Define speed, velocity, distance, displacement, and acceleration, and solve simple linear problems involving these terms.

Misjudge a stopping distance or a startup acceleration and you get a vehicle that can't brake in the space available or a drive that trips on overspeed. Mixing up these five terms is the single most common error candidates make on the exam — and the same slip undersizes a brake or misreads a ramp rate on the plant floor. Get the definitions, the graph, and the equation choice right and the arithmetic is easy.

Variable cheat-sheet — keep this open while you compute. Every symbol is taught in context below; this is your quick reference so you never scan backward mid-problem.

The structure: two contrasted pairs plus acceleration. Linear motion is movement in a straight line. Two of the terms come as scalar-vs-vector pairs — distance vs displacement, and speed vs velocity — and acceleration stands on its own as the rate of change of velocity. Hold that map in mind: pair, pair, then acceleration. It tells you which terms you must never swap.

Distance vs displacement. Distance is the total length of path travelled — magnitude only, a scalar. Displacement is the change in position relative to a reference point, with direction — a vector. Walk 500 m to the store and 500 m back: the distance is 1000 m, but the displacement is zero, because you ended where you started.

Speed vs velocity. Speed is the rate of covering distance (a scalar); velocity is speed in a stated direction (a vector). Because real journeys vary, we usually work with averages: $\text{average speed} = \frac{\text{total distance}}{\text{total time}}$ and $\text{average velocity} = \frac{\text{displacement}}{\text{time}}$. When motion is straight with no change of direction, distance equals displacement and average speed equals average velocity.

Worked example — average speed vs average velocity. A body goes 10 m east, 8 m north, then 13 m east, ending 20 m due north of its start, in 10 s.

• Distance = 10 + 8 + 13 = 31 m, so $\text{average speed} = \frac{31\ \text{m}}{10\ \text{s}} = 3.1$ m/s.
• Displacement = 20 m north, so $\text{average velocity} = \frac{20\ \text{m}}{10\ \text{s}} = 2$ m/s north.
• The two answers differ because the path was not straight — speed counts every metre walked, velocity counts only the net move.

Acceleration — the rate of change of velocity. Acceleration is how fast velocity changes per unit time, in m/s$^2$. An increase in speed is positive acceleration; a decrease is negative acceleration (deceleration). Its defining equation is $a = \frac{v - u}{t}$, where u is the initial velocity, v the final velocity, and t the time taken.

Drawing a velocity-time graph. Plot velocity up the vertical axis and time along the horizontal axis. A body moving at constant velocity draws a flat horizontal line; a body accelerating uniformly from rest draws a straight line sloping up from the origin. The slope of that line is the acceleration, and the line shows at a glance whether the body is steady, speeding up, or slowing down.

The area under the graph is the displacement. This is the one fact examiners test most on graphs: the area between the velocity line and the time axis equals the displacement. For constant velocity the area is a rectangle, $s = v \times t$. For velocity rising uniformly from zero the area is a triangle, $s = \tfrac{1}{2}\,(\text{base} \times \text{height}) = \tfrac{1}{2}\,v\,t$, which is just the average-velocity form. Keep the scales consistent — m/s with seconds, or km/h with hours — or the area comes out wrong.

Worked example — read displacement off the graph. A drive starts from rest and ramps uniformly to 30 m/s over 8 s. Sketch v (vertical) against t (horizontal): a straight line from (0, 0) up to (8 s, 30 m/s).

• The area under it is a triangle with base = 8 s and height = 30 m/s.
• $s = \tfrac{1}{2}\,(\text{base} \times \text{height}) = \tfrac{1}{2}\,(8 \times 30) = 120$ m.
• Cross-check with the average-velocity form: $\bar v = \frac{0 + 30}{2} = 15$ m/s, so $s = 15 \times 8 = 120$ m — same answer, because the triangle area IS the average velocity times time.

The four equations of uniform (constant) acceleration. When acceleration is constant, four equations link u, v, t, s and a:

$s = \frac{u + v}{2}\,t \qquad v = u + at \qquad v^2 = u^2 + 2as \qquad s = ut + \tfrac{1}{2}at^2$

They hold ONLY for constant acceleration — if a changes during the motion, none of them apply.

How to pick the right equation — match it to your knowns. Don't memorize which example uses which; choose by what you have and what you want. List your knowns, spot the one variable that is missing, then pick the equation that leaves it out.

• No acceleration given, but you have u, v and t? → $s = \frac{u + v}{2}\,t$ (average-velocity form).
• Have a, u, t and want v (or have a, u, v and want t)? → $v = u + at$.
• Have u, a, s and want v but NOT t? → $v^2 = u^2 + 2as$.
• Have u, a, t and want s? → $s = ut + \tfrac{1}{2}at^2$.

Worked example — start from rest, find distance. A body starts from rest and reaches 30 m/s in 8 s under uniform acceleration. Knowns: u = 0, v = 30 m/s, t = 8 s; want s; no a given → use the average-velocity form.

• $\bar v = \frac{u + v}{2} = \frac{0 + 30}{2} = 15$ m/s.
• $s = \bar v \times t = 15 \times 8 = 120$ m.

Worked example — a dropped object, find impact velocity. An object is dropped from rest from 20 m. Knowns: u = 0, a = g = 9.81 m/s$^2$, s = 20 m; want v; no t → use $v^2 = u^2 + 2as$.

• $v^2 = 0 + 2 \times 9.81 \times 20 = 392.4$ m$^2$/s$^2$.
• $v = \sqrt{392.4} = 19.81$ m/s.

Worked example — braking, find acceleration. A vehicle at 90 km/h brakes uniformly to rest over 50 m. First convert: $90 \div 3.6 = 25$ m/s. Knowns: u = 25 m/s, v = 0, s = 50 m; want a; no t → use $v^2 = u^2 + 2as$.

• $0 = 25^2 + 2 \times a \times 50 = 625 + 100a$.
• $a = \frac{-625}{100} = -6.25$ m/s$^2$. The negative sign shows deceleration — keep it.

Common misconceptions and exam traps.

• Treating distance and displacement as the same. They are equal only for straight-line motion with no reversal.
• Forgetting unit conversions: km/h to m/s means multiplying by 1000/3600 (divide by 3.6). 90 km/h = 25 m/s.
• On a velocity-time graph, confusing slope with area: the SLOPE is the acceleration, the AREA under the line is the displacement.
• Mixing scales on a graph — plotting km/h against seconds — which makes the area meaningless. Match m/s with seconds, or km/h with hours.
• Applying the motion equations when acceleration is not constant — they hold only for uniform acceleration.
• Dropping the negative sign on deceleration, which flips a braking answer into an impossible speed-up.

Source: PanGlobal Fourth Class, Part A, Unit A-1, Chapter 5 (Linear Velocity and Acceleration); SOPEEC 4th Class Paper 4A.