1.1 Motion, vectors and projectiles
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AS content in Unit 1 (Mechanics and Materials): specification points 1–7 and Core Practical 1. Forces, momentum, moments and energy follow on the next pages.
What this page covers
8 points
- Scalar and vector quantities, with examples, and how vectors are written
- Resolving a vector into two perpendicular components, by drawing and by calculation
- Adding two vectors: by scale drawing at any angle, by calculation at right angles
- Distance, displacement, speed, velocity and acceleration
- Displacement–time, velocity–time and acceleration–time graphs, including non-uniform acceleration
- The four equations of uniformly accelerated motion in one dimension
- Projectiles: independent horizontal and vertical motion
- Core Practical 1: determining the acceleration of a freely falling object
In one minute
7 points
- This topic gives you the tools to describe motion before asking what causes it: vectors and how to combine them, motion graphs, the four equations for constant acceleration, and projectiles treated as two independent straight-line motions.
- A vector has magnitude and direction; a scalar has magnitude only. Displacement, velocity, acceleration and force are vectors.
- A vector \(F\) at angle \(\theta\) to the horizontal has components \(F\cos\theta\) (horizontal) and \(F\sin\theta\) (vertical).
- Gradient of \(s\)–\(t\) = velocity. Gradient of \(v\)–\(t\) = acceleration; area under \(v\)–\(t\) = displacement. Area under \(a\)–\(t\) = change in velocity.
- The four equations of motion apply only when the acceleration is uniform.
- A projectile's horizontal and vertical motions are independent and share only the time.
- Free fall: \(g = 9.81\ \text{m s}^{-2}\). Plot \(h\) against \(t^2\); the gradient is \(g/2\).
Contents
14 sections
- What is the difference between a scalar and a vector?
- How do you resolve a vector into components?
- How do you add two vectors?
- Which quantities describe motion?
- What do motion graphs tell you?
- What are the equations of uniformly accelerated motion?
- How do you analyse a projectile?
- Core Practical 1: how do you measure g?
- Exam tips for 1.1
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is the difference between a scalar and a vector?
"2 km" tells you how far someone walked, but not where they ended up: for that you also need "north-east of the station". Physics separates quantities into these two kinds, because they combine in different ways. Two scalars simply add; two vectors add in a way that depends on the angle between them.
Vector: a quantity with both magnitude and direction.
| Scalars | Vectors |
|---|---|
| distance, speed | displacement, velocity |
| mass, time, temperature | acceleration, force (including weight) |
| energy, work, power | momentum, gravitational field strength |
| density, volume, pressure | electric field strength |
Vector notation. In print a vector is shown in bold (F) or with an arrow above the symbol; when you write by hand, underline it. The plain italic symbol \(F\) means the magnitude only. In one dimension a vector's direction is carried by its sign, once you have chosen which direction is positive. Two vectors are equal only if both their magnitudes and their directions are equal: a velocity of 5 m s⁻¹ east is not the same as 5 m s⁻¹ west.
The difference shows up as soon as a journey is not a straight line. In the figure a student walks 400 m east and then 300 m north. The distance walked is 700 m, but the displacement — the straight line from start to finish — is only 500 m, at 36.9° north of east.
Calling energy or work a vector because a force is involved. Work is the product of a force and a displacement along its line, and the result is a scalar — it has no direction.
How do you resolve a vector into components?
Resolving is the reverse of adding. Instead of combining two vectors into one, you replace one vector by two vectors at right angles that have the same combined effect. This is useful because perpendicular directions are independent: what happens horizontally does not affect what happens vertically.
You can also resolve by drawing: draw the vector to scale, drop perpendiculars to the two axes and measure the lengths, converting back with your scale. The drawing shows why the rule works: the vector is the hypotenuse of a right-angled triangle whose other two sides are the components.
A child pulls a sledge with a rope that makes an angle of 25° with the horizontal ground. The tension in the rope is 150 N. Calculate the horizontal and vertical components of the pull.
Step 1 The horizontal component is next to the 25° angle: \(F_{\text{h}} = 150\cos 25^\circ = \) 136 N
Step 2 The vertical component is opposite the angle: \(F_{\text{v}} = 150\sin 25^\circ = \) 63.4 N
Check: \(\sqrt{136^2 + 63.4^2} = 150\ \text{N}\) — the components rebuild the original vector.
Mark the angle on your own sketch before choosing sin or cos. If the angle is given from the vertical, the roles swap: the vertical component becomes \(F\cos\theta\) and the horizontal one \(F\sin\theta\).
How do you add two vectors?
Two vectors at right angles — by calculation
To add vectors, place them tip to tail: draw the second vector starting where the first one ends. The resultant runs from the start of the first to the tip of the second. When the two vectors are perpendicular the triangle has a right angle, so the magnitude comes from Pythagoras and the direction from \(\tan^{-1}\), measured from a reference direction that you must state.
A survey drone flies due north at 12.0 m s⁻¹ relative to the air. A wind blows due east at 5.0 m s⁻¹. Determine the velocity of the drone relative to the ground.
Step 1 The two velocities are perpendicular, so use Pythagoras: \(v = \sqrt{12.0^2 + 5.0^2} = 13.0\ \text{m s}^{-1}\)
Step 2 Direction from north: \(\theta = \tan^{-1}(5.0 \div 12.0) = 22.6^\circ\)
Velocity = 13.0 m s⁻¹ at 22.6° east of north. A velocity is a vector, so the answer is incomplete without the direction.
Two vectors at any angle — by scale drawing
When the angle between the vectors is not 90°, the triangle has no right angle and Pythagoras does not apply. At AS you add such vectors by an accurate scale drawing:
- Choose a scale and write it down (for example 1 cm = 5 N).
- Draw the first vector to scale; draw the second from its tip at the correct angle — or draw both from one point and complete the parallelogram.
- Measure the resultant with a ruler and its angle with a protractor; convert back using the scale.
Two ropes pull a crate across a smooth floor. One rope pulls with 40 N; the other pulls with 30 N at 60° to the first. Find the resultant force by scale drawing.
Step 1 Choose a scale — 1 cm = 5 N — and draw the 40 N force as an 8.0 cm line.
Step 2 From the same point draw the 30 N force as a 6.0 cm line at 60°, then complete the parallelogram.
Step 3 Measure the diagonal and its angle: diagonal ≈ 12.2 cm, so the resultant is ≈ 61 N at about 25° to the 40 N force.
A careful drawing is accurate to about ±1 mm and ±1°, so quote the result to 2 s.f.
Using Pythagoras when the two vectors are not at 90°. Pythagoras would give \(\sqrt{40^2 + 30^2} = 50\ \text{N}\) here, well below the true 61 N. At angles other than 90°, draw to scale or resolve both vectors into components first, add the components, then recombine.
Which quantities describe motion?
Each vector quantity of motion has a scalar partner that drops the direction. The difference matters when something turns round: a ball thrown up and caught has travelled a distance of twice its height but has zero displacement, and its average velocity over the flight is zero even though its average speed is not.
| Quantity | Meaning | SI unit | Type |
|---|---|---|---|
| Distance | total length of path travelled | m | scalar |
| Displacement \(s\) | distance in a stated direction from the starting point | m | vector |
| Speed | rate of change of distance | m s⁻¹ | scalar |
| Velocity \(v\) | rate of change of displacement | m s⁻¹ | vector |
| Acceleration \(a\) | rate of change of velocity | m s⁻² | vector |
What do motion graphs tell you?
Almost every motion-graph question comes down to two operations: the gradient (how fast the plotted quantity changes) and the area under the line (the plotted quantity multiplied by time).
| Graph | Gradient gives | Area under the line gives |
|---|---|---|
| displacement–time | velocity | no useful quantity |
| velocity–time | acceleration | displacement |
| acceleration–time | no useful quantity at this level | change in velocity |
The reason is the definitions. Velocity is \(\Delta s / \Delta t\), which is exactly the gradient of an \(s\)–\(t\) graph. Rearranging gives \(\Delta s = v\Delta t\): a thin strip under a \(v\)–\(t\) graph has height \(v\) and width \(\Delta t\), so its area is the small displacement, and adding all the strips gives the total. The same argument with \(a\) and \(v\) explains the \(a\)–\(t\) graph. The three graphs below describe one tram journey between two stops.
A tram accelerates uniformly from rest to 12.0 m s⁻¹ in 8.0 s, travels at constant velocity for 20.0 s, then decelerates uniformly to rest in 6.0 s (the velocity–time graph above). Determine (a) the acceleration in each stage, (b) the distance between the stops and (c) the average speed.
(a) Acceleration = gradient of each straight section: \(a_1 = 12.0 \div 8.0 = 1.5\ \text{m s}^{-2}\), \(a_2 = 0\), \(a_3 = -12.0 \div 6.0 = -2.0\ \text{m s}^{-2}\). 1.5 m s⁻², 0, −2.0 m s⁻²
(b) Distance = area under the \(v\)–\(t\) graph (a trapezium): \(\tfrac12 \times 8.0 \times 12.0 + 20.0 \times 12.0 + \tfrac12 \times 6.0 \times 12.0 = 48 + 240 + 36 = \) 324 m
(c) Average speed = total distance ÷ total time \(= 324 \div 34.0 = \) 9.53 m s⁻¹
Non-uniform acceleration
Real vehicles rarely accelerate uniformly. A curved \(v\)–\(t\) graph means the acceleration is changing. To find the acceleration at one instant, draw the tangent to the curve at that time and find its gradient with a large triangle. The displacement is still the area under the curve, found by counting squares or splitting the area into thin strips; the equations of motion cannot be used, because they assume a straight \(v\)–\(t\) line. An \(a\)–\(t\) graph that is not horizontal shows non-uniform acceleration; the area under it still gives the change in velocity.
Areas below the time axis
Velocity below the axis means motion in the negative direction, so area below the axis counts as negative displacement. Displacement = area above − area below; distance = area above + area below.
Reading the point where a \(v\)–\(t\) line crosses the time axis as "the object stops and stays at rest". It is momentarily at rest and then moves in the opposite direction; the gradient — the acceleration — has not changed.
What are the equations of uniformly accelerated motion?
When the acceleration is constant, the \(v\)–\(t\) graph is a straight line, and its gradient and area can be written as algebra. The gradient gives \(a = (v-u)/t\), which rearranges to the first equation. The area is a trapezium, \(s = \tfrac12(u+v)t\). Eliminating \(v\) or \(t\) between these two gives the other two. Each equation leaves out one of the five quantities, which is how you choose between them.
A reliable method
- Write the positive direction first ("taking upwards as positive").
- List \(s, u, v, a, t\) with signs; mark the one you want and the one you neither know nor need.
- Choose the equation that leaves out the unneeded quantity; rearrange, then substitute.
- Interpret the sign of the answer.
A distress flare is fired vertically upwards at 35.0 m s⁻¹ from the deck of a boat 4.0 m above the sea. Air resistance is negligible; \(g = 9.81\ \text{m s}^{-2}\). Calculate (a) the maximum height of the flare above the deck and (b) the time until it lands in the sea.
Step 1 Upwards positive: \(u = +35.0\ \text{m s}^{-1}\), \(a = -9.81\ \text{m s}^{-2}\).
(a) At the top \(v = 0\); \(t\) is not needed, so use \(v^2 = u^2 + 2as\): \(0 = 35.0^2 - 2 \times 9.81 \times s\), so \(s = \) 62.4 m
(b) The sea is 4.0 m below the deck, so \(s = -4.0\ \text{m}\). Find the landing velocity first: \(v^2 = 35.0^2 + 2(-9.81)(-4.0) = 1303\), so \(v = -36.1\ \text{m s}^{-1}\) (moving down). Then \(v = u + at\) gives \(t = \dfrac{-36.1 - 35.0}{-9.81} = \) 7.25 s
Finding \(v\) first and then \(t\) from \(v = u + at\) avoids solving a quadratic. Keep the unrounded value of \(v\) in your calculator for the second step and round only the final answer.
How do you analyse a projectile?
A projectile is any object moving freely under gravity after launch, such as a thrown ball or a long-jumper in the air. Air resistance is taken as negligible unless the question says otherwise. The key idea is that its horizontal and vertical motions are independent, because gravity acts only vertically:
- Horizontal: no force, so the horizontal velocity \(u_{\text{H}} = u\cos\theta\) stays constant.
- Vertical: constant acceleration \(g = 9.81\ \text{m s}^{-2}\) downwards, starting from \(u_{\text{V}} = u\sin\theta\).
- Time is the link — work it out from one direction and use it in the other.
- At the highest point the vertical velocity is zero, but the horizontal velocity is unchanged.
A neat consequence of independence: a ball dropped and a ball thrown horizontally from the same height at the same moment land together. The horizontal velocity carries the second ball sideways, but it does nothing to its vertical motion.
A player releases a basketball 2.10 m above the floor at 7.45 m s⁻¹ and 52° above the horizontal. The hoop is 3.05 m above the floor, 4.60 m horizontally from the point of release. Air resistance is negligible. (a) Calculate the time taken for the ball to travel 4.60 m horizontally. (b) Show that the ball is at the height of the hoop at this time. (c) State and explain whether the ball is rising or falling as it reaches the hoop.
Step 1 Resolve the launch velocity: \(u_{\text{H}} = 7.45\cos 52^\circ = 4.59\ \text{m s}^{-1}\), \(u_{\text{V}} = 7.45\sin 52^\circ = 5.87\ \text{m s}^{-1}\).
(a) Horizontal velocity is constant: \(t = 4.60 \div 4.587 = \) 1.00 s (1.003 s — keep this value).
(b) Vertical displacement in this time (upwards positive): \(s = 5.871 \times 1.003 - \tfrac12 \times 9.81 \times 1.003^2 = 0.954\ \text{m}\). Height \(= 2.10 + 0.954 = \) 3.054 m ≈ 3.05 m, the height of the hoop.
(c) Vertical velocity at the hoop: \(v_{\text{V}} = 5.871 - 9.81 \times 1.003 = -3.97\ \text{m s}^{-1}\). It is negative, so the ball is falling.
In a "show that" question the value is given, so the credit is for the working. Show the substitution and give your result to one more significant figure than the value quoted — here 3.054 m, not just 3.05 m.
Core Practical 1: how do you measure g?
The idea is simple: a body released from rest falls a height \(h\) in time \(t\), with \(h = \tfrac12 gt^2\) (this is \(s = ut + \tfrac12at^2\) with \(u = 0\)). Comparing with \(y = mx\), a graph of \(h\) against \(t^2\) is a straight line through the origin with gradient \(g/2\).
- Method: a small steel ball is held by an electromagnet; switching off the magnet releases the ball and starts an electronic timer; the ball falls through a trapdoor or a light gate that stops the timer.
- Measure: \(h\) from the bottom of the ball to the trapdoor with a metre rule, kept vertical with a set square; repeat each timing several times and take a mean.
- Range: at least five or six heights spread over the widest practical range, so the line is well defined.
- Why electronic timing: human reaction time (about 0.2 s) is comparable to the fall times themselves.
Use the graph above to determine \(g\).
Step 1 Gradient from a large triangle drawn on the line: \(\text{gradient} = \dfrac{1.275 - 0.193}{0.26 - 0.04} = 4.92\ \text{m s}^{-2}\)
Step 2 Gradient \(= g/2\), so \(g = 2 \times 4.92 = \) 9.84 m s⁻², within 0.3% of 9.81 m s⁻².
A systematic error that adds the same amount to every reading of \(h\) — for example measuring from the top of the ball instead of the bottom — moves the line up without changing its gradient. It shows up as an intercept, and the value of \(g\) from the gradient is unaffected. This is a good reason to use a graph rather than a single measurement.
Exam tips for 1.1
For any vector answer — displacement, velocity, acceleration, a resultant force — state the direction: a sign with the positive direction defined, an angle from a named reference, or a compass direction.
Take \(g = 9.81\ \text{m s}^{-2}\) from the data list, not 10 m s⁻². Give final answers to the number of significant figures in the least precise data, usually 2 or 3.
For a gradient, draw a triangle that covers at least half of the line and read its corners from the line, not from the data points. Show the triangle on the graph.
If an early answer is wrong but you use it correctly later, the later steps can still earn credit — but only when the substitution can be seen. A bare number gives nothing to follow.
State needs a short answer; calculate needs a numerical answer with working; show that needs every step and one extra significant figure; explain needs a reason linked to physics, not a restatement.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| At the top of its flight a projectile has zero velocity. | Only the vertical component is zero; the horizontal component is unchanged. | Velocity at the top = \(u\cos\theta\), horizontal. |
| At the top of its flight the acceleration is zero. | The weight still acts. | The acceleration is 9.81 m s⁻² downwards at every point. |
| A ball thrown horizontally stays in the air longer than one dropped from the same height. | The horizontal velocity does not affect the vertical motion. | Both land at the same time. |
| Two vectors always add by Pythagoras. | Pythagoras works only for a right-angled triangle. | At other angles, draw to scale or add components. |
| Negative acceleration means slowing down. | The sign depends on the chosen positive direction. | A body slows down when \(v\) and \(a\) have opposite signs. |
| The equations of motion work for any motion. | They are derived for a straight-line \(v\)–\(t\) graph. | Use them only for uniform acceleration; otherwise use the graph. |
How this topic connects
1.2 Forces and Newton's laws
Resolving vectors is the main tool for forces on slopes and in equilibrium, and a constant resultant force gives the constant acceleration these equations need.
1.5 Work and energy
For a speed at a given height, conservation of energy gives the answer without the time: \(v^2 = u^2 + 2gh\) for a fall through \(h\), whatever the launch angle.
Charged particles in fields (Unit 4) and practical skills (Unit 3)
A charge moving across a uniform electric field follows a parabola for the same reason a projectile does: constant velocity in one direction, constant acceleration at right angles to it. And linearising \(h = \tfrac12gt^2\) by plotting \(h\) against \(t^2\) is the pattern for many practical questions: choose axes that give a straight line, then read the constant from the gradient.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(v = u + at\) | constant \(a\); \(s\) not needed | Yes |
| \(s = \dfrac{(u+v)t}{2}\) | constant \(a\); \(a\) not needed | Yes |
| \(s = ut + \tfrac12 at^2\) | constant \(a\); \(v\) not needed | Yes |
| \(v^2 = u^2 + 2as\) | constant \(a\); \(t\) not needed | Yes |
| \(F_{\text{h}} = F\cos\theta,\ F_{\text{v}} = F\sin\theta\) | resolving a vector | No — trigonometry, learn it |
| \(R = \sqrt{A^2 + B^2},\ \theta = \tan^{-1}(B/A)\) | adding perpendicular vectors | No — trigonometry, learn it |
| \(h = \tfrac12 gt^2\) | free fall from rest (Core Practical 1) | No — it is \(s = ut + \tfrac12at^2\) with \(u = 0\) |
Check your understanding
1. Sort into scalars and vectors: momentum, speed, weight, energy, displacement.
Vectors: momentum, weight, displacement. Scalars: speed, energy.
2. A force of 80 N acts at 30° to the vertical. Calculate its horizontal and vertical components.
The angle is from the vertical, so vertical \(= 80\cos 30^\circ = \) 69 N and horizontal \(= 80\sin 30^\circ = \) 40 N.
3. A swimmer crosses a river at 1.2 m s⁻¹ straight across, while the current flows at 0.90 m s⁻¹. Find the resultant velocity.
\(\sqrt{1.2^2 + 0.90^2} = 1.5\ \text{m s}^{-1}\), at \(\tan^{-1}(0.90/1.2) = 37^\circ\) downstream from the straight-across direction: 1.5 m s⁻¹ at 37° downstream.
4. A cyclist accelerates uniformly from 3.0 m s⁻¹ to 9.0 m s⁻¹ over 48 m. Calculate the acceleration.
\(v^2 = u^2 + 2as\), so \(a = \dfrac{9.0^2 - 3.0^2}{2 \times 48} = \dfrac{72}{96} = \) 0.75 m s⁻².
5. A stone is thrown horizontally at 6.0 m s⁻¹ from a wall 1.8 m high. How far from the wall does it land?
Vertical motion starts from zero: \(t = \sqrt{2 \times 1.8 \div 9.81} = 0.61\ \text{s}\). Horizontal distance \(= 6.0 \times 0.606 = \) 3.6 m.
6. On a velocity–time graph a line crosses the time axis with a constant negative gradient. Describe the motion at that instant.
The object is momentarily at rest and reversing direction. Its acceleration (the gradient) is unchanged and non-zero, so it does not stay at rest — like the flare at the top of its flight.
Specification checklist
- (1) I can choose and use the four equations for constant acceleration in one dimension.
- (2) I can draw and read displacement–time, velocity–time and acceleration–time graphs.
- (3) I know what the gradients and areas of these graphs give, including when the acceleration is not uniform.
- (4) I can tell scalars from vectors, give examples of each and recognise vector notation.
- (5) I can resolve a vector into two perpendicular components, by drawing and by calculation.
- (6) I can add two vectors at any angle by drawing, and two perpendicular vectors by calculation.
- (7) I can treat the horizontal and vertical motion of a projectile independently.
- (11) I can describe and analyse an experiment to measure the acceleration of free fall.