DSE Physics | Kinematics Lesson-Prep Guide (Tutor Edition) | v-t Graphs, Acceleration and Stopping Distance

2026-11-01

Tutor lesson-prep guides · Secondary Physics

Students can recite v = u + at but freeze when a graph appears. This lesson ties graph, equation and words together: slope is acceleration, area is displacement, and they are three views of one idea.

Objectives — Students can: (1) read acceleration from the slope of a v-t graph and displacement from its area; (2) split stopping distance into reaction and braking parts; (3) explain why doubling speed quadruples braking distance, and why (u+v)/2 only works at constant acceleration.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Speed vs velocity	Does speed have a direction?	No; velocity does
Gradient	How do you find the slope of a line?	Rise over run
Unit conversion	What is 72 km/h in m/s?	20 m/s

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Recall: s-t vs v-t	What each graph shows	Students state the difference
8-20 min	Slope = acceleration	0 to 10 m/s	Read the slope value
20-32 min	Area = displacement	Triangle area method	Obtain 25 m
32-44 min	Stopping distance	Reaction plus braking	Obtain 56 m
44-54 min	Answer format	Given, formula, substitute, unit	Rewrite one answer
54-60 min	Exit check	Three questions, marked at once	Independent work

03 Teaching script (read it out)

Scene 1: The v-t graph: slope is acceleration

Starting from rest at 2.0 m/s squared for 5 s gives 10 m/s. The slope of a v-t graph is acceleration; the area under it is displacement: half times 5 times 10 equals 25 m. The same motion on an s-t graph is an upward curve, not a straight line.

Make them do it: tabulate speed every second for 0-5 s, plot it, and check the area against the formula.

Scene 2: Three parts of a stopping distance

During the 0.8 s reaction time the car still travels at 20 m/s, covering 16 m. Then a = -5 m/s squared, t = 20 divided by 5 = 4 s, and the braking distance is 20 squared over (2 times 5) = 40 m. Total 56 m.

Compare: doubling the speed takes braking distance from 40 m to 160 m. Let students compute it themselves before you say the word “squared”.

Scene 3: Average velocity is only an average when acceleration is constant

The whole trip is 25 m in 5 s, so average velocity is 5 m/s, and the arithmetic mean of 0 and 10 is also 5 – but only because acceleration is constant. With uneven motion the two numbers differ, and total distance over total time is the safe method.

Exam line: “Because acceleration is constant, average velocity = (u+v)/2.” That sentence earns the mark.

04 Question types and worked steps

Type	Example	Steps	Answer
Graph reading	Displacement from a v-t graph	Count squares or use area	25 m
Calculation	Final velocity	v = u + at	10 m/s
Calculation	Stopping distance	Reaction plus braking	56 m
Explanation	Why double speed means four times the distance	v squared in the numerator	160 m

Sanity check (teach this): Fix units first: divide km/h by 3.6. Area below the time axis is negative displacement. Keep the minus sign on deceleration instead of silently changing the formula.

05 Independent practice (3 questions)

Question	Answer	What to watch
Displacement after 2.0 m/s squared for 5 s	25 m	Half a t squared
Total stopping distance at 20 m/s with 0.8 s reaction	56 m	16 plus 40
Braking distance after doubling the speed	160 m	Proportional to v squared

06 Common mistakes and what to say

Mistake	Student writes	Say this
Substituting km/h directly	Treating 72 km/h as 72 m/s	Divide by 3.6 first, or you are out by a factor of 3.6.
Reading a v-t graph as s-t	Calling the slope the speed	On a v-t graph the slope is acceleration and the area is displacement.
Forgetting the reaction distance	Answering only 40 m	The car still moves before braking: 16 m in 0.8 s.
Averaging velocities at any time	Using (u+v)/2 for changing acceleration	That only holds at constant acceleration; otherwise use total distance over total time.

07 Homework (paper, section and question numbers)

DSE Physics practice paper (mechanics section) — Questions 1-6
Secondary Physics chapter index — All chapters

08 Exit check (last 5 minutes)

Question	Answer	Decision
Displacement after 2.0 m/s squared for 5 s	25 m	Three correct: move on to forces
Stopping distance at 20 m/s	56 m	One wrong: re-teach the two stages
What does the area under a v-t graph give?	Displacement	Two or more wrong: restart act one

09 Tutor FAQ

Must students draw the graph?

Yes. Graph questions appear almost every year and sketching is a scoring move, not a crutch.

Why is reaction distance not part of braking?

The car has not started to slow down yet, so it is a separate constant-speed stage.

How do I handle negative area?

It is displacement in the opposite direction; keep the sign when adding, then discuss distance as a magnitude.

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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.