Motion graphs

A speedometer needle and a straight road tell two different stories of the same journey, and a motion graph is what lets you read both from a single picture. Plot a body's position, velocity, or acceleration against time, and the shape of the curve — its slope, its curvature, the area it encloses — hands you every other quantity in the kinematics toolkit without a single extra measurement. This article stays inside the Unit 2 (Motion in a Straight Line) syllabus: reading and sketching $x$ - $t$ , $v$ - $t$ , and $a$ - $t$ graphs for uniform and non-uniform motion. We reuse the same accelerating-car and there-and-back numbers from the introductory pass, now pushed one notch further — a formal derivation of why area under a $v$ - $t$ graph equals displacement, a second worked example that exposes the distance-versus-displacement trap, and an exam-ready cheat sheet.

The core insight

On an $x$ - $t$ graph the slope gives velocity and on a $v$ - $t$ graph the slope gives acceleration, while the area enclosed by a $v$ - $t$ graph gives displacement and the area under an $a$ - $t$ graph gives the change in velocity.

The core idea

Three graphs, three pairs of slope-and-area rules, all linked by the same calculus that connects $x$ , $v$ , and $a$ .

Position-time graphs read out velocity as a slope. Because $v = dx/dt$ , the steepness of the $x$ - $t$ curve at any instant is the instantaneous velocity:

v = \frac{dx}{dt} = \text{slope of the } x\text{-}t \text{ graph}

where: $v$ is instantaneous velocity, $x$ is position, and $t$ is time. A straight line means uniform velocity; a curve means the velocity is changing; a horizontal segment means the body is momentarily or permanently at rest.

Velocity-time graphs read out acceleration as a slope, and displacement as an area. Because $a = dv/dt$ ,

a = \frac{dv}{dt} = \text{slope of the } v\text{-}t \text{ graph}, \qquad \Delta x = \int v\,dt = \text{area under the } v\text{-}t \text{ graph}

where: $a$ is instantaneous acceleration and $\Delta x$ is the displacement over the interval considered. A horizontal $v$ - $t$ line means zero acceleration; a sloped straight line means uniform acceleration; area below the time-axis subtracts from the total displacement because the body is moving in the negative direction during that stretch.

Acceleration-time graphs read out change in velocity as an area. Because $\Delta v = \int a\,dt$ , the area enclosed between an $a$ - $t$ curve and the time-axis, over an interval, is the change in velocity across that interval — not the velocity itself. These three graphs form a differentiation-and-integration chain: differentiate $x$ - $t$ to get $v$ - $t$ , differentiate again to get $a$ - $t$ ; integrate (find area under) $a$ - $t$ to recover $\Delta v$ , and integrate $v$ - $t$ to recover $\Delta x$ . Reading a graph correctly always starts by asking which of the two operations — slope or area — the question actually wants.

Figure

A velocity time graph rising in a straight line to 20 metres per second at 5 seconds, staying flat at 20 metres per second until 15 seconds, then falling in a straight line to zero at 19 seconds, with the region under the whole curve shaded to show that its area equals the total distance travelled

Derivation ADisplacement as the area under a velocity-time graph

The goal is to show, from nothing more than the definition of velocity as the slope of an $x$ - $t$ graph, that the displacement over any interval equals the geometric area enclosed between the $v$ - $t$ curve and the time-axis over that interval, and then to turn that geometric fact into a simple sum of triangle, rectangle, and trapezoid areas that can be applied directly to the piecewise-straight $v$ - $t$ graphs this syllabus draws.

Step 1. Start from a very short time step $\Delta t$ , small enough that the velocity is essentially constant over it at whatever value $v$ has at that instant. Over that tiny step the displacement is $\Delta x = v\,\Delta t$ , exactly the area of a thin rectangle of height $v$ and width $\Delta t$ sitting under the $v$ - $t$ curve.

Step 2. Add up all such thin rectangles across the full interval from $t_1$ to $t_2$ . Each rectangle's area is one small displacement, so the running sum of rectangle areas is the running sum of displacements — the total displacement $\Delta x$ over the whole interval. As $\Delta t$ shrinks, the jagged staircase of rectangle tops hugs the actual $v$ - $t$ curve, and the sum becomes the true area under it: $\Delta x = \int_{t_1}^{t_2} v\,dt$ .

Step 3. For a $v$ - $t$ graph built from straight-line segments — the only kind this syllabus asks you to handle — the area under each segment is an ordinary shape you already know how to measure: a rectangle where velocity is constant, a triangle where velocity ramps up from (or down to) zero, and a trapezoid where it ramps between two nonzero values. The total displacement is simply the sum of these piece areas, with any area that sits below the time-axis (negative velocity) subtracted rather than added.

Step 4. Apply this to the car of the worked example below: phase 1 is a triangle (velocity ramps from $0$ to $20\ \text{m/s}$ over $5\ \text{s}$ ), phase 2 is a rectangle (constant $20\ \text{m/s}$ for $10\ \text{s}$ ), phase 3 is a triangle (velocity ramps from $20\ \text{m/s}$ to $0$ over $4\ \text{s}$ ). None of the areas lie below the axis, since the car never reverses direction, so the total distance is simply their sum — confirming that "area under $v$ - $t$ " is not a formula to memorise blindly but a direct geometric consequence of $v$ being the slope of $x$ - $t$ .

Worked example

The $v$ - $t$ graph of a car shows it accelerating uniformly from rest to $20\ \text{m/s}$ in $5\ \text{s}$ , then moving at a constant $20\ \text{m/s}$ for the next $10\ \text{s}$ , then decelerating uniformly to rest in a further $4\ \text{s}$ . Find (a) the acceleration in each phase and (b) the total distance travelled.

Step 1. Phase 1, $0$ to $5\ \text{s}$ : acceleration is the slope, $a_1 = \dfrac{20 - 0}{5} = 4\ \text{m/s}^2$ .

Step 2. Phase 2, $5$ to $15\ \text{s}$ : the graph is horizontal, so $a_2 = 0$ — constant velocity.

Step 3. Phase 3, $15$ to $19\ \text{s}$ : $a_3 = \dfrac{0 - 20}{4} = -5\ \text{m/s}^2$ .

Step 4. Total distance is the area under the graph — triangle, rectangle, triangle, as derived above:

s = \tfrac12(5)(20) + (10)(20) + \tfrac12(4)(20) = 50 + 200 + 40 = 290\ \text{m}.

The car travels a total of 290 m, accelerating at $4\ \text{m/s}^2$ , cruising, then decelerating at $-5\ \text{m/s}^2$ .

Competitive stretch

An $x$ - $t$ graph shows a particle moving from $x = 0$ at $t = 0$ to $x = 12\ \text{m}$ at $t = 4\ \text{s}$ (uniform velocity), resting at $x = 12\ \text{m}$ until $t = 6\ \text{s}$ , then returning uniformly to $x = 0$ at $t = 10\ \text{s}$ . Find the average speed and the average velocity over the full 10 seconds.

Step 1. Outward velocity: $v = \frac{12 - 0}{4} = 3\ \text{m/s}$ . Return velocity: $v = \frac{0 - 12}{10 - 6} = -3\ \text{m/s}$ . The particle covers $12\ \text{m}$ out and $12\ \text{m}$ back, so total distance $= 24\ \text{m}$ , giving average speed $= \frac{24}{10} = 2.4\ \text{m/s}$ .

Step 2. Net displacement over the same 10 s is $x_{\text{final}} - x_{\text{initial}} = 0 - 0 = 0\ \text{m}$ , since the particle ends where it started.

Average speed is $2.4\ \text{m/s}$ , but average velocity is $0\ \text{m/s}$ — the two differ whenever a body reverses direction, because distance always adds while displacement can cancel.

Common traps

Confusing slope with area: slope of $x$ - $t$ gives velocity, but area under $v$ - $t$ gives displacement — mixing these two operations is the single most common graph-reading error.
Assuming a curved $x$ - $t$ graph always means speeding up — a curve only shows that velocity is changing; it may be increasing or decreasing.
Reading a downward-sloping $v$ - $t$ line as "moving backward" — it means decelerating; the body is only moving in the negative direction once the line crosses below the time-axis.
Forgetting that area below the time-axis is negative and must be subtracted, not added, when displacement (not total distance) is required.

Ask the tutor: if converting between the three graph types still trips you up, ask the AI tutor to walk through a fresh $x$ - $t$ / $v$ - $t$ / $a$ - $t$ triple, one graph at a time.

Why does area under v-t give displacement while slope of x-t gives velocity — are these not opposite operations?

They are opposite in the calculus sense, and that is exactly the point. Velocity is defined as the rate of change of position, so reading it off an $x$ - $t$ graph means taking a slope (a derivative). Displacement is the accumulated effect of velocity over time, so recovering it from a $v$ - $t$ graph means taking an area (an integral, the reverse of a derivative). Differentiating and integrating undo each other, which is why the two graphs are linked: differentiate $x$ - $t$ to get $v$ - $t$ , or integrate $v$ - $t$ to get back the displacement.

If a v-t graph is a curve rather than a straight line, can I still just measure areas of triangles and rectangles?

Only where the curve is genuinely made of straight segments, as in this syllabus. A truly curved $v$ - $t$ graph (non-uniform acceleration) needs the area under a curve, which strictly requires integration beyond this syllabus's scope. For exam purposes, motion graphs here are always built from straight-line pieces, so every area splits cleanly into rectangles, triangles, and trapezoids, and no calculus is needed to evaluate them.

How do I tell from a graph alone whether a body is speeding up or slowing down?

Look at whether the slope's magnitude is increasing or decreasing on the $v$ - $t$ graph, or check the sign of acceleration relative to the sign of velocity. If $v$ and $a$ have the same sign, the body speeds up in whatever direction it is already moving; if they have opposite signs, the body slows down. On the graph itself, a $v$ - $t$ line moving away from the time-axis means speeding up, and a line moving toward the time-axis means slowing down, regardless of which side of the axis it is on.

Motion graphs

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads