From a walking student to a falling skydiver: distance, speed, velocity and acceleration, motion graphs and the equations of motion, explained with diagrams and worked examples.
Sitting still in a train, you are not moving relative to your seat, yet you are moving at over a hundred kilometres per hour relative to the ground. Even then, the Earth carries both you and the ground around the Sun at about 30 kilometres every second. Motion is always described relative to something, and to describe it clearly physicists use a small set of precise ideas: distance, displacement, speed, velocity and acceleration. Once these are clear, you can analyse everything from a walking student to a falling raindrop and understand why speed limits exist.
Distance is the total length of the path an object travels. It is a scalar quantity: it has size only and is never negative. Displacement is the straight-line change in position from start to finish, together with its direction. It is a vector quantity: it has both size and direction.
Suppose a student walks 40 m east along a corridor and then 40 m back west. The distance travelled is 80 m, but the displacement is zero, because she finished where she started. Both are measured in metres (m).
| Scalar (size only) | Vector (size and direction) |
|---|---|
| distance | displacement |
| speed | velocity |
| mass, time, energy | force, acceleration, weight |
Speed is how fast an object covers distance.
speed = distance ÷ time (v = s ÷ t)
The SI unit is metres per second (m/s). Road speeds are often given in kilometres per hour (km/h).
km/h → m/s: divide by 3.6. m/s → km/h: multiply by 3.6.
Where does 3.6 come from? One kilometre is 1000 m and one hour is 3600 s, so 1 km/h = 1000 ÷ 3600 = 1 ÷ 3.6 m/s.
Most real journeys do not happen at one steady speed. The average speed is the total distance divided by the total time, including any stops. The instantaneous speed is the speed at one moment, which is what a car's speedometer shows.
| Moving thing | Typical speed (approximate) |
|---|---|
| Walking | 1.4 m/s (5 km/h) |
| Cycling | 5 m/s (18 km/h) |
| Car on a motorway | 30 m/s (about 110 km/h) |
| Passenger jet in cruise | 250 m/s (about 900 km/h) |
| Sound in air | 343 m/s |
| Earth around the Sun | about 30 000 m/s (30 km/s) |
| Light in a vacuum | 299 792 458 m/s (exact by definition) |
A bus travels 150 km in 2.5 hours, including a short break. Find its average speed in km/h and in m/s.
Step 1. v = s ÷ t = 150 km ÷ 2.5 h = 60 km/h.
Step 2. Convert: 60 ÷ 3.6 ≈ 16.7 m/s.
Velocity is the rate of change of displacement: velocity = displacement ÷ time. It tells you both how fast and in which direction. Two cars travelling at 20 m/s in opposite directions have the same speed but opposite velocities, which we show with signs: +20 m/s and −20 m/s.
The student in the corridor walked for 80 s in total, so her average speed was 80 m ÷ 80 s = 1 m/s, but her average velocity was 0 ÷ 80 = 0 m/s. An object moving at a constant speed around a circle has a velocity that keeps changing, because its direction keeps changing.
Relative velocity: if a car travelling at 80 km/h overtakes another travelling at 60 km/h in the same direction, the faster car moves at 20 km/h relative to the slower one. For two vehicles approaching each other at 80 km/h and 60 km/h, the closing speed is 140 km/h.
Acceleration is the rate at which velocity changes. A car speeding up, slowing down or turning a corner is accelerating.
a = (v − u) ÷ t
a = acceleration (m/s²), v = final velocity (m/s), u = initial velocity (m/s), t = time taken (s).
A negative acceleration (slowing down) is often called deceleration or retardation.
The unit m/s² means "metres per second, per second": each second the velocity changes by that many metres per second. An acceleration of 3 m/s² means that the velocity increases by 3 m/s every second.
A cyclist speeds up from 2.0 m/s to 8.0 m/s in 4.0 s. Find the acceleration.
a = (v − u) ÷ t = (8.0 − 2.0) ÷ 4.0 = 6.0 ÷ 4.0 = 1.5 m/s².
Graphs make motion easy to see. On a distance–time graph, time is on the horizontal axis and distance on the vertical axis.
On a velocity–time graph, the vertical axis shows velocity. It carries more information than a distance–time graph:
Find the acceleration in the first stage and the total distance travelled.
Step 1. Acceleration = gradient = (20 − 0) ÷ (5 − 0) = 4 m/s².
Step 2. Area of the first triangle: ½ × 5 × 20 = 50 m.
Step 3. Area of the rectangle: 10 s × 20 m/s = 200 m.
Step 4. Area of the last triangle: ½ × 5 × 20 = 50 m.
Step 5. Total distance = 50 + 200 + 50 = 300 m.
When an object moves in a straight line with constant acceleration, four equations connect the five quantities u (initial velocity), v (final velocity), a (acceleration), t (time) and s (displacement):
Want to see why these equations are true? Our companion article, Equations of Motion: Three Equations and Their Proofs, derives each one using algebra, velocity–time graphs, calculus and energy, with many more worked examples.
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t
Strategy: list what you know, identify what you want, and pick the equation that has exactly those quantities.
A car starts from rest and accelerates at 3.0 m/s² for 8.0 s. Find its final velocity and the distance covered.
Known: u = 0, a = 3.0 m/s², t = 8.0 s.
v = u + at = 0 + 3.0 × 8.0 = 24 m/s (about 86 km/h).
s = ut + ½at² = 0 + ½ × 3.0 × 8.0² = ½ × 3.0 × 64 = 96 m.
A car moving at 20 m/s brakes with a constant deceleration of 5.0 m/s² until it stops. Find the braking distance and time.
Known: u = 20 m/s, v = 0, a = −5.0 m/s².
v² = u² + 2as → 0 = 400 + 2(−5.0)s → 10s = 400 → s = 40 m.
v = u + at → 0 = 20 − 5.0t → t = 4.0 s.
An object falling with only gravity acting on it is in free fall. Near the Earth's surface, every freely falling object accelerates downward at the same rate, the acceleration due to gravity, g. The standard value is 9.80665 m/s², usually rounded to 9.8 m/s² (or 10 m/s² for quick estimates). Galileo is traditionally credited with arguing that, in the absence of air resistance, heavy and light objects fall together. In 1971 Apollo 15 astronaut David Scott demonstrated this on the Moon, where there is almost no air, by dropping a hammer and a feather; they landed at the same time.
A stone is dropped from rest and falls for 3.0 s. Ignoring air resistance, find its speed and the distance fallen (g = 9.8 m/s²).
v = u + at = 0 + 9.8 × 3.0 = 29.4 m/s.
s = ½at² = ½ × 9.8 × 3.0² = 44.1 m.
On Earth, air resistance matters. As an object speeds up, the upward air resistance grows. Eventually it equals the object's weight. The forces are then balanced, so the acceleration is zero and the object falls at a constant speed called the terminal velocity. A skydiver in a spread position reaches roughly 55 m/s (about 200 km/h). Opening a parachute greatly increases air resistance, so the skydiver slows to a new, much lower terminal velocity that is safe for landing.
A driver cannot stop instantly. The stopping distance has two parts:
A driver with a reaction time of 0.70 s is travelling at 20 m/s (72 km/h). The brakes give a deceleration of 5.0 m/s². Find the stopping distance. Then repeat it for 40 m/s (144 km/h).
At 20 m/s: thinking distance = 20 × 0.70 = 14 m; braking distance = 20² ÷ (2 × 5.0) = 40 m. Stopping distance = 54 m.
At 40 m/s: thinking distance = 40 × 0.70 = 28 m; braking distance = 40² ÷ 10 = 160 m. Stopping distance = 188 m.
Many factors make stopping distance longer: higher speed, a tired, distracted or impaired driver (longer reaction time), wet or icy roads, worn tyres and worn brakes, and a heavily loaded vehicle. The World Health Organization identifies speed as a key risk factor for road traffic injuries and recommends measures such as speed limits, seat belts and child restraints. Seat belts, airbags and crumple zones all work by increasing the time over which a person's velocity changes in a crash, which reduces the force.
A friend drops a ruler and you catch it after it has fallen 20 cm. Find your reaction time.
Step 1. From rest, s = ½gt², so t = √(2s ÷ g).
Step 2. s = 0.20 m, so t = √(2 × 0.20 ÷ 9.8) = √0.0408 ≈ 0.20 s.
A typical result is around 0.2 s for simple catching. Reaction times when driving are usually longer, because the driver must notice, decide and move a foot.
Try each question before opening the answer.
Distance = 400 m. Displacement = 0 m (she is back at the start). Average speed = 400 ÷ 50 = 8 m/s.
90 ÷ 3.6 = 25 m/s.
a = (30 − 10) ÷ 40 = 0.50 m/s².
The object is stationary (its distance is not changing).
The area is a triangle: ½ × 6 × 12 = 36 m.
At the top v = 0. 0 = 14.7 − 9.8t, so t = 14.7 ÷ 9.8 = 1.5 s.
Air resistance upward equals weight downward. The forces are balanced, so there is no resultant force and no acceleration.
Thinking distance doubles, but braking distance is proportional to the square of the speed, so it increases fourfold.
Easy experiments: time students walking and running over a measured corridor and plot distance–time graphs; roll a trolley down a ramp and use a phone video to read its positions; use the ruler-drop test for reaction times and compare results with and without a distraction; drop a flat sheet of paper and the same sheet crumpled to discuss air resistance. Never test braking with real vehicles, and keep drop experiments on the floor level or low. Numerical values are rounded; please check that they match the data booklet used by your examination board.
This article is original EDUSAMBAM writing. Definitions, constants and historical facts were checked against the sources below. Diagrams are original and were not copied from any source.
Note on the numbers: typical speeds, reaction times and braking values are rounded or illustrative. Please check them against your own syllabus or exam board specification before relying on them.
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