): x = \tfrac{1}{2}(1)t^2 = 0.5t^2. At t = 20: x = 200 m.

During Phase 2 (constant velocity 20 m/s, starting at x = 200): x = 200 + 20(t - 20). At t = 80: x = 1400 m.

During Phase 3 (a = -1 m/s^2, starting at 20 m/s from x = 1400): x = 1400 + 20(t - 80) - 0.5(t - 80)^2. At t = 100: x = 1600 m.

Position–time graph for the Delhi Metro. Phase 1: a concave-up parabola (accelerating). Phase 2: a straight line (constant velocity — constant slope). Phase 3: a concave-down parabola (decelerating). The three shapes match the three phases you feel during the ride.

All three graphs describe the same journey. The v–t graph is the most natural one to start with, because you can read both acceleration (slope) and displacement (area) directly from it. The x–t graph shows you the trajectory. The a–t graph shows you the forces.

Converting between graphs — slope down, area up

The three graphs are connected by two operations:

Slope (differentiation) takes you down: x–t \xrightarrow{\text{slope}} v–t \xrightarrow{\text{slope}} a–t.
Area (integration) takes you back up: a–t \xrightarrow{\text{area}} v–t \xrightarrow{\text{area}} x–t.

Why slope and area? Because v = \mathrm{d}x/\mathrm{d}t (velocity is the derivative of position) and x = \int v \, \mathrm{d}t (position is the integral of velocity). Differentiation and integration are inverse operations — and graphically, the derivative is the slope and the integral is the area.

To convert from an x–t graph to a v–t graph: at each instant, measure the slope of the x–t curve. That slope becomes the height on the v–t graph.

Where the x–t slope is constant (straight line), the v–t graph is flat (horizontal line).
Where the x–t slope increases (curve bends upward), the v–t graph rises.
Where the x–t slope decreases (curve bends downward), the v–t graph falls.

To convert from a v–t graph to an x–t graph: compute the running area under the v–t curve. The accumulated area at each instant becomes the height on the x–t graph (plus the initial position).

Interactive: explore the Delhi Metro journey across all three graphs

Drag the time slider below to see how the position, velocity, and acceleration are connected at every instant of the Metro ride. The vertical dashed line marks the current time on all three graphs simultaneously.

Drag the red point to sweep through the Delhi Metro's journey. Watch how the three graphs are synchronized: when $a$ is positive (bottom graph), $v$ increases (middle graph) and the $x$–$t$ curve bends upward (top graph). When $a = 0$, the velocity is constant and the position grows linearly. When $a$ is negative, $v$ decreases and the $x$–$t$ curve bends downward.

Common confusions

"A rising x–t graph means the object is accelerating." Not necessarily. A straight rising line means constant velocity — no acceleration at all. Only a curved rising line means acceleration. The curvature, not the rise, tells you about acceleration.
"If the v–t graph is below the time axis, the object has moved backward to a negative position." No — a negative velocity means the object is moving in the negative direction at that instant. Its position depends on where it started and the total area (including positive portions). The object could be at a positive position while moving in the negative direction.
"Area under the curve means area of the geometric shape I see on screen." Be careful with signs. Area above the time axis is positive displacement; area below is negative. If the v–t curve dips below the axis, you must subtract that region. The total displacement is the net area (positive minus negative), not the total shaded region.
"The a–t graph for constant acceleration is a straight line." No — it is a horizontal line (a special case of a straight line, but an important distinction). A sloped line on the a–t graph means the acceleration itself is changing linearly with time.
"You cannot find position from an acceleration–time graph." You can — it just takes two steps. First, find the velocity by computing the area under the a–t curve (you need the initial velocity as a starting point). Then find the position by computing the area under the resulting v–t curve (you need the initial position).

Worked examples

Example 1: Reading a three-phase $x$–$t$ graph — autorickshaw in traffic

An autorickshaw's position–time graph has three phases:

Phase 1 (0 to 5 s): x = t^2 (accelerating from rest)
Phase 2 (5 to 15 s): x = 25 + 10(t - 5) (constant velocity)
Phase 3 (15 to 20 s): x = 125 + 10(t - 15) - (t - 15)^2 (decelerating to a stop)

Sketch the v–t and a–t graphs.

The autorickshaw's $x$–$t$ graph: a concave-up parabola (Phase 1), then a straight line (Phase 2), then a concave-down parabola (Phase 3).

Step 1. Find velocity in each phase by differentiating.

Phase 1: v = \mathrm{d}x/\mathrm{d}t = 2t. At t = 0: v = 0. At t = 5: v = 10 m/s.

Phase 2: v = \mathrm{d}x/\mathrm{d}t = 10 m/s (constant).

Phase 3: v = \mathrm{d}x/\mathrm{d}t = 10 - 2(t - 15). At t = 15: v = 10 m/s. At t = 20: v = 0.

Why: velocity is the slope of the x–t graph, which is the derivative of the position function. In Phase 1, the slope of t^2 is 2t — it grows linearly. In Phase 2, the slope of a straight line is its coefficient, 10. In Phase 3, the slope of 125 + 10(t - 15) - (t - 15)^2 decreases linearly.

Step 2. Sketch the v–t graph.

The autorickshaw's $v$–$t$ graph: a rising line (Phase 1), a flat line (Phase 2), a falling line (Phase 3). This is a trapezoid whose area equals the total displacement.

Step 3. Find acceleration in each phase by differentiating velocity.

Phase 1: a = \mathrm{d}v/\mathrm{d}t = 2 m/s^2 (constant).

Phase 2: a = 0 (constant velocity, no acceleration).

Phase 3: a = \mathrm{d}v/\mathrm{d}t = -2 m/s^2 (constant deceleration).

Why: acceleration is the slope of the v–t graph. In Phase 1, the slope of the rising line is (10 - 0)/(5 - 0) = 2. In Phase 2, the line is flat, so slope is 0. In Phase 3, the slope is (0 - 10)/(20 - 15) = -2.

Step 4. Sketch the a–t graph.

The autorickshaw's $a$–$t$ graph: three horizontal segments. The step changes at $t = 5$ and $t = 15$ correspond to the transitions between phases.

Result: The v–t graph is a trapezoid (rise–flat–fall). The a–t graph is a step function (+2, 0, −2). You obtained both by taking successive slopes of the graph above: slope of x–t gave v–t, and slope of v–t gave a–t.

What this shows: Converting from x–t downward to v–t and a–t is a slope operation at each step. A parabola in x–t becomes a straight line in v–t and a horizontal line in a–t — each differentiation reduces the degree of the curve by one.

Example 2: Calculating displacement and acceleration from a $v$–$t$ triangle — a braking cricket ball

A cricket ball is rolled across a pitch. Its velocity–time graph is a triangle: the ball starts at v = 0, accelerates uniformly to 8 m/s at t = 3 s, then decelerates uniformly back to v = 0 at t = 7 s. Find: (a) the acceleration in each phase, and (b) the total displacement.

The $v$–$t$ triangle for the cricket ball. The ball speeds up for 3 seconds, then slows down for 4 seconds. The area of this triangle is the total displacement.

Step 1. Find the acceleration in Phase 1 (0 to 3 s).

a_1 = \frac{\Delta v}{\Delta t} = \frac{8 - 0}{3 - 0} = \frac{8}{3} \approx 2.67 \text{ m/s}^2

Why: the slope of a straight line on the v–t graph is the acceleration. The velocity rises from 0 to 8 m/s over 3 seconds — a positive slope means positive acceleration.

Step 2. Find the acceleration in Phase 2 (3 to 7 s).

a_2 = \frac{\Delta v}{\Delta t} = \frac{0 - 8}{7 - 3} = \frac{-8}{4} = -2 \text{ m/s}^2

Why: the velocity drops from 8 to 0 over 4 seconds. The negative slope means negative acceleration — the ball is decelerating. The deceleration (2 m/s^2) is less severe than the acceleration (2.67 m/s^2), which is why it takes longer to stop than it took to speed up.

Step 3. Find the displacement — the area of the triangle.

The v–t graph forms a triangle with base 7 s and height 8 m/s.

\text{displacement} = \text{area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 8 = 28 \text{ m}

Why: the area under the v–t curve equals the displacement. This triangle has its peak at t = 3 s, but the area formula for a triangle does not depend on where the peak is — only on the base and height.

Step 4. Verify by computing each phase separately.

Phase 1 displacement: \frac{1}{2} \times 3 \times 8 = 12 m.

Phase 2 displacement: \frac{1}{2} \times 4 \times 8 = 16 m.

Total: 12 + 16 = 28 m. Confirmed.

Why split and re-check? Because in more complex problems, the two phases might have different signs (if velocity goes negative). Breaking it into phases ensures you handle signs correctly. Here both phases have positive velocity, so both displacements are positive.

Result: Phase 1 acceleration: +2.67 m/s^2. Phase 2 acceleration: -2 m/s^2. Total displacement: 28 m.

The $a$–$t$ graph for the cricket ball. Constant acceleration of $+2.67$ m/s$^2$ for 3 s, then constant deceleration of $-2$ m/s$^2$ for 4 s. The area under the positive segment is $3 \times 2.67 = 8$ m/s (the velocity gained). The area under the negative segment is $4 \times (-2) = -8$ m/s (the velocity lost). Net change: $0$ — the ball started and ended at rest.

What this shows: The area under the v–t triangle gives displacement, and the slopes of its two sides give the accelerations. The a–t graph confirms that the total change in velocity is zero — the positive area (+8 m/s) exactly cancels the negative area (-8 m/s), consistent with the ball starting and ending at rest.

If you are comfortable with the basics — slope, area, and the six canonical shapes — the ideas below take you deeper into what motion graphs can reveal.

Non-uniform acceleration and curved v–t graphs

All the examples above used constant acceleration, which produces straight-line v–t graphs. But real motion often involves acceleration that changes with time. A car engine pushes harder at low speeds and less at high speeds. A falling object with air resistance experiences a net force that decreases as it approaches terminal velocity.

When the v–t graph is curved, the slope at each point gives the instantaneous acceleration, and the area under the curve gives the displacement — but you can no longer compute the area using triangles and rectangles. You need calculus.

For example, if v(t) = 10(1 - e^{-t/5}) (an object approaching terminal velocity of 10 m/s with exponentially decreasing acceleration):

a(t) = \frac{\mathrm{d}v}{\mathrm{d}t} = 2e^{-t/5}

x(t) = \int_0^t v \, \mathrm{d}t' = 10t + 50(e^{-t/5} - 1)

The v–t curve rises steeply at first, then flattens. The a–t curve starts at 2 m/s^2 and decays exponentially toward zero. The x–t curve is concave up initially (accelerating), then becomes nearly linear as the velocity saturates.

Jerk — when acceleration itself changes

The slope of the a–t graph has a name: jerk (or sometimes "jolt"). It measures how rapidly the acceleration changes, in units of m/s^3.

You feel jerk every time the Delhi Metro transitions between phases — the sudden onset of acceleration (positive jerk) and the sudden onset of braking (negative jerk). Engineers design the Metro's acceleration profile to limit jerk, because high jerk is what throws standing passengers off balance.

In the idealised step-function a–t graph above, the jerk is infinite at t = 20 and t = 80 (instantaneous changes in acceleration). In reality, the transitions are smooth — the a–t graph has short ramps instead of vertical steps, and the jerk is finite.

The area under x–t — absement

For completeness: the area under the x–t graph also has a physical meaning. It is called absement (a portmanteau of "absence" and "displacement"), measured in metre-seconds. Absement quantifies the total "position-hours" — how far away something was and for how long. It is used in some engineering contexts (fluid flow through a valve that opens and closes), but it rarely appears in JEE-level physics.

Negative velocity and negative displacement

When the v–t graph dips below the time axis, the area below the axis represents negative displacement — motion in the negative direction. The total displacement is the algebraic sum of the positive and negative areas. The total distance is the sum of the absolute values of both areas.

For example, if a ball is thrown straight up at 20 m/s with g = 10 m/s^2, its v–t graph is a straight line: v = 20 - 10t. It crosses zero at t = 2 s. From t = 0 to t = 2, the area is \frac{1}{2} \times 2 \times 20 = 20 m (upward displacement). From t = 2 to t = 4, the area is \frac{1}{2} \times 2 \times (-20) = -20 m (downward displacement). Total displacement: 0 (the ball returns to your hand). Total distance: 40 m (20 m up + 20 m down).

Where this leads next

Free Fall — the most important special case of constant acceleration. The graphs of a freely falling object are exactly the canonical shapes you learned here, with a = g = 9.8 m/s^2.
Position, Distance, and Displacement — the conceptual foundation for understanding what the x-axis of these graphs actually measures.
Calculus in Physics — Differentiation — the formal mathematics behind "slope gives velocity" and "slope gives acceleration."
Calculus in Physics — Integration — the formal mathematics behind "area gives displacement" and "area gives velocity change."
Uniformly Accelerated Motion — the SUVAT equations, which are the algebraic counterpart to everything you read off these graphs.