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A.1 Kinematics · Part 1 of 3 · student investigation · 80 minutes

From position to the equations of motion

Start with a race you predict, learn to read motion from graphs, build the four equations of motion yourself, and then use them to settle the race. Part 2 takes these ideas into free fall and projectiles.

1 Predict · 7
2 Describe · 8
3 s–t graphs · 10
4 Tangents · 8
5 Acceleration · 7
6 Build suvat · 14
7 Choose · 10
8 Apply · 10
9 Show + reflect · 6

Learning goals: by the end you can…

tell distance from displacement and average speed from average velocity.
read velocity from the gradient of a position–time graph.
explain how a tangent gives the instantaneous velocity.
use the signs of v and a to say if an object speeds up or slows down.
derive the four equations of motion from a velocity–time graph.
choose the right equation and solve multi-step problems.

How to work in this lesson

How to work. Pairs discuss and agree, but each of you writes. Pairs → individual discuss the method, then write your own answer. Individual no discussion until your teacher says so. Your typed answers are saved on this device only.

Everything you need is on this page: each stage has its own simulation, and some stay locked until you have made your prediction. Extra help after class: the A.1 visual guide. ATL key: Thinking = predict / analyse / explain · Communication = explain / share · Social = collaborate · Self-management = plan / monitor / reflect · Research = observe / use evidence.

1. Start with a prediction

7 minPairs

ATL Thinking: predict and justify. Social: compare your prediction with a partner.

The race. Car A passes a start line at a steady 10 m s⁻¹. At that moment car B starts from rest at the same line and speeds up steadily at 2.5 m s⁻².

1.1Think alone first: who is in front after 4.0 s?

A. Car A B. Car B C. They are level

Commit to an answer. You will check it in stage 8.

Check your answer

Car A. After 4.0 s, A has gone 40 m and B only 20 m (you will calculate this in stage 8).

1.2Will car B ever catch car A? If yes, guess when.

Check your answer

Yes. B keeps getting faster, so it must catch A in the end. In stage 8 you will find it happens at t = 8.0 s, 80 m from the line.

1.3What information did you use to make your guess? What would you need to be sure?

What a good answer looks like

We used the speeds and the acceleration. To be sure we need an equation for the position of each car at time t, then we can find when the positions are equal.

2. Describing motion

8 minPairs

ATL Thinking: apply definitions. Communication: agree on wording with your partner.

A jogger on a straight path runs from the origin to s = +300 m, turns, and stops at s = +100 m. The whole run takes 100 s (steady 5.0 m s⁻¹).

2.0Predict: for the whole run, which is bigger?

the average speed the size of the average velocity they are equal

Simulation · Follow the jogger move the time slider or press Play

time t 0 s

position s—

distance so far—

displacement so far—

average speed so far—

average velocity so far—

2.1Total distance =

2.2Displacement =

2.3Average speed =

m s⁻¹

2.4Average velocity =

m s⁻¹

Back check

2.bCheck the time a second way. The jogger runs 300 m out and 200 m back at a steady 5.0 m s⁻¹. Time out + time back =

2.5Write one sentence that explains why your answers to 2.3 and 2.4 are different.

They are different because ________ , so ________ .

What a good answer looks like

They are different because the jogger turned back, so the distance (500 m) is much bigger than the displacement (100 m). Speed uses distance; velocity uses displacement.

3. Reading a position–time graph

10 minPairs → individual

ATL Thinking: analyse a graph. Research: take evidence from data.

Simulation · Read the position–time graph drag the time marker

time t 2.0 s

position s—

velocity (gradient here)—

distance travelled so far—

Each small square is 1 s across and 5 m up. Read the gradient of each straight part.

3.1Velocity from 0 to 4 s =

m s⁻¹

3.2Velocity from 4 to 6 s =

m s⁻¹

3.3Velocity from 6 to 10 s =

m s⁻¹

3.4Average speed for the whole 10 s =

m s⁻¹

Back check

3.bReverse 3.3. Start at s = 20 m at t = 6 s and move at your velocity from 3.3 for 4 s. Position at t = 10 s =

3.5Describe the whole motion in words, as if to someone who cannot see the graph.

What a good answer looks like

The object moves away from the origin at a steady 5.0 m s⁻¹ for 4 s, reaching 20 m. It stops there for 2 s. Then it comes back at a steady 5.0 m s⁻¹ and reaches the origin at 10 s.

4. From average to instantaneous

8 minPairs

ATL Thinking: spot a pattern and generalise. Research: use the chord simulation as evidence.

A trolley's position is s = t² (s in m, t in s). Find its average velocity from t = 3.0 s to a later time, making the interval smaller each time.

4.0Predict: as the time interval gets smaller and smaller, the average velocity…

falls towards zero settles on one value keeps growing

Simulation · Shrink the chord on s = t² the chord starts at t = 3.0 s

end of the interval t₂ 5.0 s

∆t—

∆s—

average velocity ∆s/∆t—

Interval	Δs / m	Average velocity / m s⁻¹
3.0 → 5.0 s	25 − 9 = 16
3.0 → 4.0 s	16 − 9 = 7
3.0 → 3.5 s	12.25 − 9 = 3.25
3.0 → 3.1 s	9.61 − 9 = 0.61

4.1What value is the average velocity getting closer to? What is it called?

m s⁻¹

4.2On an s–t graph, what line has this gradient?

What a good answer looks like

The tangent to the curve at t = 3.0 s. As the interval shrinks, the chord between the two points turns into the tangent, so instantaneous velocity = gradient of the tangent.

5. Acceleration and signs

7 minIndividual

ATL Thinking: apply a rule to new cases. Self-management: check your own reasoning.

5.1A ball rolls with v = −6.0 m s⁻¹. 2.0 s later v = −2.0 m s⁻¹. Its acceleration =

m s⁻²

5.2Is the ball speeding up or slowing down?

Speeding up, because a is positive Slowing down, because v and a have opposite signs Neither: a positive acceleration with negative velocity is impossible

Hinge question

Choose your answer. You will see at once if it is right, and why.

A velocity–time graph is a straight line that goes down from +12 m s⁻¹ at t = 0, crosses zero at t = 3.0 s and reaches −6.0 m s⁻¹ at t = 4.5 s. Between t = 3.0 s and t = 4.5 s the object is…

A. slowing down, because the acceleration is negative B. speeding up while moving backwards C. slowing down while moving backwards D. not moving, because the velocity is negative

Simulation · Test it: velocity and acceleration with signs right is positive

starting velocity u −6.0 m s⁻¹acceleration a +2.0 m s⁻²

time—

velocity v—

speed—

the ball is…—

5.3A friend says: "Negative acceleration always means slowing down." Write a short reply with an example.

What a good answer looks like

Not always. A ball dropped from rest, with up positive, has v negative and a = −9.8 m s⁻²: both negative, so it speeds up. The sign of a gives the direction of the change in velocity; you must compare it with the sign of v.

6. Build the equations of motion

14 minPairs → individual

ATL Thinking: derive general results from one example. Communication: explain each step.

A cart speeds up steadily from u = 4.0 m s⁻¹ to v = 16 m s⁻¹ in t = 6.0 s.

Simulation · Areas under the velocity–time graph start with u = 4.0, v = 16, t = 6.0

u 4.0 m s⁻¹v 16 m s⁻¹t 6.0 s

a = gradient = (v − u)/t—

rectangle u × t—

triangle ½ × t × (v − u)—

total area = s—

6.1Acceleration a = (v − u)/t =

m s⁻²

6.2Displacement = area under the graph (a trapezium, ½(u + v)t) =

6.3Now split the area: rectangle (u × t) = m, triangle (½ × t × at) = m

The general versionThe triangle's height is v − u = at, so its area is ½ × t × at = ½at². Adding the rectangle: s = ut + ½at². Then take t = (v − u)/a and put it into s = ½(u + v)t: s = (v − u)(v + u)/2a, so v² = u² + 2as.

6.4Test v² = u² + 2as with this cart: v = √(4.0² + 2 × 2.0 × 60) =

m s⁻¹

Back check

6.bCheck 6.2 another way. With constant acceleration, the average velocity is halfway between u and v. Average velocity = s/t from 6.2 =

m s⁻¹

6.5All four equations need one condition to be true. What is it, and why?

What a good answer looks like

The acceleration must be constant. We used a straight-line v–t graph (constant gradient) to get the trapezium and triangle. If a changes, the graph curves and these shapes are wrong.

7. Choose the right equation

10 minIndividual

ATL Thinking: choose a strategy. Self-management: check each answer before moving on.

📘 From the data booklet (A.1, constant acceleration)

s = ((u + v)/2)tv = u + ats = ut + ½at²v² = u² + 2as

You do not need to memorize these: they are given in the data booklet in the exam. Your job is to choose the right one. Each equation leaves out one of s, u, v, a, t.

Known: ___ · Wanted: ___ · Not involved: ___ → use the equation without it.

For each question: first choose the equation from the list, then solve.

7.1u = 5.0 m s⁻¹, a = 3.0 m s⁻², t = 4.0 s. Find v.

Equation:

m s⁻¹

7.2From rest, a = 2.0 m s⁻² over 25 m. Find v.

Equation:

m s⁻¹

7.3Slows from 20 m s⁻¹ to rest in 5.0 s. Find s.

Equation:

7.4u = 3.0 m s⁻¹, a = 2.0 m s⁻², t = 6.0 s. Find s.

Equation:

Back check

7.bCheck 7.2 in two steps. Find the time with v = u + at, then use s = ½(u + v)t. s =

Find the mistake

A student wrote:

“A car brakes from 20 m s⁻¹ to rest with a deceleration of 4.0 m s⁻². v² = u² + 2as → 0 = 20² + 2 × 4.0 × s → s = −50 m. The minus sign means the car went backwards.”

What is wrong?

Nothing is wrong: a negative distance just means the car moved backwards. The acceleration should be −4.0 m s⁻²: the car slows down, so a is opposite to v. Then s = +50 m. The student should have used s = ut + ½at², because v² = u² + 2as only works when speeding up.

8. Apply: settle the race

10 minPairs → individual

ATL Thinking: build a model and test a prediction. Research: compare with your stage 1 guess.

Simulation · Run the race car A: steady 10 m s⁻¹ · car B: from rest, 2.5 m s⁻²

time t 0.0 s

position of A—

position of B—

speed of B—

in front—

Worked startSame start line and same clock. Car A: s_A = 10t. Car B: s_B = ½ × 2.5 × t² = 1.25t².

8.1Positions at t = 4.0 s: s_A = m, s_B = m

8.2Set s_A = s_B and solve. B catches A at t =

8.3Speed of B at that moment =

m s⁻¹

Back check

8.bBack check 8.2: put t = 8.0 s into s_B = 1.25t². s_B =

8.4New problem: a driver at 25 m s⁻¹ has a reaction time of 0.60 s and brakes at 5.0 m s⁻². Total stopping distance =

What if? · stretch

8.wA new race. Car B starts from rest at 2.5 m s⁻², but only 2.0 s after car A (steady 10 m s⁻¹) passes the line. Measure t from when A passes the line. B catches A at t =

8.5Look back at stage 1. Was your prediction right? What changed your thinking?

What a good answer looks like

A good answer compares the guess with the result (A leads at 4 s, B catches up at 8.0 s and 80 m) and names the idea that helped, for example: "writing a position equation for each car".

9. Show it, then reflect

6 minIndividual

ATL Communication + Self-management: choose how to show your understanding, retrieve without notes, plan your next step.

Success criteria: your answer must…

link the gradient of s–t to v and the gradient of v–t to a;
say that the area under v–t is the displacement;
state the condition for the equations of motion;
use one number from today.

Option A · Written

One paragraph: "How the race can be solved from a graph and from equations."

Option B · Sketch

Sketch s–t and v–t graphs for both cars on the same axes and label where B catches A.

Option C · Spoken

Explain stage 6 to your partner in 1 minute. Your partner ticks the success criteria.

9.1Write your paragraph or your notes.

What a good answer looks like

Model answer (Option A): On an s–t graph, car A is a straight line (constant gradient = 10 m s⁻¹) and car B is a curve that gets steeper (its velocity increases). They meet where the graphs cross, at 8.0 s and 80 m. On a v–t graph, A is horizontal and B is a line with gradient a = 2.5 m s⁻²; the area under each line is the displacement, and the areas are equal (80 m) at 8.0 s. The equations of motion give the same result, because B's acceleration is constant.

Exit ticket (no notes)

1A walk: 40 m east, then 10 m west. Displacement in m (east +) =

2Gradient of an s–t line from (0 s, 5 m) to (5 s, 30 m), in m s⁻¹ =

3From rest, a = 4.0 m s⁻² for 3.0 s. Distance in m =

4Brakes from 12 m s⁻¹ to rest in 24 m. Size of deceleration in m s⁻² =

5The area under a velocity–time graph gives the…

acceleration displacement average speed

6Which idea from today is still the least clear to you?

What a good answer looks like

There is no single right answer. A good answer names one idea exactly, for example: "I'm not sure when to use v² = u² + 2as."

7Feedforward: Next time I will check ________ before I ________.

What a good answer looks like

A good feedforward names an action: "Next time I will write s, u, v, a, t with signs before I choose an equation."

TOK · 2 minutes

Theory of knowledge: how do we know? There is no single right answer. Write two or three sentences.

TOKThe equations of motion are exact only when the acceleration is constant, and in real life it never is perfectly constant. Why do physicists still trust and use a model that they know is not exactly true?

Points you could make

A model is a simplification that keeps the most important physics and leaves out small effects.
It is useful when its error is smaller than the uncertainty of our measurements; we state the assumption (constant a) and check it.
We stop using it where it fails, for example when air resistance is large.

Go to Part 2: falling, flying and drag Read the notes