HookThe skydiver who broke the sound barrier with his body
On 14 October 2012, Felix Baumgartner stepped off a capsule slung beneath a helium balloon nearly \(39\ \text{km}\) above New Mexico, and fell. In the thin upper air there was almost nothing to slow him, and about \(50\) seconds later he was travelling at \(1{,}357.6\ \text{km h}^{-1}\) — roughly \(377\ \text{m s}^{-1}\), faster than the speed of sound, the first human to break the sound barrier with his own body. Then, as he dropped into denser air, he began to slow. His speed rose and then fell during a single fall, and that one fact carries an exact meaning: his acceleration was not constant.
That distinction runs through the whole of kinematics — the description of how a body moves, before we ask what force causes it. In this section you fix the language (position, displacement, distance, velocity, speed and acceleration), read the two motion graphs that carry most of the intuition, master the five suvat equations that solve any motion with constant acceleration, switch to calculus for the far more common case where acceleration varies with time (Baumgartner's fall among them), and finish with motion under gravity and projectiles, where a single clean idea splits every launch into two simple problems. Each is the same handful of quantities seen from a different angle.
ModelThe language of motion — and why distance is not displacement
Kinematics is precise about its words, and the marks begin with getting them right. Position is where a body is relative to a fixed origin. Displacement is the change in position: a vector, with a magnitude and a direction, or on a line a sign. Distance travelled is the scalar total length of the path, which ignores direction and never decreases. These two part company the instant a body turns round.
Velocity is the rate of change of displacement, a vector measured in \(\text{m s}^{-1}\); speed is its magnitude, a scalar. Acceleration is the rate of change of velocity, a vector in \(\text{m s}^{-2}\). Because velocity is a vector, a body can accelerate by speeding up, by slowing down, or by merely changing direction at constant speed — a point that catches people out. When velocity and acceleration point the same way the body speeds up; when they point in opposite ways it slows, a case often called deceleration or retardation.
Along a straight line all of this reduces to a sign. Fix one direction as positive; then a negative displacement is a position on the other side of the origin, a negative velocity is motion in the negative direction, and a negative acceleration might be a genuine slowing (if velocity is positive) or a speeding up in the negative direction (if velocity is already negative). A sprinter who runs \(100\ \text{m}\) up the track and jogs \(100\ \text{m}\) back has travelled a distance of \(200\ \text{m}\) but has a displacement of zero — the same start and finish — which is why the average velocity over that trip is zero while the average speed is not.
ModelReading motion graphs — gradients and areas
Two graphs carry most of the intuition in kinematics, and each hides its information in a different feature. On a displacement–time graph the gradient is the velocity: a straight line is constant velocity, a horizontal line is a body at rest, a steepening curve is a body speeding up, and a negative gradient is motion back towards the origin. The height tells you where the body is; the slope tells you how fast.
On a velocity–time graph two features matter. The gradient is the acceleration, so a straight sloping line is constant acceleration and a horizontal line is constant velocity. And the area between the line and the time axis is the displacement, because displacement is velocity accumulated over time, which is exactly what an area totals. This is the graph examiners lean on hardest, because a journey in several phases becomes a set of triangles and rectangles whose areas you simply add. Watch the sign: area below the time axis counts as negative displacement (motion the other way), so for total distance you add the sizes of the areas, but for displacement you subtract the ones below the axis.
A car starts from rest, accelerates uniformly to \(20\ \text{m s}^{-1}\) in \(5\ \text{s}\), holds that speed for \(10\ \text{s}\), then brakes uniformly to rest in a further \(5\ \text{s}\). Find the acceleration in the first phase and the total distance travelled.
The velocity–time graph is a triangle, then a rectangle, then a triangle. The first-phase acceleration is the gradient: \(a=\dfrac{20-0}{5}=4\ \text{m s}^{-2}\). The total distance is the area under the whole graph: the rising triangle is \(\tfrac12\times 5\times 20=50\ \text{m}\), the rectangle is \(20\times 10=200\ \text{m}\), and the falling triangle is \(\tfrac12\times 5\times 20=50\ \text{m}\). Adding them, the car travels \(50+200+50=300\ \text{m}\). Splitting a motion graph into these standard shapes is faster and less error-prone than any single equation for a multi-stage journey.
MechanismThe suvat equations — five tools for constant acceleration
When acceleration is constant, five equations connect the five quantities of straight-line motion: \(s\) (displacement), \(u\) (initial velocity), \(v\) (final velocity), \(a\) (acceleration) and \(t\) (time) — the suvat equations. They are \(v=u+at\); \(s=ut+\tfrac12 at^2\); \(s=vt-\tfrac12 at^2\); \(s=\tfrac12(u+v)t\); and \(v^2=u^2+2as\). Each omits exactly one of the five variables, and the whole skill of using them is choosing the equation that omits the quantity you neither know nor want.
They are not arbitrary. \(v=u+at\) is just 'final velocity = start velocity + (rate of change)(time)', and \(s=\tfrac12(u+v)t\) is the area of the velocity–time trapezium — average velocity times time. Combining those two eliminates \(v\) or \(t\) to give the rest, and \(v^2=u^2+2as\) is the one to reach for whenever time is neither given nor wanted. The non-negotiable condition is that acceleration is constant throughout the interval, which is exactly why suvat cannot describe Baumgartner's whole fall. In two dimensions the same equations hold as vector statements — for example \(\mathbf{v}=\mathbf{u}+\mathbf{a}t\) — applied to each component separately.
The Highway Code gives the stopping distance for a car at \(30\ \text{mph}\) (\(13.4\ \text{m s}^{-1}\)) as \(23\ \text{m}\). Reconstruct it, taking a driver reaction time of \(0.67\ \text{s}\) and a firm dry-road braking deceleration of \(6.4\ \text{m s}^{-2}\).
Stopping distance is thinking distance plus braking distance. During the reaction time the car moves at constant speed, so the thinking distance is \(13.4\times 0.67\approx 9\ \text{m}\). For the braking distance the car decelerates from \(u=13.4\) to \(v=0\), and time is not given, so use \(v^2=u^2+2as\): \(0=13.4^2+2(-6.4)s\), giving \(s=\dfrac{13.4^2}{2\times 6.4}=\dfrac{179.56}{12.8}\approx 14\ \text{m}\). The total is \(9+14=23\ \text{m}\), exactly the published figure. Notice how the choice of \(v^2=u^2+2as\) fell straight out of the fact that the question never mentioned time.
MechanismCalculus in kinematics — when the acceleration changes
When acceleration is not constant, the suvat equations are useless and calculus takes over. Displacement, velocity and acceleration form a ladder joined by differentiation going one way and integration the other. Velocity is the derivative of displacement, \(v=\dfrac{ds}{dt}\); acceleration is the derivative of velocity, \(a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}\). Running the ladder upward, velocity is the integral of acceleration and displacement the integral of velocity, with a constant of integration fixed each time by a known value such as the starting position or speed.
This reframes the graph facts as calculus: 'gradient of displacement–time is velocity' is differentiation, and 'area under velocity–time is displacement' is integration, \(s=\int v\,dt\). The tell-tale that a problem needs calculus rather than suvat is an expression for velocity or acceleration that still contains \(t\) — if \(a\) depends on time, it is not constant. Two landmarks fall straight out of the derivatives: the body is instantaneously at rest when \(v=0\), and its velocity is momentarily greatest or least when \(a=\dfrac{dv}{dt}=0\). In two dimensions you differentiate or integrate each \(\mathbf{i}\) and \(\mathbf{j}\) component independently.
A particle moves on a line so that its displacement from the origin is \(s=t^3-6t^2+9t\) metres at time \(t\) seconds. Find its velocity and acceleration, and the times when it is instantaneously at rest.
Differentiate for velocity: \(v=\dfrac{ds}{dt}=3t^2-12t+9\). Differentiate again for acceleration: \(a=\dfrac{dv}{dt}=6t-12\). The particle is at rest when \(v=0\): factorise \(3t^2-12t+9=3(t^2-4t+3)=3(t-1)(t-3)\), so \(v=0\) at \(t=1\ \text{s}\) and \(t=3\ \text{s}\). Between those times \(v\) is negative, so the particle has reversed direction — behaviour the suvat equations could never produce, because here \(a=6t-12\) plainly changes with time. As a bonus, the acceleration is zero at \(t=2\ \text{s}\), the instant midway between the two stops, where the particle is moving fastest in the negative direction.
CaseMotion under gravity and projectiles — splitting one launch into two
Close to the Earth's surface, and ignoring air resistance, a body moves under gravity with a constant downward acceleration \(g\approx 9.8\ \text{m s}^{-2}\) — and because it is constant, the suvat equations apply again. For purely vertical motion, choose a positive direction (up, say), so a thrown ball has \(a=-9.8\ \text{m s}^{-2}\) throughout, including at the top of its flight, where the velocity is momentarily zero but the acceleration is not.
A projectile — a body given an initial velocity and then left to gravity — is handled by the single most important idea in this section: the horizontal and vertical motions are independent and share only the time. Horizontally there is no force in the model, so the horizontal velocity is constant and the horizontal displacement is simply \(x=(u\cos\theta)t\). Vertically the body is in free fall, so you run vertical suvat with \(a=-g\). You solve the vertical motion for a time — time to reach the ground, or time to the top where the vertical velocity is zero — and feed that same time into the horizontal equation. This is where Baumgartner's jump stops being a textbook projectile: over most of his fall air resistance was enormous, his acceleration varied, and the tidy suvat picture, valid only without air resistance, breaks down.
A stone is kicked horizontally at \(15\ \text{m s}^{-1}\) from the top of a \(20\ \text{m}\) cliff. Find how far from the base it lands and its speed on impact. Take \(g=9.8\ \text{m s}^{-2}\) and ignore air resistance.
Treat the two directions separately. Vertically, the stone starts with zero vertical velocity and falls \(20\ \text{m}\): using \(s=\tfrac12 g t^2\), \(20=\tfrac12(9.8)t^2\), so \(t^2=\dfrac{20}{4.9}=4.08\) and \(t=2.02\ \text{s}\). Horizontally, the velocity stays \(15\ \text{m s}^{-1}\), so the range is \(x=15\times 2.02=30.3\ \text{m}\). For the impact speed, find the vertical velocity gained: \(v_y=gt=9.8\times 2.02=19.8\ \text{m s}^{-1}\). The stone's speed is the magnitude of its velocity vector, \(\sqrt{15^2+19.8^2}=\sqrt{225+392}=\sqrt{617}\approx 24.8\ \text{m s}^{-1}\). The horizontal speed never changed; all the extra came from the vertical fall — which is the whole point of treating the components on their own.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Kinematics is method-marked, so structure beats speed. For any suvat question, write down the five letters \(s,u,v,a,t\) and fill in what you know, including a sign for direction; the missing-and-unwanted variable then names the equation to use, which is why 'no time mentioned' should trigger \(v^2=u^2+2as\) automatically. Fix a positive direction at the start and stick to it — most lost marks are sign slips where an upward launch and downward gravity are given the same sign.
Decide early whether the motion is constant or variable acceleration: an acceleration or velocity written as a function of \(t\) means calculus, not suvat, and you differentiate to go down the ladder or integrate (with a constant fixed by initial conditions) to go up it. For projectiles, resolve into horizontal and vertical from the first line and never mix them: horizontal has \(a=0\) and constant velocity, vertical has \(a=-g\), and the shared time is the bridge between them. On graph questions, remember that on a velocity–time graph the gradient is acceleration and the area is displacement — and quote the \(g\) value the paper specifies, keeping it consistent throughout.