Module 3: Kinematics · Theory · Prerequisites: K5, Foundation 9
These graphs carry two quantities, not one. The slope is a rate; the area is a total.
Mahavakya
Do not memorise which feature of which graph gives which quantity. Ask what derivative or integral connects them, and the rule becomes inevitable.
On this page
1. Height and slope, again
On a velocity–time graph the two features carry different quantities, exactly as on the position–time graph, and the same confusion arises.
The height is the velocity. How far the curve sits from the axis is how fast the body is moving.
The slope is the acceleration. How steeply the curve rises is how fast the velocity is changing.
So a steep velocity–time graph does not mean a fast body. It means a rapidly changing velocity, which is a different claim. A body can be moving at 200 m/s with a perfectly flat graph, and at 0.1 m/s with a very steep one.
One consequence catches everyone at least once. A horizontal velocity–time line at v = 5 m/s means a = 0 — not v = 0. The body is moving steadily. It is only at rest if the horizontal line sits on the axis.
2. The area, and why it is displacement
The velocity–time graph offers something the position–time graph does not: an area with a physical meaning.
area under v–t = Δx
It is worth being careful about why, because the usual justification is not quite a justification.
Check the dimensions: velocity times time is (L T−1)(T) = L, a length. That is reassuring, and it is not a proof. Plenty of quantities have the dimensions of length without being this particular displacement, and dimensional analysis can only ever rule things out.
The actual reason is Foundation 9’s. Since v = dx/dt, we have dx = v dt, and summing those small contributions across the interval is precisely what the integral — the area — does.
The general habit is worth more than this instance. When you meet a graph and want to know what its slope or its area means, do not try to recall a rule. Ask which derivative or integral connects the two quantities on the axes. The slope of y against x is dy/dx; the area is ∫y dx. Whether either is useful depends on whether that combination is a quantity physics cares about.
Apply it to the position–time graph and you find something worth knowing: the area under an x–t graph is ∫x dt, which is not a standard kinematic quantity at all. Students sometimes compute it. There is nothing to compute.
3. Signed area
Area below the axis counts as negative — because the velocity there is negative, and the body is moving backwards.
The crossing at t = 4 s is where the two answers part company.
displacement = net signed area = 16 − 4 = 12 m
distance = total unsigned area = 16 + 4 = 20 m
The whole difference between the two answers lives in that crossing. Before t = 4 the body moves forward; after it, backward. Find the crossing first, split there, and both quantities follow.
Not every zero is a crossing. A velocity graph can touch the axis and return the same way — v = (t − 2)² is the standard case, vanishing at t = 2 and positive on both sides. No reversal, no splitting needed, and displacement equals distance throughout. This is K4’s repeated-root point in graphical form.
4. The acceleration–time graph
One more graph, and the same two readings apply.
area under a–t = Δv
slope of a–t = jerk, the rate of change of acceleration
The area is the useful one, and it is worth being precise about what it is not. The area under an acceleration–time graph gives neither the velocity, nor the position, nor the distance. It gives the change in velocity, and to get the velocity itself you must add the value you started with:
v(t) = v0 + area under a–t up to t
That v0 is Foundation 9’s constant of integration, and it is not an algebraic leftover — it is the initial condition, the piece of information the graph cannot contain. Two bodies with identical acceleration graphs and different starting velocities have different motions.
5. Walking the chain both ways
You now have three graphs and can move between them in either direction.
x → slope→ v → slope→ a
a → area→ v → area→ x
Going down takes slopes; going up takes areas, and needs an initial value at each step.
There is a free check available on the way. Each differentiation lowers the degree of a polynomial by one, so each integration raises it by one.
Integrate a linear acceleration and the position must be cubic.
Use this before checking any arithmetic. Constant acceleration gives linear velocity and quadratic position. Linear acceleration gives quadratic velocity and cubic position. If you have integrated a linear a(t) and arrived at a quadratic x(t), there is an error, and you know it without substituting a single number.
6. An equation falls out of the geometry
Take a body with constant acceleration, so the velocity–time graph is a straight line from u to v over a time t. The area beneath it is a trapezium:
s = ½(u + v)t
And the slope of that line is the acceleration, so a = (v − u)/t, giving t = (v − u)/a. Substitute:
s = ½(u + v)(v − u)/a
2as = (u + v)(v − u) = v² − u²
v² = u² + 2as
One of the standard equations of motion, obtained from the area of a trapezium and the difference of two squares. No calculus, no memory.
That is worth noticing for what it says about the equation rather than about the derivation. K8 will reach the same result by integrating, and there is a third route through the acceleration–time graph. Three independent paths to one equation means the equation is a consequence — of constant acceleration and nothing else — rather than a fact to be stored.
7. What a jump means on each graph
K5 asked which position–time graphs are possible. The same question here has a different answer, and the contrast is instructive.
A jump in the x–t graph is impossible. The body would be in one place and then, with no time elapsed, elsewhere. Nothing physical does that.
A jump in the v–t graph is an idealisation. The velocity changes instantaneously, which requires infinite acceleration — and position stays perfectly continuous throughout. This is what a textbook draws for a collision.
So the same visual feature means quite different things depending on the axes. And the two are connected: a jump in v–t is exactly what produces a corner in x–t. One event, two graphs, two appearances.
Both are the impulse approximation — a very short interval of very large acceleration, redrawn as an instant. It is a modelling convenience, and like every such convenience it is worth knowing you have made it.
Worked example
A body starts from rest. Its acceleration is +3 m/s² for 4 s, then zero for 2 s, then −2 m/s² for 3 s. Find the velocity at each stage boundary and the total displacement.
Velocities — running areas under the a–t graph, added to v0 = 0:
v(4) = 0 + (3)(4) = 12 v(6) = 12 + 0 = 12 v(9) = 12 − (2)(3) = 6 m/s
Displacement — areas under the v–t graph, now that we know its shape:
triangle ½(4)(12) = 24 rectangle (2)(12) = 24 trapezium ½(12 + 6)(3) = 27
total = 75 m
The velocity never becomes negative, so the body never reverses and the distance is also 75 m. Note the two-step structure: the acceleration graph gave the velocity graph, and the velocity graph gave the displacement. Each step was an area, and each needed the value carried forward from the step before.
Contemplation
Three graphs of the same motion, and each makes something easy that the others make hard. The reversal is a crossing on one and a turning point on another. The acceleration is a curvature on the first graph and a slope on the second, and curvature is far harder to judge by eye than slope.
Nothing has been added by moving between them. The motion was fully specified by any one of the three. What changes is which features are visible — and a physicist choosing a graph is choosing what to make obvious, in the same way that Foundation 3’s choice of coordinates decided what an equation would look like.
That is worth remembering when a problem seems difficult. Sometimes the difficulty belongs to the physics. Often it belongs to the representation, and the second kind can simply be exchanged for another.
Common misconceptions
1. A steeper v–t graph means a faster body
Height is velocity; slope is acceleration. A steep graph means the velocity is changing fast, which says nothing about how large it is.
2. A horizontal v–t graph means the body is at rest
It means a = 0. Rest requires the line to lie on the axis.
3. The area under a–t gives the velocity
It gives the change in velocity. Without v0 the graph cannot tell you the velocity at all — and it gives neither the position nor the distance either.
4. Area is always displacement
Only under a v–t graph. Under a–t it is a change in velocity, and under x–t it is nothing physics has a use for.
5. Every zero of the velocity graph splits the distance calculation
Only the crossings do. A graph touching the axis and returning the same way, like (t − 2)², involves no reversal and needs no split.
Reflection
- What does the area under a jerk–time graph give? Answer it using the habit in section 2 rather than by recalling anything.
- Two bodies have identical acceleration–time graphs. What can differ between their motions, and what cannot?
- The trapezium derivation of v² = u² + 2as used no calculus. Where did the assumption of constant acceleration enter?
Key takeaways
- On a v–t graph, height is velocity and slope is acceleration.
- The area under v–t is displacement, because dx = v dt — not because the dimensions work out.
- Net signed area gives displacement; total unsigned area gives distance. Split at crossings only.
- The area under a–t is the change in velocity, and needs v0 added.
- Slopes walk down the chain; areas walk up it and need an initial value at each step.
- Each integration raises polynomial degree by one — a check available before any arithmetic.
- A jump in x–t is impossible; a jump in v–t is an idealised impulse.
What comes next
Section 6 obtained an equation of motion from a trapezium. That was a special case — the acceleration was constant, so the velocity graph was straight and the area was a shape school geometry can handle. K7 takes the simplest case of all, where the acceleration is zero, and K8 develops the constant-acceleration case properly.
Prerequisites: K5 and Foundation 9 · Reading: 16 min · Practice: 30 min · Difficulty: Intermediate