Speed, Velocity and Acceleration

Speed, velocity and acceleration are the first calculation-heavy ideas in 0625, and they feed directly into graphs, forces and momentum. Examiners test them every session because the definitions sound similar but mean different things, and the unit conversions catch students out.

What is the difference between speed and velocity?

Speed is the distance travelled per unit time. Velocity is speed in a given direction, so velocity is a vector and speed is a scalar. A car doing 30 m/s north and a car doing 30 m/s south have the same speed but different velocities. Extended students must use this distinction: a body moving in a circle at constant speed has a changing velocity because its direction changes.

Quantity	Symbol	Equation (words)	Equation (symbols)	Unit
Speed	$v$	speed = distance ÷ time	$v = \dfrac{s}{t}$	m/s
Average speed	$v$	total distance ÷ total time	$v = \dfrac{s}{t}$	m/s
Acceleration	$a$	change in velocity ÷ time taken	$a = \dfrac{\Delta v}{\Delta t}$	m/s²

Acceleration is the change in velocity per unit time. A negative value means deceleration. Core students calculate acceleration from speed-time data; Extended students must also state that acceleration is a vector and link $F = ma$ to it.

How do you convert km/h to m/s?

Divide by 3.6. One kilometre is 1000 m and one hour is 3600 s, so $72\ \text{km/h} = 72\,000\ \text{m} \div 3600\ \text{s} = 20\ \text{m/s}$. This single conversion appears in roughly one paper in three, often hidden inside a longer question.

Worked Exam Question

A train accelerates uniformly from 8.0 m/s to 20.0 m/s in 30 s. (a) Calculate the acceleration of the train. [3] (b) State what is meant by the term acceleration. [1]

Solution (a). Equation: $a = \dfrac{\Delta v}{\Delta t}$. Substitute: $a = (20.0 - 8.0) \div 30 = 12.0 \div 30$. Answer: $a = 0.40\ \text{m/s}^2$.

Solution (b). Acceleration is the change in velocity per unit time.

Mark scheme

• M1: $a = \dfrac{\Delta v}{\Delta t}$ or $\dfrac{v - u}{t}$ stated or used.
• M1: correct substitution, $(20.0 - 8.0)/30$.
• A1: $0.40\ \text{m/s}^2$ with unit.
• B1 (b): change in velocity per unit time (not “speeding up”).

Common Mistakes

• Using 20.0 instead of the change in velocity. That gives $0.67\ \text{m/s}^2$ and loses two marks. Fix: always subtract the initial velocity first.
• Writing m/s for acceleration. Fix: acceleration is always m/s²; check the unit before moving on.
• Defining acceleration as “getting faster”. Examiners reject this. Fix: write “change in velocity per unit time”.
• Mixing km/h with seconds. Fix: convert to m/s (divide by 3.6) before substituting.
• Calling deceleration impossible with positive speed. A car slowing from 30 m/s to 10 m/s has negative acceleration. Fix: keep the sign of $\Delta v$.

Exam Technique Tip

For every motion calculation, write four lines: equation in symbols, substitution with numbers, the rearrangement if needed, then the answer with its unit. Even with a wrong final answer, the stated equation and substitution usually secure the M1 marks, typically two of the three marks on offer.

How This Is Examined

This subtopic spans Core and Extended. Papers 1 and 2 test it with MCQs on average speed and unit choice. Papers 3 and 4 set 3-4 mark calculations like the one above, and Paper 4 adds vector reasoning: scalar versus vector classification and changing velocity in circular motion. Paper 6 brings it back through ticker-tape or light-gate experiments. Worth knowing for Malaysian candidates: highway speed signs in km/h make ideal conversion practice. The PLUS limit of 110 km/h is 30.6 m/s. Sitting Extended? Expect speed, velocity and acceleration to combine with $F = ma$ in multi-step questions.