Motion graphs and the SUVAT equations

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.

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.

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.

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.

Describe the drawing

The 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 with no energy loss, so the loop repeats exactly. The oddest feature of the v-t graph is just the bounce, seen in graph form.

The bouncing ball is a common exam example. 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 bounce returning to full height; for a real ball each bounce reaches 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. Choosing the appropriate equation is the main step in most questions.

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

The acceleration must be constant. A falling object with air resistance has an acceleration that decreases as it falls, so the SUVAT equations do not apply. 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.

Where the four come from

None of the four is a new law. All four follow by rearrangement from the definitions of velocity and acceleration, in four steps.

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

Two. Displacement is average velocity multiplied by time, straight from v = Δs/Δt$v = \Delta s/\Delta t$v = Δs/Δt. 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, 12(u + v)$\tfrac{1}{2}(u + v)$(1/2)(u + v), giving the second equation s = 12(u + v)t$s = \tfrac{1}{2}(u + v)t$s = (1/2)(u + v)t. That is also the area of the trapezium under the velocity-time graph, so the graphical and algebraic routes give the same result.

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 + 12at²$s = ut + \tfrac{1}{2}at^{2}$s = ut + (1/2)at^2.

Four. This time eliminate t rather than v. Rearranging the first result gives t = v - ua$t = \frac{v - u}{a}$t = (v - u)/a, and substituting that into the second gives s = 12(u + v) × v - ua$s = \tfrac{1}{2}(u + v) \times \frac{v - u}{a}$s = (1/2)(u + v) × (v - u)/a. Multiply both sides by 2a and the right-hand side is a difference of two squares, 2as = (v - u)(v + u)$2as = (v - u)(v + u)$2as = (v - u)(v + u), which multiplies out to 2as = v² - u²$2as = v^{2} - u^{2}$2as = v^2 - u^2, or v² = u² + 2as$v^{2} = u^{2} + 2as$v^2 = u^2 + 2as with the terms moved across.

The condition enters once, at step two, where the average velocity is set equal to the mean of the first and last values. That step is valid only while the acceleration is constant, and the same restriction applies to all four equations.