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

Curves, tangents and drag

Real motion rarely has constant acceleration. Analyse a falling paper cone whose acceleration keeps changing, find instantaneous acceleration from a tangent and distance from an area, then throw a stone below the horizontal.

1 Predict · 6
2 Observe · 10
3 Tangents · 12
4 Areas · 10
5 Why it curves · 8
6 Below the horizontal · 14
7 Show it · 12
8 Reflect · 8

Learning goals: by the end you can…

tell uniform from non-uniform acceleration on a velocity–time graph.
find instantaneous acceleration from the gradient of a tangent, and average acceleration from a chord.
find displacement from the area under a curved graph, and explain why the equations of motion do not apply.
explain, with forces, why the acceleration of a falling object decreases to zero.
solve problems with projectiles launched below the horizontal.

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. Use g = 9.8 m s⁻². 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 notes and the 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

6 minPairs

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

Warm-up · 3 minutes · no notes

From the last lesson. Answer from memory, then check.

W1A ball is thrown straight up at 9.8 m s⁻¹. Maximum height =

W2A ball rolls off a table 1.8 m high at 2.0 m s⁻¹. Horizontal distance to where it lands =

W3A skydiver falls at terminal speed and then opens her parachute. Just after it opens, her acceleration is…

zero, because she is at terminal speed upwards, because drag is now bigger than her weight 9.8 m s⁻² downwards

1.1The velocity–time graph of a car is a curve that rises but gets flatter and flatter. What is happening to the car?

A. It is slowing down B. It is speeding up, but its acceleration is decreasing C. It is moving with constant acceleration

Commit to an answer. You will test it in stage 2.

Check your answer

B. The velocity is still increasing (the graph rises), so the car speeds up. The graph gets flatter, so the gradient, which is the acceleration, gets smaller.

1.2Could you use v = u + at to find the car's velocity after 3 s? Why or why not?

Check your answer

No. The equations of motion only work when the acceleration is constant. Here the acceleration changes, so there is no single value of a to use. You need the graph instead.

2. Observe: a falling paper cone

10 minPairs

ATL Research: take measurements from data. Thinking: find a pattern.

A paper cone is dropped from rest. A motion sensor records its velocity every 0.10 s. Run the simulation to watch the sensor collect the data in the table.

Simulation · Drop the paper cone the sensor records v every 0.10 s

time—

velocity—

acceleration now—

t / s	v / m s⁻¹	average acceleration in the next 0.10 s / m s⁻²
0.00	0.00	7.7
0.10	0.77	4.8
0.20	1.25
0.30	1.54
0.40	1.72
0.50	1.83	—
0.60	1.89	—
0.80	1.96	—

Back check

2.bBack check the table: add the five accelerations 7.7, 4.8, 2.9, 1.8 and 1.1 m s⁻², each lasting 0.10 s. Total gain in velocity =

m s⁻¹

2.1Describe how the acceleration changes. Was your prediction in 1.1 right?

What a good answer looks like

The acceleration decreases all the time (7.7 → 4.8 → 2.9 → 1.8 → 1.1 m s⁻²) while the velocity still increases. This is non-uniform acceleration, like the car in 1.1.

2.2The velocity seems to level off. Estimate the speed it is heading towards, in m s⁻¹.

3. Explain: tangents and chords

12 minPairs → individual

ATL Thinking: tell two kinds of acceleration apart. Communication: use the words tangent and chord correctly.

The ruleOn a curved v–t graph, the instantaneous acceleration at one moment is the gradient of the tangent at that point: a straight line that just touches the curve. The average acceleration between two times is the gradient of the chord: the straight line joining the two points on the curve.

Simulation · Tangent and chord on the cone’s v–t graph move P and the end of the chord

time of point P 0.20 schord from 0 to 0.40 s

tangent: two marked points—

gradient of the tangent (a at P)—

gradient of the chord (average a)—

3.1Use the two marked points on the dashed tangent. The instantaneous acceleration at P (t = 0.20 s) =

m s⁻²

3.2Average acceleration from t = 0 to t = 0.40 s (use the table: the chord from (0, 0) to (0.40, 1.72)) =

m s⁻²

Back check

3.bIs 3.7 m s⁻² reasonable? Use a short chord centred on P: from t = 0.10 s to 0.30 s in the table. Gradient =

m s⁻²

3.3Your answers to 3.1 and 3.2 are different. Explain why both are correct.

What a good answer looks like

3.1 is the acceleration at one moment (t = 0.20 s). 3.2 is the average over 0.40 s: it includes the large accelerations near the start, so it is bigger. They would only be equal if the acceleration were constant (a straight graph).

3.4Exam tip: when you draw a tangent yourself, why should you make it long and read two points far apart?

What a good answer looks like

Points far apart give a large rise and run, so small reading errors make only a small percentage error in the gradient. Never use the point P itself and the origin: that gives a chord, not the tangent.

4. Build: area under a curved graph

10 minIndividual

ATL Thinking: break a hard area into simple shapes. Self-management: check that an answer is reasonable.

Hinge question

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

On the cone’s curved v–t graph, a student draws a straight line from the origin to the point at t = 0.30 s and finds its gradient. This gradient is…

A. the instantaneous acceleration at t = 0.30 s B. the average acceleration from t = 0 to t = 0.30 s C. the distance fallen in the first 0.30 s D. the velocity at t = 0.30 s

The displacement is still the area under the v–t graph, even when it is curved. Split the area from 0 to 0.40 s into four strips 0.10 s wide and treat each one as a trapezium: area = ½(v₁ + v₂) × 0.10.

Simulation · Area in strips the area under v–t is the distance fallen

number of strips from 0 to 0.40 s 4

sum of the strips—

true area under the curve—

area under the straight line ½(u + v)t—

4.1Strip areas, in m: 0–0.1 s: 0.0385 · 0.1–0.2 s: 0.101 · 0.2–0.3 s: · 0.3–0.4 s:

4.2Distance fallen in the first 0.40 s =

Back check

4.bIs 0.44 m reasonable? The cone is never faster than 1.72 m s⁻¹ in the first 0.40 s. Largest possible distance = 1.72 × 0.40 =

4.3A student uses s = ½(u + v)t = ½(0 + 1.72) × 0.40 = 0.34 m. Explain why this is too small.

What a good answer looks like

That equation assumes constant acceleration: a straight line from 0 to 1.72 m s⁻¹. The real curve lies above that line (the cone speeds up quickly at first), so the real area, and the real distance, is larger.

On a printed graph you can also count squares: find what one small square stands for (for example 0.1 m s⁻¹ × 0.05 s = 0.005 m) and multiply by the number of squares under the curve, counting part-squares as halves.

5. Why does the graph curve?

8 minPairs

ATL Thinking: explain a graph with forces. Research: test your answers with the simulation.

5.1At the very start (t = 0) the cone is not moving yet. Its acceleration is…

zero g, because there is no drag yet less than g

Simulation · Forces on the falling cone weight is constant; drag grows with speed

time—

drag ÷ weight—

acceleration—

5.2Explain, using forces, why the acceleration decreases and the graph levels off at about 2.0 m s⁻¹.

As the speed increases, the drag ________, so the resultant force ________. When drag equals ________, …

What a good answer looks like

As the speed increases, the drag increases, so the resultant force (weight − drag) decreases and so does the acceleration: the graph gets flatter. When drag equals the weight (in air, buoyancy is negligible), the resultant force is zero, the acceleration is zero and the speed stays at the terminal speed, about 2.0 m s⁻¹: the graph is horizontal.

5.3Describe the shape of the acceleration–time graph for the cone.

What a good answer looks like

It starts at 9.8 m s⁻² at t = 0 and decreases, quickly at first and then more slowly, getting closer and closer to zero. It never goes below zero, because the cone never slows down.

6. Apply: a launch below the horizontal

14 minPairs → individual

ATL Thinking: adapt a known method to a new case. Self-management: choose a sign convention and keep it.

The ruleMeasure the angle from the horizontal as usual. Below the horizontal, the vertical component points down. Taking down as positive for the whole fall keeps every number positive: uᵧ = +u sin θ and a = +9.8 m s⁻².

A stone is thrown from the top of a wall 15 m high at 6.0 m s⁻¹, 40° below the horizontal. Air resistance is negligible.

6.0Predict: compared with a stone thrown horizontally at the same speed, this stone lands…

sooner and closer to the wall sooner and further from the wall at the same time and place

Simulation · Throw from the wall wall 15 m high · down is positive for the fall

speed u 6.0 m s⁻¹angle (− below, + above the horizontal) −40°

uₓ—

uᵧ (down +)—

time to the ground—

distance from the wall—

landing speed—

6.1Components: uₓ = m s⁻¹, uᵧ (down) = m s⁻¹

6.2Time to reach the ground =

6.3Horizontal distance from the wall =

6.4Speed on landing (use v² = u² + 2gh for the speed, or combine the components) =

m s⁻¹

Back check

6.bCheck 6.4 with the components. Downward velocity on landing: vᵧ = uᵧ + gt with t from 6.2. vᵧ =

m s⁻¹

Find the mistake

A student wrote:

“Down is positive. The stone is thrown 40° below the horizontal, so uᵧ = −6.0 sin 40° = −3.86 m s⁻¹. 15 = −3.86t + 4.9t² → t = 2.19 s.”

What is wrong?

Nothing is wrong: the vertical component of a throw is always negative. With down positive, the downward component is +3.86 m s⁻¹. The student mixed two sign rules. Then t = 1.40 s. g should be −9.8 m s⁻², so the equation should be 15 = −3.86t − 4.9t².

6.5Thrown horizontally at the same 6.0 m s⁻¹, the stone takes 1.75 s and lands 10.5 m away, with the same landing speed. Explain each difference and the similarity.

What a good answer looks like

Thrown downwards, the stone already has a downward velocity, so it reaches the ground sooner (1.40 s instead of 1.75 s). It also has a smaller horizontal component (4.60 instead of 6.0 m s⁻¹) for less time, so it lands closer. The landing speed is the same because it starts with the same speed and falls the same height: the same kinetic energy is gained.

What if? · stretch

6.wHow high would the wall have to be so that the stone thrown at 6.0 m s⁻¹, 40° below the horizontal, lands 10.5 m from the wall (as far as the horizontal throw from 15 m)? Height =

7. Show it

12 minIndividual

ATL Communication + Thinking: choose how to show your understanding.

Success criteria: your answer must…

use the tangent for instantaneous values and the area for displacement;
say clearly when the equations of motion can and cannot be used;
give numbers with units, or an explanation with a force argument.

Option A · Challenge

A tangent to a curved v–t graph at t = 3.0 s passes through (1.0 s, 2.0 m s⁻¹) and (5.0 s, 10 m s⁻¹). Find the acceleration at 3.0 s, then explain why you cannot use it to find the velocity at 6.0 s with v = u + at.

Option B · Explain

Sketch the v–t and a–t graphs of a skydiver from jumping to terminal speed, and explain each shape with forces.

Option C · Compare

Two stones leave a cliff at the same speed: one 30° above and one 30° below the horizontal. Compare their times, landing distances and landing speeds, with reasons.

7.1Your answer.

What a good answer looks like

Option A: a = (10 − 2.0)/(5.0 − 1.0) = 2.0 m s⁻². This is the acceleration at one moment only; the graph is curved, so the acceleration changes and v = u + at would give a wrong answer. Use the graph instead.
Option B: v–t: starts with gradient g, gets flatter, levels off at terminal speed. a–t: starts at g, decreases towards zero. Drag grows with speed, so the resultant force falls until it is zero (drag = weight, as buoyancy in air is negligible).
Option C: the stone thrown below lands sooner and closer (same horizontal speed, less time); the one thrown above rises first, so it is in the air longer. Both land with the same speed: same starting speed and same drop in height.

8. Reflect

8 minIndividual

ATL Self-management: retrieve without notes, then plan your next step.

Exit ticket (no notes)

1A tangent to a curved v–t graph passes through (1.0 s, 2.0 m s⁻¹) and (5.0 s, 14 m s⁻¹). Instantaneous acceleration in m s⁻² =

2Each small square under a v–t curve stands for 0.5 m. There are about 30 squares. Displacement in m =

3A ball is thrown at 10 m s⁻¹, 30° below the horizontal. Downward component of its velocity in m s⁻¹ =

4A v–t graph rises and gets flatter. The acceleration is…

decreasing increasing zero

5When the acceleration changes, the equations of motion…

still work if you use the average acceleration for everything cannot be used: use gradients and areas of the graph only work for the distance

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 am not sure how to draw a tangent at the right angle."

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

What a good answer looks like

A good feedforward names an action: "Next time I will check whether the graph is straight before I use the equations of motion."

TOK · 2 minutes

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

TOKTwo students draw tangents to the same curve by eye and get slightly different velocities. Can knowledge that depends on human judgement still be scientific?

Points you could make

Some judgement cannot be avoided in measurement.
Science reduces its effect with methods: a large gradient triangle, repeats, averages, or a calculation instead of a drawing.
It is scientific when the judgement is stated, the uncertainty is given and other people can check and repeat it.

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