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Motion graphs and the SUVAT equations

Motion graphs encode everything through two operations, gradient and area, and they work for any motion at all. The four SUVAT equations are faster, but they carry one condition, and ignoring it produces confident wrong answers.

Year 12AQA 3.4.1.3 CIE 2.1.1, 2.1.2, 2.1.3, 2.1.4, 2.1.5, 2.1.6, 2.1.7 OCR A 3.1.1(a), 3.1.1(b), 3.1.1(c), 3.1.1(d), 3.1.2(a)(i), 3.1.2(b)(i), 3.1.2(c)EDEXCEL 9, 10, 11

The maths behind it: Kinematics with constant acceleration on InkMaths.

IN THIS TOPIC

Use v = Δs/Δt and a = Δv/Δt, and tell an average value from an instantaneous one.
Read gradients and areas off displacement-time, velocity-time and acceleration-time graphs, curved ones included.
Choose the constant-acceleration equation that omits the quantity you neither have nor want.
Derive all four constant-acceleration equations from the definitions of velocity and acceleration.
Spot when the acceleration is not constant, and abandon SUVAT for the area under a graph.

COMMON MISCONCEPTION

SUVAT always works.

The quantities and their definitions

Motion runs on three quantities. Displacement s is the vector version of distance. Velocity is the rate of change of displacement, and acceleration the rate of change of velocity. Both defining equations are printed for you.

v = \frac{\Delta s}{\Delta t} \qquad a = \frac{\Delta v}{\Delta t}

ON THE AQA DATA SHEET

Applied over an interval these give average values. Shrink the interval and they become instantaneous ones, which is exactly what the gradient of a graph reads off at a point.

Reading the graphs

Two operations decode every motion graph in the course, gradient and area. On a displacement-time graph the gradient is the velocity. A curve whose gradient steepens is speeding up, and the gradient of the tangent gives the instantaneous velocity at that moment.

FIG. 1A displacement-time curve. The gradient of the tangent at a point is the instantaneous velocity there.

On a velocity-time graph, the gradient is the acceleration and the area beneath is the displacement. The area is the more valuable half, because it works for any shape of graph, including motions no equation describes. On an acceleration-time graph, the area gives the change in velocity.

FIG. 2A velocity-time graph. Gradient reads the acceleration; the shaded area beneath the line is the displacement.

Signs carry meaning throughout. Negative velocity is motion the other way, and area below the time axis subtracts from the displacement, which is how a ball thrown up and caught again ends with zero displacement despite plenty of distance travelled.

FIG. 3The ball falls, bounces, rises, and its velocity-time graph draws itself alongside: straight sloping lines of constant gradient, minus g all the way, with an instant vertical flip from the most negative to the most positive at every impact. Idealised here with no energy loss so the shape reads cleanly. The near-vertical jumps look odd at first sight, but each one is just a bounce, seen in graph form.

The bouncing ball is a common exam example, and it is worth a careful look. During every flight, up or down, the only force is gravity, so every sloped section has the same gradient, −g. The bounce itself is the near-vertical jump. The figure is drawn lossless, every jump returning to full height; for a real ball each jump ends a little lower than the last as speed is lost.

The four equations, and the condition

For constant acceleration only, four equations connect the five quantities s, u, v, a and t. The AQA, OCR A and Edexcel booklets print all four; CIE prints s = ut + ½at² and v² = u² + 2as only, so on 9702 the other two are recall items. Everywhere the real work is choosing the right one, not recalling it.

v = u + at

ON THE AQA DATA SHEET

s = \tfrac{1}{2}(u + v)t

ON THE AQA DATA SHEET

s = ut + \tfrac{1}{2}at^{2}

ON THE AQA DATA SHEET

v^{2} = u^{2} + 2as

ON THE AQA DATA SHEET

Each equation omits exactly one of the five quantities, which is how you choose between them.

WORKED EXAMPLE

Choosing the equation

A car brakes from 24 m s⁻¹ to rest over 60 m. Find the deceleration.

List what is known. u = 24, v = 0, s = 60, and a is wanted. The quantity not involved anywhere is t, so reach for the equation without t in it, $v^{2} = u^{2} + 2as$ .

0 = 24² + 2a × 60, so a = −576/120 = −4.8 m s⁻². The minus sign is the deceleration doing its job, and it should be left in.

The condition is the part people skip. Acceleration must be constant. A falling object with air resistance has an acceleration that shrinks as it goes, so SUVAT is flatly invalid there. Treat it as a velocity-time graph problem and take the displacement from the area instead. Free fall with no air resistance does qualify, with a = g downward.

GUIDED PRACTICE

Straight up, and briefly stopped

A ball is thrown straight up at 15 m s⁻¹. Using v = 0 at the top and your own sign convention, find the maximum height and the time to reach it.

Show the working

Taking up as positive, a = −9.81. The t-free equation gives the height, 0 = 15² − 2 × 9.81 × s, so s = 225/19.62 = 11.5 m.

Then v = u + at gives the time, 0 = 15 − 9.81t, so t = 1.5 s. That sign convention, set in the first line, is the only reason both answers came out positive and meaningful.

INDEPENDENT PRACTICE

A ride in two stages

A cyclist accelerates from rest at 1.2 m s⁻² for 5.0 s, then holds a steady speed for a further 8.0 s. Find the total distance. Treat the two stages separately.

Show the working

Stage one gives v = 1.2 × 5.0 = 6.0 m s⁻¹ and s = ½ × 1.2 × 5.0² = 15 m.

Stage two has no acceleration at all, so SUVAT is not wanted. s = 6.0 × 8.0 = 48 m.

Adding them, 15 + 48 = 63 m. Two stages, two different rule-sets, and the boundary between them sits exactly where the acceleration changed.

Where the four come from

None of the four is a new law. All four are the two definitions at the top of this page, rearranged, and the whole derivation runs to four short steps that are worth being able to write out from cold.

One. Constant acceleration means the average rate of change of velocity and the instantaneous one are the same number, so the definition $a = \Delta v/\Delta t$ can be applied across the whole interval with no approximation at all, as $a = \frac{v - u}{t}$ . Multiply through by t and add u to both sides, and the first equation follows as $v = u + at$ .

Two. Displacement is average velocity multiplied by time, straight from $v = \Delta s/\Delta t$ . Under constant acceleration the velocity climbs at a steady rate, so its average over the interval is the plain mean of the two end values, $\tfrac{1}{2}(u + v)$ , giving the second equation $s = \tfrac{1}{2}(u + v)t$ . That is also the area of the trapezium under the velocity-time graph, so the graph and the algebra are one fact seen twice.

Three. Substitute the first result into the second. Replacing v with u + at gives s = ½(u + u + at)t = ½(2u + at)t, and multiplying that out separates the two terms into $s = ut + \tfrac{1}{2}at^{2}$ .

Four. This time eliminate t rather than v. Rearranging the first result gives $t = \frac{v - u}{a}$ , and feeding that into the second gives $s = \tfrac{1}{2}(u + v) \times \frac{v - u}{a}$ . Multiply both sides by 2a and the right-hand side is a difference of two squares, $2as = (v - u)(v + u)$ , which multiplies out to $2as = v^{2} - u^{2}$ , or $v^{2} = u^{2} + 2as$ with the terms moved across.

Notice where the condition entered. It was used once, at step two, when the average velocity was set equal to the mean of the first and last values. That step is false the moment the acceleration varies, which is why “constant acceleration only” is a condition on the physics and not a footnote on the algebra.

ASSESSMENT FOCUS

OCR words braking questions in road-safety language. Thinking distance is speed times reaction time, braking distance comes from work done against the braking force or from SUVAT, and stopping distance is their sum. The physics has not changed, so answer in the vocabulary the question used.
Before touching SUVAT, ask whether the acceleration is constant. Air resistance disqualifies it, so does an engine easing off, and so does any curved velocity-time graph. Questions test exactly this.
Write the SUVAT list, fill in knowns, mark the wanted quantity, and pick the equation missing the one you neither have nor need. It is faster than remembering which equation “does” which problem.
CIE asks for the derivation itself, not only the use, so practise the four steps until they write themselves: a = (v − u)/t rearranged, average velocity ½(u + v) times t, one substituted into the other, then t eliminated between them.
Set a sign convention in the first line and keep it. An object thrown upward with downward g positive is where most sign errors are born.
“Instantaneous” means the gradient of the tangent at a point; “average” means total change over total time. Questions ask for the difference in words and expect it in method.
On any graph question, say what the gradient and the area mean for those axes before calculating. That statement frequently scores before any arithmetic does.

CHECK YOURSELF

A car brakes uniformly from 24 m s⁻¹ to rest over 60 m. How long does the braking take?

Show a hint

You know u, v and s. The quickest route uses the equation without a.

Show the answer

Using $s = \tfrac{1}{2}(u + v)t$ , 60 = 12 × t, so t = 5.0 s.

You can also get there through v = u + at, once a = −4.8 m s⁻² comes out of v² = u² + 2as. Two routes agreeing is the check the four equations provide.

Gradient and area read every graph.

SUVAT needs constant acceleration.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the motion graphs and the suvat equations questions page.

7 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Required practical 3: measuring g by free fall puts this topic in the lab, and the written papers ask about it.
Gradients and areas under graphs is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.

Use v = Δs/Δt and a = Δv/Δt, and tell an average value from an instantaneous one.
Read gradients and areas off displacement-time, velocity-time and acceleration-time graphs, curved ones included.
Choose the constant-acceleration equation that omits the quantity you neither have nor want.
Derive all four constant-acceleration equations from the definitions of velocity and acceleration.
Spot when the acceleration is not constant, and abandon SUVAT for the area under a graph.

Open the full revision checklist to track your progress across the whole unit.

Motion graphs and the SUVAT equations

The quantities and their definitions

Reading the graphs

The four equations, and the condition

Where the four come from

Question workbook

Question workbook

Answer book