S3 Science | Forces and Motion: Tutor Prep Guide

S3 Science | Forces and Motion: Tutor Prep Guide | Speed, Graphs and Net Force

2026-11-13

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Students can say “fast” but not convert it: 15 metres per second and 54 kilometres per hour are the same speed, and many treat them as two different answers. This lesson lands three things at once, units, graph slope and the meaning of net force.

Objectives — Students can: (1) divide distance by time to get speed and convert 15 m/s into 54 km/h by multiplying by 3.6; (2) read a distance-time graph and state that the slope is the speed, so a horizontal line means at rest rather than constant speed; (3) decide motion from the net force: 30 N pushing against 12 N of friction leaves 18 N, so a 6 kg mass accelerates at 3 m/s squared, while zero net force means the velocity does not change, not that the object stops.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Units	How many metres in a kilometre, and seconds in an hour?	1,000 m and 3,600 s; a wrong answer here breaks every conversion
Prior idea	On a distance-time graph, what separates constant speed from rest?	Constant speed is a rising line, rest is horizontal
Vocabulary	What is the difference between mass and weight?	Mass is in kilograms, weight is a force in newtons

02 Lesson flow (45 minutes)

Time	Stage	Tutor does	Student does
0-5 min	Vocabulary check	Ask what “fast” means and write the student wording on the board	Give their own version, errors allowed
5-13 min	Speed and units	Write 120 m and 8 s, let them divide, then multiply by 3.6	Get 15 m/s and 54 km/h
13-22 min	Slope of the graph	Draw a rising, a horizontal and a steep line, ask the speed for each	State slope equals speed and horizontal means rest
22-31 min	Net force	30 N push, 12 N friction, 6 kg: net force first, then acceleration	Get 18 N and 3 m/s squared
31-40 min	Stopping distance	0.8 s reaction plus 4 s braking, calculated in two parts	Get 12 m plus 30 m, so 42 m
40-45 min	Exit check	Collect three answers and mark correct, partial or wrong	Hand in three answers

03 Teaching script (read it out)

Scene 1: Act 1: speed is a division, not a feeling

“120 metres in 8 seconds is 120 divided by 8, so 15 metres per second. Put that into everyday language: multiply by 3.6 and you get 54 kilometres per hour, a town driving speed. Where does 3.6 come from? An hour holds 3,600 seconds, so 15 metres each second is 54,000 metres in an hour. Do not memorise the factor without being able to rebuild it.”

Leave the line on the board: 15 m/s equals 54 km/h, with 3,600 s beside it.

Scene 2: Act 2: the slope of the graph is the speed

“On a distance-time graph the horizontal axis is time and the vertical is distance. A steeper line covers more distance in the same time, so the slope is the speed. A horizontal line means the distance is not changing, which is rest, not constant speed. If the line falls, the object has turned back, so the velocity is negative or the direction must be stated. When you see a graph, do not calculate first: ask which part is steepest, which is flat and whether anything turned back.”

Have students trace the line with a finger, naming each speed, flat means rest last.

Scene 3: Act 3: net force decides acceleration, not how fast something goes

“A 30 N push against 12 N of friction leaves 18 N to the right. With a mass of 6 kg the acceleration is 18 divided by 6, so 3 metres per second squared. Emphasise that net force tells you whether the velocity changes, not how large it is. Zero net force can mean an object at rest, or an object cruising at 20 m/s: both have zero net force. So never write ‘no force, so it stops’. Write ‘zero net force, so the velocity stays the same’.”

Write the pair on the board: zero net force leaves velocity unchanged, non-zero changes it.

04 Question types and worked steps

Type	Example	Steps	Answer
Speed calculation	A car covers 120 m in 8 s. Average speed, and in km/h?	1) distance over time 2) unit m/s 3) multiply by 3.6 for km/h	15 m/s, which is 54 km/h
Graph reading	What does a horizontal section of a distance-time graph show?	1) check the vertical axis 2) no change in distance 3) state rest, not constant speed	At rest, speed zero
Net force	30 N push, 12 N friction, mass 6 kg. Acceleration?	1) net force (add same direction, subtract opposite) 2) F equals ma 3) state the direction	Net force 18 N; 3 m/s squared in the direction of the push
Multi-stage	Reaction time 0.8 s and 4 s of braking from 15 m/s. Total stopping distance?	1) reaction distance is speed times reaction time 2) braking distance uses average speed 3) add the two	12 m plus 30 m, so 42 m

Sanity check (teach this): Sanity checks: 120 divided by 8 must give 15, and 1.5 is a decimal slip; 15 times 3.6 must give 54, while 5.4 or 540 is a place-value error; 500 g is 0.5 kg and weighs 5 N, so 500 N forgets the conversion; 20 m/s to rest in 5 s is minus 4 m/s squared, and plus 4 gets the direction wrong; 18 N over 6 kg gives 3, and 48 means friction was never subtracted; 42 m is the stopping distance, and 30 m drops the reaction distance.

05 Independent practice (3 questions)

Question	Answer	What to watch
A car covers 120 m in 8 s. What is its average speed in m/s?	15 m/s	Was the division used and the unit written?
Convert that speed into km/h.	54 km/h	Multiplied by 3.6, not divided?
30 N push, 12 N friction, mass 6 kg: what is the acceleration?	3 m/s squared	Was the net force found before dividing by mass?

06 Common mistakes and what to say

Mistake	Student writes	Say this
Mixing up the units	“15 m/s means 15 km/h”	“15 metres every second is 54,000 metres every hour, so 54 km/h. The factor 3.6 is derived, not memorised.”
Reading a flat line as constant speed	“It is flat here, so it is moving at constant speed”	“Flat means the distance is not changing, so it has stopped. Constant speed still has to rise, just with a smaller slope.”
Treating zero net force as no motion	“Net force is zero, so it stops”	“Zero net force means the velocity does not change. Standing still is one case, cruising at 20 m/s is another.”
Forgetting the reaction distance	“The braking distance is the stopping distance”	“The car already moved before the brakes bit: 0.8 s at 15 m/s is 12 m. Leaving it out loses more than a tenth of the answer.”

07 Homework (paper, section and question numbers)

Science chapter index: redo one speed and one net force question from this chapter — Forces and Motion chapter, practice questions 1 and 2
Science chapter index: draw a distance-time graph and label the speed for three sections — Forces and Motion chapter, practice question 3

08 Exit check (last 5 minutes)

Question	Answer	Decision
What is 15 m/s in km/h?	54 km/h	Answering 15 or 150 means the conversion is not secure
What does a horizontal line on a distance-time graph show?	Rest, with zero speed	Answering constant speed is a concept error
Does zero net force always mean an object is at rest?	No, the velocity is unchanged and it can cruise	Answering yes confuses net force with speed

09 Tutor FAQ

Which year is this for?

S3, and a useful warm-up for S4 physics. The focus is speed, slope and net force, not a list of formulas.

Is 45 minutes enough for three acts?

Yes, roughly 8 to 9 minutes each, leaving time for stopping distance and the exit check.

Is there a practice set?

Yes, the chapter index links the chapter practice with answers on the page.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.

Practice papers　Paper 1 (basics)　·　Paper 2 (intermediate)　·　Paper 3 (challenge)　·　Paper 4 (error clinic)