DP Physics SL A.1 Kinematics

DP Physics SL A.1 Kinematics - Full notes

RevisionPrep

DP Physics SL | A.1 Kinematics

DP PHYSICS · THEME A: SPACE, TIME AND MOTION

A.1

Kinematics

Revision notes for SL and HL · IB-style

Contents

1	Describing motion
2	Motion graphs
3	Equations of motion
4	Free fall
5	Projectile motion
6	Air resistance and terminal speed
7	Common misconceptions and links
8	Exam-style question with markscheme
9	Check your understanding
10	Summary and topic map

Work through one section per sitting. Every section ends with a Quick check: if you can answer it without looking back, move on.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe motion using displacement, velocity and acceleration, and tell distance apart from displacement
find average and instantaneous speed, velocity and acceleration, including from graphs
sketch and interpret s–t, v–t and a–t graphs for uniform and non-uniform acceleration
use the equations of motion for uniform acceleration, and recognise when I can't
analyse projectiles by splitting the motion into horizontal and vertical components
describe how air resistance affects a projectile's path, time of flight, velocity, acceleration, range and terminal speed
convert between s–t, v–t, a–t and v–s graphs for the same motion
measure g and find its uncertainty from a graph

There is no additional HL content in A.1. SL and HL students sit the same material.

Guiding questions

How can the motion of a body be described quantitatively and qualitatively? How can the position of a body in space and time be predicted? How can the analysis of motion in one and two dimensions be unified?

How this topic is examined

Kinematics turns up across the papers: quick multiple-choice items, data-based questions built on graphs or experimental results, and longer structured questions (often the opening parts of a mechanics question). These command terms come up most:

Command term	What the examiner wants
Calculate / Determine	a numerical answer with working and units; determine often means the method isn't obvious
Sketch	correct shape and key features (intercepts, turning points, gradients), axes labelled, no scale needed
Estimate	an approximate value, e.g. area by counting squares or a tangent drawn by eye
Show that	every step shown; the answer is given, so marks come only from working; give one more significant figure than shown
Outline / Explain	a short account / a reasoned chain linking cause and effect (e.g. drag → acceleration → height)

How to use these notes

You'll see these boxes throughout. Each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer is in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to see the physics
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Describing motion

Before you calculate anything, decide where zero is and which way is positive.

Think about it

Your fitness app says you ran 5 km this morning. You started and finished at your front door. What was your displacement? What was your average velocity?

Kinematics describes motion without asking what causes it (forces come in A.2). You need four quantities, a clear origin and one habit: decide which direction is positive before writing any numbers.

1.1 Position and frame of reference

Position is always measured from a chosen origin, along a chosen positive direction. Change either and the numbers change, though the physics doesn't. In most questions you pick the origin at the starting point. (Motion measured by observers who are themselves moving belongs to A.5, Galilean relativity.)

1.2 Key terms

Term	Meaning	Scalar or vector	Unit
Distance	Total length of the path travelled.	scalar	m
Displacement, s	Change in position: straight-line distance from start to finish, in a stated direction.	vector	m
Speed	Rate of change of distance.	scalar	m s−1
Velocity, v	Rate of change of displacement.	vector	m s−1
Acceleration, a	Rate of change of velocity.	vector	m s−2
Uniform acceleration	Acceleration constant in size and direction.	vector	m s−2

1.3 Distance and displacement

Figure 1. The path length is the distance; the straight arrow from start to finish is the displacement.

•One lap of a 400 m track: distance 400 m, displacement 0. Average speed is large, average velocity is zero.
•A ball thrown up 5 m and caught at the same height: distance 10 m, displacement 0.
•Distance never decreases. Displacement can go up, down and become negative.

Key point

If the question asks for a velocity or a displacement, the answer needs a direction (or a sign). A bare number is a speed or a distance, and loses the mark.

1.4 Average and instantaneous values

Both are in the data booklet. Used over a whole interval they give average values:

The instantaneous value is what is happening at one moment. It is the gradient of the tangent to the graph at that moment (section 2.4). A speedometer shows instantaneous speed; an average-speed camera measures Δs/Δt between two points.

WORKED EXAMPLE 1 Speed or velocity?

A student walks 30 m east in 20 s, then 10 m west in 10 s. Find (a) the distance, (b) the displacement, (c) the average speed, (d) the average velocity.

Solution

(a) distance = 30 + 10 = 40 mA1

(b) displacement = 30 − 10 = 20 m east (taking east as positive)A1

(d) average velocity = 20 ÷ 30 = 0.67 m s−1 eastA1

Leaving out "east" in (d) costs the mark even with the right number.

In the real world

Usain Bolt's 100 m world record of 9.58 s (Berlin, 2009) works out at an average speed of about 10.4 m s−1. His top instantaneous speed was nearer 12 m s−1. The first few seconds spent accelerating from the blocks pull the average down.

Common mistake

Averaging the stage speeds: (1.5 + 1.0) ÷ 2 = 1.25 m s−1. This is only right if the stages last the same time. Go back to total distance ÷ total time.

1.5 Signs of velocity and acceleration

Acceleration is any change in velocity, not just speeding up. Taking up (or forwards) as positive:

Situation	v	a	What is happening
Car moving forwards, speeding up	+	+	speed increasing
Car moving forwards, braking	+	−	speed decreasing
Car reversing, speeding up	−	−	speed increasing (backwards)
Ball rising after being thrown up	+	−g	slowing down
Ball falling	−	−g	speeding up
Ball at the top of its flight	0	−g	instantaneously at rest

Rule of thumb: v and a with the same sign means speeding up; opposite signs means slowing down. "Negative acceleration" does not automatically mean slowing down (look at the reversing car).

Key point

At the top of a vertical throw v = 0 but a is still g downwards. Writing a = 0 at the top is one of the most common errors in this topic.

1.6 Working with vectors

Displacement, velocity and acceleration are vectors, so they add and split like arrows. Two skills are needed in this topic: adding vectors tip to tail, and resolving one vector into two perpendicular components.

Figure 2. (a) Adding two perpendicular displacements. (b) Resolving a velocity into horizontal and vertical components.

•The component next to the angle uses cos; the one opposite the angle uses sin. If θ is measured from the vertical, they swap.
•Components of a vector along the same line simply add, with signs. That is why projectile problems become easy once everything is resolved.

WORKED EXAMPLE 2 Adding and resolving

(a) A hiker walks 3.0 km north then 4.0 km east. Find her displacement. (b) A ball is kicked at 25 m s−1 at 40° above the horizontal. Find the horizontal and vertical components of its velocity.

Solution

(a) size = √(3.02 + 4.02) = 5.0 kmA1

direction: θ = tan−1(4.0 ÷ 3.0) = 53° east of north (or a bearing of 053°)A1

(b) vx = 25 cos 40° = 19 m s−1; vy = 25 sin 40° = 16 m s−1A1 A1

Check: the components are both smaller than 25, and √(19.22 + 16.12) = 25 m s−1.

Common mistake

Adding vector sizes as if they were numbers: 3.0 + 4.0 = 7.0 km. Only vectors along the same line add like that. At right angles, use Pythagoras; state the direction as well.

Quick check

1. A car drives once round a circular track of radius 50 m. What are the distance and the displacement?

2. Can speed ever be negative? Can velocity?

3. A ball moves upwards (positive) with a = −9.8 m s−2. Is it speeding up or slowing down?

4. Back to the fitness app: what were your displacement and average velocity?

Answers: 1. 314 m (2πr), 0 2. speed no, velocity yes 3. slowing down (opposite signs) 4. both zero

2 Motion graphs

Gradient and area: two tools that unlock every motion graph.

Think about it

A dashcam loses its video but keeps its speed log. From the speed–time data alone, could you work out how far the car went, when it braked hardest, and whether it ever reversed?

Graph questions appear on almost every paper. Learn the shapes first, then two tools: gradient and area under the line.

2.1 Displacement–time graphs

Figure 3. Typical s–t shapes. The gradient at any point is the velocity.

•Horizontal line: at rest. Straight sloping line: constant velocity.
•Curve getting steeper: speeding up. Curve levelling off: slowing down.
•A maximum or minimum on an s–t graph is where the object stops and turns round (gradient zero, so v = 0).

2.2 Velocity–time graphs

Figure 4. Typical v–t shapes. The gradient is the acceleration; where the line crosses the t-axis the object changes direction.

•Horizontal line: constant velocity (a = 0). Straight sloping line: uniform acceleration.
•A line crossing the time axis means the velocity changes sign: the object reverses. It does not mean it has returned to the start.
•A curved v–t graph means the acceleration is changing (non-uniform).

Try this

Stand in front of an ultrasonic motion sensor (or use a free phone app such as phyphox) and try to walk so that you "draw" a given s–t graph. Walking away gives a positive gradient, walking back a negative one, standing still a flat line. Then try a v–t graph: it's much harder.

2.3 Gradients and areas

Figure 5. Uniformly accelerated motion shown as (a) s–t, (b) v–t and (c) a–t graphs.

Graph	Gradient gives	Area under the graph gives
s–t	velocity	nothing useful
v–t	acceleration	displacement
a–t	(not needed)	change in velocity

Units check: area under v–t is (m s−1) × s = m, which is why it gives displacement. Use this on any unfamiliar graph: multiply the axis units for area, divide them for gradient.

2.4 Instantaneous values from a curve

Figure 6. The chord gives an average value; the tangent at a point gives the instantaneous value.

1.Mark the point on the curve at the time asked for.
2.Place a ruler so it just touches the curve at that point, with equal "gaps" either side. Draw it long.
3.Choose two points on your tangent that are far apart (use the whole line, not the curve). Gradient = rise ÷ run, with units from the axes.

Common mistake

Drawing a line from the origin to the point and calling its gradient the instantaneous velocity. That gives the average from t = 0. The instantaneous value needs a tangent at that one point, drawn across at least half the grid so your reading is accurate.

WORKED EXAMPLE 3 Reading a v–t graph

A car starts from rest and moves as shown below. Calculate (a) the total distance travelled, (b) the average speed, (c) the acceleration between t = 10 s and t = 13 s.

Figure 7. Velocity–time graph for the car.

Solution

(a) Split the area into a triangle, a rectangle and a triangle:

½ × 4 × 12 + 12 × 6 + ½ × 3 × 12 = 24 + 72 + 18M1

distance = 114 mA1

(b) average speed = 114 ÷ 13 = 8.8 m s−1A1

Not 6 m s−1: the car spent most of the journey at 12 m s−1.

= −4.0 m s−2 (negative: the car is slowing down)A1

2.5 Area below the time axis

Figure 8. Ball thrown up at 14.7 m s⁻¹ (up positive). The two areas are equal and opposite.

Area below the axis is negative displacement. The ball rises 11 m then falls 11 m:

•displacement after 3.0 s = +11 − 11 = 0 (back at the hand)
•distance travelled = 11 + 11 = 22 m

For displacement, add areas with their signs. For distance, add their sizes.

2.6 Non-uniform acceleration

When the v–t graph is curved, the equations of motion don't apply. You still use the graph:

•acceleration at an instant: gradient of the tangent
•displacement: area under the curve, by counting squares (full squares, plus part squares paired up) or by splitting it into trapezium strips

WORKED EXAMPLE 4 Estimating from a curved graph

The graph shows the velocity of a cyclist starting from rest. Estimate (a) the distance travelled in the first 4.0 s, (b) the acceleration at t = 2.0 s.

Figure 9. Velocity–time graph for the cyclist, with 1 s trapezium strips and the tangent at t = 2.0 s.

Solution

(a) Read v at 1, 2, 3, 4 s: 3.9, 6.3, 7.8, 8.6 m s−1

area ≈ 1.0 × [½(0 + 8.6) + 3.9 + 6.3 + 7.8] = 22 mM1 A1

The curve bulges above each strip's straight top, so this is a slight underestimate (the true value is about 23 m).

(b) Tangent passes through (0.3, 3.2) and (3.7, 9.5):

gradient = (9.5 − 3.2) ÷ (3.7 − 0.3) ≈ 1.8 m s−2M1 A1

An answer in the range 1.6 to 2.0 m s−2 would normally be accepted for a hand-drawn tangent.

2.7 Sketching one graph from another

If the s–t graph is…	the v–t graph is…	the a–t graph is…
a horizontal line	zero	zero
a straight sloping line	a horizontal line	zero
a parabola (curving up)	a straight line sloping up	a positive horizontal line
a parabola with a maximum	a straight line crossing the axis at the maximum	a negative horizontal line

Work from left to right: the value on each graph is the gradient of the graph before it.

WORKED EXAMPLE 5 A lift journey

A lift starts from rest at the ground floor. It accelerates upwards uniformly at 1.5 m s−2 for 2.0 s, moves at constant velocity for 8.0 s, then decelerates uniformly to rest in 3.0 s. (a) Sketch the v–t and a–t graphs. (b) Find the height of the top floor.

Figure 10. (a) Velocity–time and (b) acceleration–time graphs for the lift.

Solution

(a) Top speed = 1.5 × 2.0 = 3.0 m s−1. Deceleration = 3.0 ÷ 3.0 = 1.0 m s−2, so the a–t graph shows +1.5, then 0, then −1.0 m s−2.A1 A1

Sharp steps on the a–t graph appear wherever the v–t graph has a corner.

(b) height = area under v–t = ½ × 2.0 × 3.0 + 8.0 × 3.0 + ½ × 3.0 × 3.0M1

= 3.0 + 24 + 4.5 = 31.5 m (about 32 m)A1

Link to A.2: during the first 2 s the passengers feel heavier, and during the last 3 s lighter. The a–t graph tells you exactly when.

2.8 Converting between graphs: the graph atlas

The same motion can be drawn four ways. Every graph below was calculated from the motion itself, not sketched by eye, so use them to check your own sketches. Read each row from left to right:

•s–t → v–t: the value on the v–t graph is the gradient of the s–t graph.
•v–t → a–t: the value on the a–t graph is the gradient of the v–t graph.
•v–t → s–t: the area under the v–t graph up to time t gives s at that time.
•v–t and s–t → v–s: pair up v and s at the same instant, so time disappears. The arrows on the v–s graphs show which way the point moves as time passes.

Figure 11. Graph atlas, part 1: four motions with uniform acceleration.

Figure 12. Graph atlas, part 2: a reversal, air resistance, changing acceleration and motion in the negative direction.

What to notice in the atlas

•A straight v–t line (uniform a) always gives a horizontal a–t line and a parabola on the s–t graph. It gives a curve on the v–s graph, never a straight line (unless a = 0).
•Where the v–t graph crosses zero, the s–t graph has a maximum or minimum, and the v–s graph doubles back on itself: s can't keep increasing while v is negative.
•Air resistance: v–t levels off at terminal speed, a–t falls to zero, and s–t becomes a straight line (constant velocity).
•When a ∝ t, the a–t graph is a sloping line, v–t is a parabola and s–t grows with t3. The equations of motion don't apply.
•Motion in the negative direction: s–t and v–t are below the axis; on the v–s graph the point moves left and down.

Key point

Build the set in order: v–t first (it holds the most information), then its gradient for a–t and its area for s–t. Only then pair v with s for the v–s graph.

2.9 Extension: velocity–displacement (v–s) graphs

v–s graphs aren't one of the listed graph types for A.1, but they turn up in multiple-choice and data questions built on v2 = u2 + 2as. Treat this as stretch material: it deepens your understanding and heads off some common traps.

Figure 13. How to read v–s graphs. Arrows show the direction of increasing time.

The gradient of a v–s graph is dv/ds, not the acceleration. Using the chain rule:

•acceleration = velocity × gradient of the v–s graph. Two points with the same gradient can have very different accelerations if their speeds differ.
•Uniform acceleration gives a curve, v = √(u2 + 2as). From rest it starts vertical at the origin and then flattens, even though a is constant (panel b).
•A straight sloping v–s line means a = kv: the acceleration grows as the object speeds up. It is not uniform acceleration (panel c).
•To get a straight line for uniform acceleration, plot v2 against s: gradient 2a, intercept u2 (panel f).
•The area under a v–s graph has no useful meaning. The area under a graph of 1/v against s does: it is the time taken.

Common mistake

Using a = v ÷ s, or reading the gradient of a v–s graph as the acceleration. Both are wrong: a = v × gradient. Be careful with infographics online: several popular ones draw a straight v–s line for "uniform acceleration from rest" and a flattening curve for "decreasing acceleration". It is the other way round.

WORKED EXAMPLE 6 Acceleration from a v–s graph

On the v–s graph of a braking car, the tangent at the point where v = 8.0 m s−1 has a gradient of −0.50 s−1. Find the car's acceleration at that point.

Solution

a = v × gradient = 8.0 × (−0.50)M1

= −4.0 m s−2A1

Units check: (m s−1) × (s−1) = m s−2. The gradient alone (−0.50 s−1) can't be an acceleration: its unit is wrong.

WORKED EXAMPLE 7 Finding a from a graph of v² against s

A trolley is pushed and then rolls down a ramp. A light gate measures its speed v after it has travelled a distance s from the release point. Use the results to find the trolley's acceleration and its initial speed.

s / m	0.20	0.40	0.60	0.80	1.00
v / m s−1	0.755	0.943	1.10	1.24	1.36
v2 / m2 s−2	0.570	0.889	1.21	1.54	1.85

Figure 14. Graph of v² against s for the trolley.

Solution

v2 = u2 + 2as, so a graph of v2 against s is a straight line (form y = c + mx)R1

gradient = 1.60 m s−2 = 2a, so a = 0.80 m s−2M1 A1

intercept = 0.25 m2 s−2 = u2, so u = 0.50 m s−1A1

Plotting v against s would give a curve, which is much harder to analyse. Choosing what to plot is the skill being tested.

Quick check

1. What does a horizontal line on a v–t graph tell you?

2. What are the units of the area under an a–t graph, and what does it represent?

3. An s–t graph has a maximum at t = 4 s. What is the velocity at that moment?

4. On a v–t graph, the line is below the time axis. Which way is the object moving?

5. For uniform acceleration from rest, which is a straight line: v against s, or v2 against s?

6. On a v–s graph, what do you get if you multiply the gradient by v?

Answers: 1. constant velocity (a = 0) 2. m s−1, change in velocity 3. zero (it turns round) 4. in the negative direction 5. v2 against s 6. the acceleration

3 Equations of motion

Four equations, five quantities, one condition: the acceleration must be constant.

Think about it

Why does doubling your speed much more than double your stopping distance? (The stopping-distance example below answers it.)

Five quantities, four equations. Each equation leaves one quantity out, and that is how you choose. All four are in the data booklet.

If you don't know, and don't need…	use
s	v = u + at
a	s = ½(u + v)t
v	s = ut + ½at2
t	v2 = u2 + 2as

3.1 Where they come from

Figure 15. Uniform acceleration from u to v in time t.

•Gradient: a = (v − u)/t, which rearranges to v = u + at.
•Area of the trapezium: s = ½(u + v)t.
•Split the area into a rectangle (ut) and a triangle (½ × t × at): s = ut + ½at2.
•Eliminate t between the first two: t = (v − u)/a, so s = (v + u)(v − u)/2a, giving v2 = u2 + 2as.

A "show that" question can ask for any of these steps.

3.2 Method

1.Draw a quick sketch and mark the positive direction.
2.Write down s, u, v, a, t. Fill in the three you know, with signs.
3.Pick the equation that leaves out the quantity you neither know nor want.
4.Check the answer is sensible (a car doesn't brake to rest in 400 m from 25 m s−1).

Key point

These equations only apply when acceleration is constant. If air resistance matters, or the v–t graph is curved, you can't use them.

WORKED EXAMPLE 8 Stopping distance

A driver travelling at 25 m s−1 sees a hazard. Her reaction time is 0.60 s, after which the car decelerates uniformly at 6.5 m s−2. Calculate the total stopping distance.

Solution

Thinking distance: brakes not yet applied, so constant speed. s = 25 × 0.60 = 15 mA1

Braking: u = 25, v = 0, a = −6.5, t not needed, so use v2 = u2 + 2as

0 = 252 + 2(−6.5)sM1

s = 625 ÷ 13 = 48 mA1

total stopping distance = 15 + 48 = 63 mA1

Notice: doubling the speed quadruples the braking distance (s ∝ u2), but only doubles the thinking distance.

In the real world

Road-safety stopping-distance charts use the same two ideas: constant speed while you react, then uniform deceleration. The braking part grows with u2, so at 40 m s−1 (144 km h−1) it is 16 times the braking distance at 10 m s−1.

Figure 16. Stopping distances for a 0.6 s reaction time and a deceleration of 6.5 m s⁻².

WORKED EXAMPLE 9 Catching up

A car is at rest at traffic lights. As the lights change, a truck passes it at a constant 15 m s−1. At the same instant the car sets off with a uniform acceleration of 2.5 m s−2. Find (a) the time for the car to catch the truck, (b) the distance travelled by then, (c) the speed of the car at that moment, (d) the greatest distance between them before the car catches up.

Figure 17. Displacement–time graphs for the car and the truck.

Solution

(a) Same displacement at the same time: 15t = ½ × 2.5 × t2M1

t = 15 ÷ 1.25 = 12 s (the root t = 0 is the start)A1

(b) s = 15 × 12 = 180 mA1

(d) The gap grows while the truck is faster and shrinks once the car is faster, so it is greatest when the speeds are equal: 2.5t = 15, t = 6.0 sR1

gap = 15 × 6.0 − 1.25 × 6.02 = 90 − 45 = 45 mA1

WORKED EXAMPLE 10 Two-stage motion: a model rocket

A model rocket is launched vertically from rest. Its engine gives it a uniform upward acceleration of 12 m s−2 for 5.0 s, then the fuel runs out. Ignoring air resistance, find (a) its speed and height when the engine stops, (b) the maximum height, (c) the total time until it hits the ground, (d) its speed on impact.

Figure 18. Velocity–time graph for the rocket (up positive).

Solution

Two stages with different accelerations: treat them separately. The end of stage 1 is the start of stage 2.

(a) v = 0 + 12 × 5.0 = 60 m s−1; s = ½ × 12 × 5.02 = 150 mA1 A1

(b) Stage 2: u = 60, v = 0, a = −9.8. Extra height = 602 ÷ (2 × 9.8) = 184 mM1

maximum height = 150 + 184 = 330 mA1

Fall from rest through 334 m: t = √(2 × 334 ÷ 9.8) = 8.3 sM1

total time = 5.0 + 6.1 + 8.3 = 19 sA1

(d) v = √(2 × 9.8 × 334) = 81 m s−1A1

Note the rocket keeps rising for 6.1 s after the engine stops. "Engine off" does not mean "starts falling".

Quick check

1. Which equation of motion does not contain t?

2. A car accelerates uniformly from rest to 20 m s−1 in 8.0 s. How far does it travel?

3. Can the equations of motion be used for a skydiver before she reaches terminal speed?

Answers: 1. v2 = u2 + 2as 2. 80 m 3. no: her acceleration is changing

Common mistake

Applying the deceleration in the stopping-distance example during the reaction time as well, or dropping the minus sign on a. Without the sign you get v2 = 625 + 13s, which has no sensible solution.

4 Free fall

Take away the air and everything falls with the same acceleration.

Think about it

In 1971, Apollo 15 astronaut David Scott dropped a hammer and a feather at the same moment while standing on the Moon. Which one reached the ground first?

An object in free fall has only gravity acting on it. Near the Earth's surface every object falls with the same acceleration, g = 9.8 m s−2 downwards (use the data booklet value), whatever its mass.

•Thrown up, dropped or thrown down: with up as positive, a = −g for the whole flight, including the top.
•Without air resistance, the time up equals the time down to the same level, and the object comes back at its launch speed.

Figure 19. Ball thrown straight up at 14.7 m s⁻¹, up positive: (a) s–t, (b) v–t, (c) a–t. The dotted line marks the top of the flight.

Read the three graphs together: the maximum of (a) happens where (b) crosses zero, and (c) is flat at −g throughout, including at the top.

Two results worth being able to derive (not in the data booklet):

H is the maximum height above the launch point and t the time to reach it.

WORKED EXAMPLE 11 Ball thrown up from a balcony

A ball is thrown vertically upwards at 14 m s−1 from a balcony 20 m above the ground. It misses the balcony on the way down. Find (a) its maximum height above the balcony, (b) the time until it hits the ground, (c) its speed as it hits the ground.

Solution

Up positive, origin at the balcony: a = −9.8 m s−2 and the ground is at s = −20 m.

(a) At the top v = 0: 0 = 142 + 2(−9.8)HM1

H = 196 ÷ 19.6 = 10 mA1

(b) One equation covers the whole flight because a is constant: −20 = 14t − 4.9t2M1

4.9t2 − 14t − 20 = 0, so t = [14 + √(196 + 392)] ÷ 9.8 = 3.9 sA1

speed = 24 m s−1 (downwards)A1

Common mistake

Writing s = +20 m for the ground because it is 20 m away. Displacement is measured from the launch point; with up positive, below the launch point is negative.

WORKED EXAMPLE 12 Where do they meet?

Ball A is dropped from the top of a 45 m tower. At the same instant ball B is thrown vertically upwards from the ground directly below at 15 m s−1. When and where do they pass each other?

Figure 20. Height against time for both balls.

Solution

Use one origin (the ground) and up as positive for both balls:

A: y = 45 − 4.9t2 B: y = 15t − 4.9t2M1

They meet when the heights are equal: 45 − 4.9t2 = 15t − 4.9t2M1

The t2 terms cancel: t = 45 ÷ 15 = 3.0 sA1

height = 45 − 4.9 × 3.02 = 0.9 m above the groundA1

Why the t2 terms cancel: both balls have the same acceleration, so relative to each other they move at a constant 15 m s−1 and close a 45 m gap in 3.0 s.

Answer: hammer and feather

The hammer and the feather landed together. The Moon has no atmosphere, so there was no air resistance, and both fell with the Moon's gravitational acceleration of about 1.6 m s−2. On Earth the feather is slowed by the air; the physics of falling isn't different, the air is.

Try this

Measure your reaction time with a 30 cm ruler. A friend holds it vertically with the zero mark between your open finger and thumb, then lets go without warning. Read the distance d it falls before you catch it. Your reaction time is t = √(2d / g): a catch at 20 cm means about 0.20 s. Do five catches and average.

4.1 A bouncing ball

Figure 21. Velocity–time graph of a ball dropped onto a hard floor (up positive).

•Every sloping line has the same gradient, −g: between bounces the ball is in free fall whether it is going up or down.
•At each bounce the velocity jumps from negative to positive in a very short time. That near-vertical line is a huge upward acceleration while the ball is squashed against the floor.
•Each rebound speed is lower than the impact speed, so each bounce reaches a lower height (energy is transferred to the surroundings as internal energy and sound).
•Area of each triangle above the axis = height of that bounce.

4.2 Measuring g

Figure 22. Measuring g: apparatus (left) and a graph of drop height against time squared (right).

A steel ball is released from an electromagnet and falls a height h onto a trapdoor; an electronic timer measures the fall time t. Repeat for several heights and take a mean time for each.

From s = ut + ½at2 with u = 0:

so a graph of h against t2 is a straight line through the origin with gradient g/2. Plotting a straight line lets you spot anomalies and systematic errors, which a single calculation can't.

WORKED EXAMPLE 13 Finding g from data

h / m	0.20	0.40	0.60	0.80	1.00
t / s	0.203	0.285	0.351	0.403	0.452
t2 / s2	0.0412	0.0812	0.1232	0.1624	0.2043

Use the data to determine g.

Solution

Plot h against t2 (Figure 22). Best-fit gradient = 4.91 m s−2M1

g = 2 × gradient = 9.8 m s−2A1

The line passes very close to the origin, so there is no significant systematic error (e.g. a delay in the magnet releasing the ball would shift the line sideways).

Uncertainty in g

Each time has an uncertainty of about ±0.005 s, so each t2 has an uncertainty of about ±2tΔt (percentage uncertainty doubles when you square). Draw these as horizontal error bars, then draw the steepest and shallowest lines that still pass through all the bars.

Figure 23. Error bars with best-fit, steepest and shallowest lines.

steepest gradient = 5.11 m s−2; shallowest gradient = 4.72 m s−2

uncertainty in gradient = ½(5.11 − 4.72) = 0.20 m s−2M1

g = 2 × (4.91 ± 0.20) = 9.8 ± 0.4 m s−2A1

The accepted value, 9.81 m s−2, lies within this range, so the result is consistent with it.

Improving the experiment: use a small dense ball so air resistance is negligible; measure h to the bottom of the ball; use a wide range of heights; repeat each drop and average.

4.3 Data-based question: video analysis of a projectile

Video analysis software gives the position of a ball frame by frame. Here the camera recorded a ball thrown across a lab at 10 frames per second. The origin is the launch point, up positive.

t / s	0.10	0.20	0.30	0.40	0.50	0.60
x / m	0.24	0.48	0.72	0.96	1.20	1.44
y / m	0.251	0.404	0.459	0.416	0.275	0.036
(y/t) / m s−1	2.51	2.02	1.53	1.04	0.55	0.06

WORKED EXAMPLE 14 Extracting g from video data

(a) Show that the horizontal velocity is constant and find it. (b) Explain why a graph of y/t against t is a straight line, and use it to find g and the initial vertical velocity. (c) Find the launch speed and angle.

Figure 24. Graphs from the video data: (a) x against t, (b) y/t against t.

Solution

(a) x increases by 0.24 m every 0.10 s, so x–t is a straight line through the origin: ux = 2.4 m s−1A1

(b) y = uyt − ½gt2, so dividing by t: y/t = uy − (g/2)t, the form y = c + mxR1

intercept = uy = 3.0 m s−1; gradient = (0.06 − 2.51) ÷ (0.60 − 0.10) = −4.9 m s−2M1

g = 2 × 4.9 = 9.8 m s−2A1

Linearising a quadratic like this is a skill the data-based questions test often: rearrange so the unknowns become a gradient and an intercept.

Quick check

1. A stone is dropped from rest. How fast is it moving after 3.0 s?

2. How far has it fallen in that time?

3. What is the acceleration of a ball at the very top of a vertical throw?

4. In the h against t2 experiment, what is the gradient equal to?

Answers: 1. 29 m s−1 2. 44 m 3. g downwards 4. g/2

5 Projectile motion

Split it into horizontal and vertical, solve each on its own, and let time connect them.

Think about it

A bullet is fired horizontally from a rifle. At exactly the same moment, a second bullet is dropped from the same height. Ignoring air resistance, which hits the ground first?

A projectile is launched and then moves under gravity alone. The key idea: horizontal and vertical motions are independent. The only thing they share is the time.

Figure 25. Multiflash images: a dropped ball and a ball launched horizontally at the same instant.

The two balls are at the same height at every instant. Horizontal motion has no effect on vertical motion. The horizontally launched ball also covers equal horizontal distances in equal times: its horizontal velocity doesn't change.

	Horizontal	Vertical
acceleration	0	−g
velocity	constant, ux = u cos θ	starts at uy = u sin θ, then vy = uy − gt
displacement	x = uxt	y = uyt − ½gt2
controls	how far it travels	how long it is in the air

Key point

Nearly every projectile question works like this: use one direction to find t, then use t in the other direction.

5.1 Launched horizontally

•uy = 0, so vertically the object behaves exactly as if dropped.
•Time in the air depends only on the height: h = ½gt2. Launching faster makes it land further away, not later.

WORKED EXAMPLE 15 Ball rolling off a table

A ball leaves a table 1.25 m high horizontally at 3.0 m s−1. Find (a) the time to reach the floor, (b) how far from the table it lands, (c) its velocity as it lands.

Solution

(a) Vertically (down positive): uy = 0. 1.25 = ½ × 9.8 × t2M1

t = 0.51 sA1

(b) Horizontally: x = 3.0 × 0.505 = 1.5 mA1

v = √(3.02 + 4.952) = 5.8 m s−1, at tan−1(4.95 ÷ 3.0) = 59° below the horizontalA1 A1

5.2 Launched at an angle

Figure 26. Velocity vectors along the path. The horizontal component is the same everywhere; the vertical component shrinks, reverses and grows.

Figure 27. Velocity components against time for the same projectile.

Resolve first: ux = u cos θ, uy = u sin θ. For level ground and no air resistance:

•Time of flight: y = 0 when uyt − ½gt2 = 0, so T = 2u sin θ / g.
•Maximum height: vy = 0 at the top, so 0 = (u sin θ)2 − 2gH.
•Range: R = uxT = u cos θ × 2u sin θ / g = u2 sin 2θ / g (using 2 sin θ cos θ = sin 2θ).

These are not in the data booklet, and only work for level ground. Be ready to derive them rather than quote them.

Figure 28. Paths for the same launch speed at different angles.

•Maximum range at θ = 45°. Angles adding to 90° give the same range, but the steeper one goes higher and stays up longer.
•At the top vy = 0 and vx is unchanged. The speed there is u cos θ, not zero.
•The path is a parabola: x ∝ t while y depends on t2.

Velocity at any moment: combine the components.

WORKED EXAMPLE 16 Free kick

A football is kicked from the ground at 18 m s−1 at 35° above the horizontal, towards a goal 28 m away. The crossbar is 2.44 m high. Ignoring air resistance, find (a) the time of flight and range on level ground, (b) the maximum height, (c) whether the ball passes under the crossbar.

Solution

ux = 18 cos 35° = 14.7 m s−1; uy = 18 sin 35° = 10.3 m s−1M1

(a) T = 2 × 10.3 ÷ 9.8 = 2.1 s; R = 14.7 × 2.11 = 31 mA1 A1

(b) H = 10.32 ÷ (2 × 9.8) = 5.4 mA1

Height then: y = 10.3 × 1.90 − 4.9 × 1.902 = 1.9 mA1

1.9 m < 2.44 m (and above zero), so the ball passes under the bar, already on its way down (top at 1.05 s).R1

WORKED EXAMPLE 17 Two angles for the same range

A ball is launched from level ground at 20 m s−1 to land on a target 30 m away. Find the two possible launch angles and the time of flight for each.

Figure 29. Two paths that reach the same target.

Solution

R = u2 sin 2θ / g: sin 2θ = 30 × 9.8 ÷ 202 = 0.735M1

2θ = 47.3° or 180° − 47.3° = 132.7°, so θ = 24° or 66°A1 A1

T = 2u sin θ / g: 1.6 s at 24°, 3.7 s at 66°A1

Both answers are physically correct. The question's context decides which to give (e.g. a low, fast shot to get under a ceiling).

In the real world: why not 45°?

Long jumpers take off at around 20°, not 45°. The 45° result assumes the launch speed is the same at every angle, but a sprinter can't keep their run-up speed while driving steeply upwards, and they land with their centre of mass lower than at take-off. Real sport rarely matches the textbook assumptions exactly.

5.3 Launched at an angle from a height

The shortcuts for T, H and R no longer work because the landing point is below the launch point. Go back to the component equations and use the full vertical displacement.

WORKED EXAMPLE 18 Stone thrown from a cliff

A stone is thrown from the top of a 40 m cliff at 15 m s−1, 30° above the horizontal. Find (a) the time until it hits the sea, (b) the horizontal distance from the cliff, (c) its speed on impact.

Figure 30. Sketch of the cliff throw. Always draw one.

Solution

ux = 15 cos 30° = 13.0 m s−1; uy = 15 sin 30° = 7.5 m s−1. Up positive, sea at y = −40 m.M1

(a) −40 = 7.5t − 4.9t2 → 4.9t2 − 7.5t − 40 = 0M1

t = [7.5 + √(7.52 + 4 × 4.9 × 40)] ÷ 9.8 = 3.7 sA1

(b) x = 13.0 × 3.72 = 48 mA1

Quick check: v2 = u2 + 2gh = 225 + 784 = 1009, so v = 32 m s−1. The launch angle doesn't affect the impact speed.

5.4 The equation of the path

Eliminating t between x = u cos θ t and y = u sin θ t − ½gt2 gives the trajectory directly:

It is not in the data booklet, but it is a quick derivation: substitute t = x/(u cos θ) into the y equation. It is useful when a question gives a horizontal distance and asks for a height, or the reverse, without mentioning time. The x2 term with a negative coefficient is why the path is a downward parabola.

WORKED EXAMPLE 19 Basketball shot

A player releases a basketball 2.0 m above the floor at 45° to the horizontal. The hoop is 3.05 m above the floor and 6.0 m away horizontally. Find the launch speed needed for the ball to pass through the hoop (ignore air resistance).

Figure 31. Sketch for the basketball shot.

Solution

Origin at the release point: the hoop is at x = 6.0 m, y = 3.05 − 2.0 = 1.05 mA1

1.05 = 6.0 tan 45° − 9.8 × 6.02 ÷ (2u2 cos2 45°)M1

1.05 = 6.0 − 352.8 ÷ u2 → u2 = 352.8 ÷ 4.95 = 71.3M1

u = 8.4 m s−1A1

Classic demonstration: the monkey and the hunter

Aim a launcher straight at a target that is released and starts to fall at the very instant you fire. It hits, whatever the launch speed (as long as it reaches that far). Without gravity the projectile would travel along the straight line to the target; with gravity, both the projectile and the target fall the same distance below that line in the same time.

Common mistake

Using the full 18 m s−1 (or 15 m s−1) in the horizontal equation instead of u cos θ, or using R = u2 sin 2θ / g when the landing height differs from the launch height. Resolve on your first line and only use the shortcut formulae on level ground.

Keep unrounded values in your calculator through multi-step working and round only the final answer.

Quick check

1. What is the horizontal acceleration of a projectile with no air resistance?

2. Which launch angle gives the greatest range on level ground?

3. A ball is thrown horizontally at 10 m s−1 from a height of 20 m. How long is it in the air?

4. Back to the bullets: which lands first?

Answers: 1. zero 2. 45° 3. 2.0 s 4. they land together; the vertical motions are identical

6 Air resistance and terminal speed

Drag grows with speed. That one fact explains terminal speed and every change to a projectile's path.

Think about it

Raindrops fall from clouds a kilometre or more above you. With no air resistance, a drop falling 1 km would arrive at about 140 m s−1. Why do real raindrops hit you at under 10 m s−1?

Air resistance (drag) opposes the motion and increases with speed. In A.1 you describe and explain its effects; you don't calculate them. The acceleration is no longer constant, so the equations of motion can't be used.

6.1 Forces on a moving object

Figure 32. Forces on a ball rising, falling, and falling at terminal speed.

•Rising: drag and weight both point down. The resultant is bigger than the weight, so the deceleration is greater than g.
•Falling: drag points up, against the weight. The resultant is smaller than the weight, so the acceleration is less than g.
•Terminal speed: drag has grown until it equals the weight. Resultant zero, acceleration zero, constant velocity.

What affects the size of the drag force?

Factor	Effect on drag	Example
speed	drag increases as speed increases (much faster than in proportion at high speed)	a cyclist feels far more drag at 40 km h−1 than at 20 km h−1
cross-sectional area	larger area, more drag	a skydiver spreads out to fall more slowly
shape	streamlined shapes have less drag	racing cyclists crouch; cars are tested in wind tunnels
density of the fluid	denser fluid, more drag	a ball falls far more slowly through water than air

Mass does not change the drag force at a given speed, but it changes the weight that drag has to balance. That is why mass affects terminal speed.

6.2 Terminal speed: the skydiver

Figure 33. Velocity–time graph for a skydiver (down positive). The parachute opens at about 18 s.

Stage	What is happening	Why
A	steep, nearly straight line; a ≈ g	just after leaving the plane, speed is low so drag is small
B	curve flattens	speed rises, drag rises, resultant force and acceleration fall
C	horizontal: terminal speed	drag = weight, resultant force zero
D	very steep drop in v	parachute opens: drag suddenly exceeds weight, resultant force upwards, so she decelerates
E	new, lower terminal speed	drag falls as she slows until drag = weight again

Common mistake

Saying the skydiver "goes up" when the parachute opens. Her velocity stays downward throughout stage D; only her acceleration points upwards. The upward movement on a video is the camera operator falling faster.

Terminal speed is higher for objects with more weight for a given size and shape (a dense ball falls faster than a feather-light one of the same size), and lower for larger, less streamlined shapes.

In the real world

In 2012 Felix Baumgartner jumped from about 39 km above the Earth and became the first person to break the sound barrier in free fall. The air at that height is so thin that drag is tiny, so his speed kept rising far beyond a normal skydiver's terminal speed of roughly 55 m s−1. As he fell into thicker air, drag increased and slowed him down before he opened his parachute.

6.3 A vertical throw with air resistance

Figure 34. Velocity–time graph for a ball thrown straight up at 20 m s⁻¹, with and without air resistance (up positive).

•At the start the graph is steeper than −g: drag and weight both act downwards.
•The graph crosses the axis earlier: the ball reaches a lower maximum height sooner.
•At the very top, v = 0 so drag = 0 and the acceleration is exactly g for an instant. The gradient there matches the dashed line.
•On the way down the graph is less steep than −g and it lands with a smaller speed than it was thrown with. The area below the axis equals the area above (it returns to the same height), so the falling part lasts longer.

6.4 Projectiles with air resistance

Figure 35. The same launch with and without air resistance.

	No air resistance	With air resistance
path	symmetrical parabola	not symmetrical; comes down more steeply than it went up
maximum height	(u sin θ)2 / 2g	lower
range	u2 sin 2θ / g	shorter; best launch angle is less than 45°
time up / time down	equal	time down is longer than time up
time of flight	2u sin θ / g	usually shorter overall
horizontal velocity	constant	decreases throughout
acceleration	g downwards, constant	changes in size and direction: greater than g rising, less than g falling
landing speed (level ground)	same as launch speed	less than launch speed

Key point

Rising, drag adds to gravity; falling, drag works against it. Every row of the table follows from this one idea.

Common mistake

Answering "explain why the maximum height is lower" with just "because of air resistance". Give the chain: drag acts downwards on the way up, so the deceleration is greater than g, so the vertical velocity reaches zero sooner, so the ball rises less high.

Quick check

1. What is the resultant force on an object falling at terminal speed?

2. Which has the higher terminal speed: a sheet of paper falling flat or the same sheet crumpled into a ball?

3. With significant air resistance, is the time going up or the time coming down longer?

4. Back to the raindrops: why do they arrive so slowly?

Answers: 1. zero 2. the crumpled ball (smaller area) 3. coming down 4. they reach a low terminal speed within a few metres, when drag equals their tiny weight

7 Common misconceptions and links

Students often think…	Actually…
acceleration means speeding up	acceleration is any change of velocity, including slowing down or changing direction
negative acceleration means slowing down	only if the velocity is positive; a negative acceleration speeds up an object moving in the negative direction
at the top of a throw, a = 0	v = 0 there, but a = g downwards throughout
heavier objects fall faster	without air resistance, all objects fall with acceleration g
the area under an s–t graph means something	it doesn't; use its gradient
a v–t line crossing the axis means the object is back at the start	it means the object changes direction; the start is where the total signed area is zero
a projectile's speed is zero at the top	only the vertical component is zero; the speed is u cos θ
throwing a projectile faster horizontally keeps it in the air longer	time of flight depends only on the vertical motion
the equations of motion always work	only for constant acceleration
an object at terminal speed has no forces acting on it	forces act but they balance; the resultant is zero
opening a parachute makes the skydiver move upwards	she keeps moving down but decelerates
the gradient of a v–s graph is the acceleration	a = v × gradient; for uniform a the v–s graph is a curve, and v2 against s is the straight line

How A.1 connects to the rest of the course

Topic	Connection
A.2 Forces and momentum	Newton's second law explains why objects accelerate; drag and terminal speed are analysed with forces there
A.3 Work, energy and power	v2 = u2 + 2as is closely linked to energy conservation (½mv2 gained = mgh lost in free fall)
A.4 Rigid body mechanics (HL)	the same equations reappear for rotation, with angular displacement, velocity and acceleration
A.5 Galilean and special relativity	velocities depend on the observer's frame of reference
D.1 Gravitational fields	g is the gravitational field strength; projectiles become orbits at very high speeds
C.1 Simple harmonic motion	displacement, velocity and acceleration graphs for oscillations use the same gradient and area ideas

8 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

Figure 36. A tennis serve (not to scale).

A tennis player serves the ball horizontally at 30 m s−1 from a height of 2.6 m. The net is 12 m away and 0.91 m high. The service line is 18.3 m from the player. Ignore air resistance unless told otherwise.

(a)Calculate the time taken for the ball to reach the net. [1]
(b)Show that the ball clears the net by about 0.9 m. [3]
(c)Determine whether the ball lands before the service line (i.e. "in"). [3]
(d)Calculate the speed of the ball as it hits the ground. [2]
(e)In reality air resistance acts on the ball. Explain its effect on the point where the ball lands. [3]

Markscheme

Part	Answer	Marks
(a)	t = 12 ÷ 30 = 0.40 s	A1
(b)	vertical drop in 0.40 s = ½ × 9.8 × 0.402 = 0.784 m	M1
	height at the net = 2.6 − 0.784 = 1.82 m	A1
	clearance = 1.82 − 0.91 = 0.91 m (≈ 0.9 m)	A1
(c)	time to fall 2.6 m: t = √(2 × 2.6 ÷ 9.8) = 0.73 s	M1
	horizontal distance = 30 × 0.728 = 22 m	A1
	22 m > 18.3 m, so the ball lands beyond the service line: the serve is out	A1
(d)	vy = 9.8 × 0.728 = 7.1 m s−1	M1
	v = √(302 + 7.12) = 31 m s−1	A1
(e)	drag opposes the motion, reducing the horizontal velocity	M1
	the ball covers less horizontal distance in the time it takes to fall	R1
	so it lands closer to the server (possibly in, but the net clearance also decreases)	A1

Examiner's comments. In (b), a "show that" answer must give the value to more precision than the question (0.91 m, not just 0.9 m). In (c), many candidates stop at 22 m without comparing with 18.3 m and saying what it means. In (e), "air resistance slows it down" alone scores nothing: link the reduced horizontal velocity to the reduced horizontal distance.

9 Check your understanding

Section A: multiple choice

1.A ball is thrown vertically upwards. Ignoring air resistance, which is correct at the highest point?

A. v = 0, a = 0 B. v = 0, a = g down C. v = 0, a = g up D. v maximum, a = 0

2.A car accelerates uniformly from rest and travels 20 m in the first 2.0 s. How far does it travel in the first 4.0 s?

A. 40 m B. 60 m C. 80 m D. 160 m

3.What does the area under an acceleration–time graph represent?

A. displacement B. change in velocity C. change in acceleration D. distance

4.For a projectile on level ground with no air resistance, which quantity stays constant?

A. speed B. velocity C. horizontal component of velocity D. vertical component of velocity

5.From the same height, ball X is dropped and ball Y is thrown horizontally at the same moment. Ignoring air resistance:

A. X lands first B. Y lands first C. they land together at the same speed D. they land together, Y faster

6.An object's velocity falls uniformly from 10 m s−1 to 0 in 5.0 s. What is its displacement in that time?

A. 2.0 m B. 25 m C. 50 m D. 10 m

7.A stone dropped from rest falls a distance d in time t. How far does it fall in time 2t?

A. d B. 2d C. 4d D. 8d

8.An object falling at terminal speed has

A. no drag acting B. drag equal to its weight C. acceleration g D. a resultant force equal to its weight

9.Two projectiles are launched from level ground at the same speed, one at 30° and one at 60°. Ignoring air resistance, which quantity is the same for both?

A. maximum height B. time of flight C. range D. speed at the highest point

10.A ball is thrown vertically upwards and air resistance is significant. At the highest point its acceleration is

A. zero B. less than g C. equal to g D. greater than g

Section B: structured

11.A cyclist accelerates uniformly from 3.0 m s−1 to 11 m s−1 in 5.0 s. Calculate (a) the acceleration, (b) the distance travelled. [3]
12.A stone is dropped from a bridge and hits the water 2.2 s later. Ignoring air resistance, calculate the height of the bridge and the speed of the stone as it hits the water. [3]
13.A ball is thrown horizontally at 8.0 m s−1 from the top of a 45 m cliff. How far from the base of the cliff does it land? [3]
14.A projectile is launched at 20 m s−1 at 60° above the horizontal from level ground. Calculate its maximum height and its range. [4]
15.A ball is dropped onto a hard floor and bounces twice. Sketch its velocity–time graph, taking up as positive. [3]
16.Explain why, when air resistance is significant, a ball thrown upwards takes longer to fall than to rise. [3]
17.A train accelerates uniformly from rest at 0.50 m s−2 to 20 m s−1, travels at this speed for 2.0 minutes, then decelerates uniformly at 0.80 m s−2 to rest. Sketch the v–t graph and determine the total distance and the average speed. [5]
18.Show that a ball thrown horizontally at 12 m s−1 from a height of 1.8 m lands about 7 m away. [3]
19.A stone is thrown vertically upwards at 20 m s−1. Calculate the two times at which it is 15 m above the launch point, and explain why there are two answers. [4]
20.A paper cone is dropped and reaches a terminal speed. A second, identical cone is placed inside it and the pair is dropped. Explain how the terminal speed changes. [3]

Answers

1. B 2. C (s ∝ t2, so 4 × 20 m) 3. B 4. C 5. D 6. B (area = ½ × 10 × 5.0) 7. C 8. B 9. C 10. C (at the top v = 0, so drag = 0)

11. (a) a = (11 − 3.0) ÷ 5.0 = 1.6 m s−2 [M1 A1] (b) s = ½(3.0 + 11) × 5.0 = 35 m [A1]

12. h = ½ × 9.8 × 2.22 = 24 m [M1 A1]; v = 9.8 × 2.2 = 22 m s−1 [A1]

13. 45 = ½ × 9.8 × t2, t = 3.0 s [M1 A1]; x = 8.0 × 3.03 = 24 m [A1]

14. uy = 17.3 m s−1 [M1]; H = 17.32 ÷ 19.6 = 15 m [A1]; R = 202 sin 120° ÷ 9.8 = 35 m [M1 A1]

15. Straight lines all with the same negative gradient [1]; near-vertical jumps from negative to positive at each bounce [1]; each rebound velocity smaller than the impact velocity before it [1]. Compare with the bouncing-ball graph in section 4.1.

16. Falling, drag acts upwards against weight, so the downward acceleration is less than g [1]. Rising, drag and weight both act downwards so the ball decelerates faster than g and reaches a lower height [1]. Falling from rest through that height with the smaller acceleration takes longer than the rise [1].

17. Stage 1: t = 40 s, s = ½ × 20 × 40 = 400 m [M1]; stage 2: 20 × 120 = 2400 m [A1]; stage 3: t = 25 s, s = 250 m [A1]; total = 3050 m ≈ 3.1 km [A1]; average speed = 3050 ÷ 185 = 16 m s−1 [A1]

18. t = √(2 × 1.8 ÷ 9.8) = 0.606 s [M1 A1]; x = 12 × 0.606 = 7.3 m ≈ 7 m [A1]

19. 15 = 20t − 4.9t2 [M1]; t = [20 ± √(400 − 294)] ÷ 9.8 [M1]; t = 1.0 s and 3.1 s [A1]; it passes 15 m once going up and once coming down [R1]

20. The pair has twice the weight but the same shape and area [1]; terminal speed is reached when drag = weight, so drag must be larger [1]; drag increases with speed, so the terminal speed is higher [1]

10 Summary and topic map

Equation	When to use it
v = Δs/Δt, a = Δv/Δt	average values; gradients of graphs
v = u + at	no s involved
s = ½(u + v)t	no a involved
s = ut + ½at2	no v involved
v2 = u2 + 2as	no t involved
ux = u cos θ, uy = u sin θ	first line of any angled projectile
T = 2u sin θ/g, H = (u sin θ)2/2g, R = u2 sin 2θ/g	level ground, no drag (derive; not in the data booklet)
h = ½gt2 (plot h against t2)	measuring g: gradient = g/2

Graph	Gradient	Area	Key features
s–t	v	none	max/min: object turns round
v–t	a	s (below axis negative)	crossing the axis: object reverses
a–t	none	Δv	free fall: horizontal line at −g
v–s	a/v (so a = v × gradient)	none	uniform a gives a curve; plot v2 against s for a straight line

Before the exam, make sure you…

•state your positive direction and give every s, u, v and a a sign
•remember that a point below the launch point has negative s when up is positive
•find average speed from total distance ÷ total time, never by averaging stage speeds
•use a tangent (not a chord from the origin) for instantaneous values on a curve
•subtract areas below the time axis for displacement, add them for distance
•don't use the equations of motion when the acceleration isn't constant
•resolve launch velocities into u cos θ and u sin θ straight away, and only use the T, H, R shortcuts on level ground
•know that the speed at the top of a projectile's path is u cos θ, not zero
•give velocities with a direction, stating what any angle is measured from
•convert between s–t, v–t, a–t and v–s graphs for the same motion, starting from the v–t graph
•explain drag effects as a chain: direction of drag, effect on acceleration, effect on height, range or time

Kinematics on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Topic map for A.1 Kinematics.

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