where $v$ is speed in $\text{m/s}$, $s$ is distance in $\text{m}$ and $t$ is time in $\text{s}$.

Velocity is defined in terms of speed plus a direction, which makes it a vector while speed is a scalar. Numerically they are the same; the difference is that a velocity is incomplete without saying which way.

In the lab, speed and acceleration are rarely measured with a stopwatch alone — a ticker-timer pulls a paper tape through a vibrating arm that stamps a dot at a fixed rate (commonly $50$ dots per second, so $0.02\text{s}$ between dots). Evenly spaced dots mean constant speed; dots that spread further apart mean the object is accelerating.

A more modern alternative uses light gates: an infrared beam connected to a data logger times how long a card of known length blocks the beam as it passes through, giving $\text{speed}=\frac{\text{length of card}}{\text{time blocked}}$. Two light gates a measured distance apart let a computer calculate acceleration automatically, which is why they are standard equipment in school laboratories and in real motorsport timing systems.

An object moving in a circle at a steady rate has constant speed but continually changing velocity, because its direction changes at every instant.

Since acceleration is defined as change of velocity per unit time, that object is accelerating — and by Newton's laws an acceleration requires a resultant force. This is why a car on a bend needs friction from the road, and why the passengers feel thrown outwards.

The word total in both places is what makes this equation work, and it is where marks are lost. Average speed is not the average of the individual speeds unless each stage takes the same time.

Example. A cyclist travels $30\text{km}$ at $30\text{km/h}$, then $30\text{km}$ at $10\text{km/h}$.

Averaging the two speeds would give $20\text{km/h}$, which is wrong. The cyclist spends three times as long going slowly, so the slow stage dominates the average.

where $a$ is in ${\text{m/s}}^2$, $\Delta v$ is the change in velocity in $\text{m/s}$ and $\Delta t$ is the time taken in $\text{s}$.

The symbol $\Delta$ means 'change in', so $\Delta v=v-u$, the final velocity minus the initial velocity. Writing $\frac{v}{t}$ instead of $\frac{v-u}{t}$ is the commonest error in the topic, and gives the right answer only when the object started from rest.

A deceleration is simply a negative acceleration, and this must be carried through the arithmetic.

Example. A car slows from $20\text{m/s}$ to $5\text{m/s}$ in $3.0\text{s}$.

The answer may be quoted either as an acceleration of $-5.0{\text{m/s}}^2$ or as a deceleration of $5.0{\text{m/s}}^2$. What is not acceptable is 'a deceleration of $-5.0{\text{m/s}}^2$', which says the same thing twice and means the opposite.

The definitions $v=\frac{s}{t}$ and $a=\frac{\Delta v}{\Delta t}$ both involve time. Sometimes a question gives a distance but not a time — for example, a braking distance — and asks for a final speed, or the other way round. For an object moving with constant acceleration, combining those two equations (eliminating $t$ between them) gives a third, very useful relationship that has no time in it at all:

where $u$ is the initial speed, $v$ is the final speed, $a$ is the (constant) acceleration and $s$ is the distance moved, all in consistent SI units. It applies only while the acceleration stays constant — exactly the condition needed for a straight-line section of a speed-time graph.

Example. A car accelerates from rest at $2.5{\text{m/s}}^2$ over a distance of $40\text{m}$. Find its final speed.

Remember to take the square root at the end — a common slip is to give $200$ as the final speed instead of $\sqrt{200}$. For a decelerating object, $a$ is negative, and the equation works exactly as before provided the sign is carried through consistently.

On a distance-time graph, distance is on the vertical axis and time on the horizontal, so the gradient is $\frac{\text{distance}}{\text{time}}$ — which is the speed.

For a curve, the speed at an instant is found by drawing a tangent at that point and taking the gradient of the tangent.

A speed-time graph carries twice as much information as a distance-time graph, because both the gradient and the area under the line mean something:

Both come straight from the axis units, which is the reliable way to remember them: gradient divides the vertical by the horizontal, area multiplies them.

Split the area into rectangles and triangles and add them:

Example. A car accelerates uniformly from rest to $20\text{m/s}$ in $8.0\text{s}$, then holds $20\text{m/s}$ for $12\text{s}$.

A speed-time curve that rises steeply and then flattens towards a horizontal line describes an object whose acceleration is decreasing to zero. It is still speeding up throughout, but by less and less.

When the line finally becomes horizontal, the acceleration is zero and the speed is constant. That is exactly the shape produced by an object reaching terminal velocity, which section 9 explains.

A horizontal line means at rest on one graph and constant speed on the other — opposite conclusions from an identical shape. There is no way to tell them apart from the shape alone.

So before interpreting any motion graph, read the vertical axis label first. It takes a second and it eliminates the single most common source of lost marks in this topic.

A short memory device some students find useful: "D-T graph: Gradient gives Speed. S-T graph: Gradient gives Acceleration, Area gives Distance" — sometimes shortened to SAD for the three quantities (Speed, Acceleration, Distance) you can read off as you move from the D-T graph's gradient to the S-T graph's gradient and then its area.

Whatever memory aid you use, it only helps once you have checked which graph you are looking at — the letters D and S at the start of the mnemonic are there to remind you of that too.

Near the Earth's surface, the acceleration of free fall $g$ is approximately constant and approximately $9.8{\text{m/s}}^2$. Many questions use $10{\text{m/s}}^2$ to simplify the arithmetic — use whichever value the question gives.

In the absence of air resistance, all objects fall with this same acceleration regardless of their mass. A feather and a hammer dropped together in a vacuum land together.

This surprises people, and it is worth seeing why. A heavier object is pulled down by a larger weight — but it also has more mass to accelerate, and the two effects cancel exactly:

The mass cancels. Doubling the mass doubles the force and doubles the inertia, so the acceleration is unchanged.

In everyday life a feather does fall more slowly than a hammer — but that is air resistance, not gravity, and it is the subject of the next section.

Terminal velocity depends on an object's weight and its surface area, which is why it varies so much in practice. A small raindrop has so little weight that it reaches its (low) terminal velocity almost immediately, which is why rain does not keep accelerating all the way from the clouds to the ground. A skydiver in a spread-eagle, belly-to-earth position reaches a terminal velocity of roughly $55\text{m/s}$ (about $200\text{km/h}$) — considerably faster if they instead fall head-first to reduce their surface area.

Because a smaller drag force takes longer to grow to match the object's weight, a streamlined object reaches a higher terminal (or maximum) velocity than a bulkier one of the same weight. This is the same physics as the head-first skydiver above, generalised: a cyclist crouching low, a car's sloped windscreen, and the pointed nose of a high-speed train are all shaped to reduce drag so that a given driving force can push the object to a greater top speed.

The reverse is also used deliberately: a parachute is shaped to do the opposite of streamlining, maximising drag so that terminal velocity is as low as possible.

Opening a parachute makes a sharp change to the story, and it is a favourite exam question:

Note that the skydiver slows down — they do not move upwards. A resultant force upwards on a downward-moving object produces deceleration, not reversal.

Mark schemes here look for forces, not for descriptions of speed. The three statements that score are:

Saying 'it stops accelerating because it cannot go any faster' describes the outcome without the mechanism and earns nothing. And note that at terminal velocity the forces are balanced — not absent. Weight still acts, and so does air resistance; they simply cancel.

Question. A runner covers $400\text{m}$ in $50\text{s}$, rests for $30\text{s}$, then covers a further $200\text{m}$ in $40\text{s}$. Calculate the average speed for the whole journey.

Marking. 1 mark per point. The rest period is the trap: it adds no distance but it adds time, so leaving it out inflates the answer.

Question. A train travelling at $30\text{m/s}$ brakes to $12\text{m/s}$ in $6.0\text{s}$. (a) Calculate its acceleration. (b) State what the sign tells you. (c) Describe the shape of the speed-time graph for this motion.

Marking. 1 mark per point. Use $v-u$, not $v$ alone; using $\frac{12}{6.0}$ would give $2.0{\text{m/s}}^2$ and score nothing.

Question. A cyclist accelerates uniformly from rest to $12\text{m/s}$ in $6.0\text{s}$, travels at $12\text{m/s}$ for $10\text{s}$, then decelerates uniformly to rest in $4.0\text{s}$. (a) Calculate the acceleration in the first stage. (b) Calculate the total distance travelled.

Marking. 1 mark per point. Stating that distance is the area under the graph is itself a marking point — write it down before calculating.

Question. A skydiver jumps from a plane. Explain, in terms of the forces acting, how she reaches terminal velocity, and what happens when she opens her parachute.

Marking. 1 mark per point. Every mark is tied to a statement about forces — answers written purely in terms of speed do not score.

For a speed–time graph, a horizontal line above zero means the object keeps moving at constant speed, not that it is stationary. An upward straight slope means a constant positive acceleration; a downward straight slope means constant negative acceleration. A curved section means the acceleration itself changes because the gradient is not constant.

When a question asks for distance, split the area into recognisable shapes and include units: a rectangle gives $\text{base}\times \text{height}$ and a triangle gives $\frac{1}{2}\times \text{base}\times \text{height}$. The units confirm the interpretation: $\text{s}\times \text{m/s}=\text{m}$.

Balanced forces mean zero acceleration, not zero speed. A car at constant speed and a book resting on a table both have zero resultant force, but only the book has zero speed.