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

1. 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.

2. A vector has magnitude and direction; a scalar has magnitude only. Displacement, velocity, acceleration and force are vectors.

3. A vector F at angle θ to the horizontal has components Fcos θ (horizontal) and Fsin θ (vertical).

4. Gradient of s–t = velocity. Gradient of v–t = acceleration; area under v–t = displacement. Area under a–t = change in velocity.

5. The four equations of motion apply only when the acceleration is uniform.

6. A projectile's horizontal and vertical motions are independent and share only the time.

7. Free fall: g = 9.81 m s⁻². Plot h against t²; the gradient is g/2.

"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 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.

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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Step 1 The horizontal component is next to the 25° angle: Fₕ = 150cos 25° = 136 N

Step 2 The vertical component is opposite the angle: Fᵥ = 150sin 25° = 63.4 N

Check: √(136² + 63.4²) = 150 N — the components rebuild the original vector.

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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Step 1 The two velocities are perpendicular, so use Pythagoras: v = √(12.0² + 5.0²) = 13.0 m s⁻¹

Step 2 Direction from north: θ= tan⁻¹(5.0 ÷12.0) = 22.6°

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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

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.

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.

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).

The reason is the definitions. Velocity is Δs / Δt, which is exactly the gradient of an s–t graph. Rearranging gives Δs = vΔt: a thin strip under a v–t graph has height v and width Δt, 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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) Acceleration = gradient of each straight section: a₁ = 12.0 ÷8.0 = 1.5 m s⁻², a₂ = 0, a₃ = -12.0 ÷6.0 = -2.0 m s⁻². 1.5 m s⁻², 0, −2.0 m s⁻²

(b) Distance = area under the v–t graph (a trapezium): ½ ×8.0 ×12.0 + 20.0 ×12.0 + ½ ×6.0 ×12.0 = 48 + 240 + 36 = 324 m

(c) Average speed = total distance ÷ total time = 324 ÷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.

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 = ½(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

1. Write the positive direction first ("taking upwards as positive").

2. List s, u, v, a, t with signs; mark the one you want and the one you neither know nor need.

3. Choose the equation that leaves out the unneeded quantity; rearrange, then substitute.

4. Interpret the sign of the answer.

WORKED EXAMPLE · QUESTION

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 m s⁻². Calculate (a) the maximum height of the flare above the deck and (b) the time until it lands in the sea.

SOLUTION

Step 1 Upwards positive: u = +35.0 m s⁻¹, a = -9.81 m s⁻².

(a) At the top v = 0; t is not needed, so use v² = u² + 2as: 0 = 35.0² - 2 ×9.81 ×s, so s = 62.4 m

(b) The sea is 4.0 m below the deck, so s = -4.0 m. Find the landing velocity first: v² = 35.0² + 2(-9.81)(-4.0) = 1303, so v = -36.1 m s⁻¹ (moving down). Then v = u + at gives t = (-36.1 - 35.0)/(-9.81) = 7.25 s

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_H = ucos θ stays constant.

Vertical: constant acceleration g = 9.81 m s⁻² downwards, starting from u_V = usin θ.

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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Step 1 Resolve the launch velocity: u_H = 7.45cos 52° = 4.59 m s⁻¹, u_V = 7.45sin 52° = 5.87 m s⁻¹.

(a) Horizontal velocity is constant: t = 4.60 ÷4.587 = 1.00 s (1.003 s — keep this value).

(b) Vertical displacement in this time (upwards positive): s = 5.871 ×1.003 - ½ ×9.81 ×1.003² = 0.954 m. Height = 2.10 + 0.954 = 3.054 m ≈ 3.05 m, the height of the hoop.

(c) Vertical velocity at the hoop: v_V = 5.871 - 9.81 ×1.003 = -3.97 m s⁻¹. It is negative, so the ball is falling.

The idea is simple: a body released from rest falls a height h in time t, with h = ½ gt² (this is s = ut + ½at² with u = 0). Comparing with y = mx, a graph of h against t² 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.

WORKED EXAMPLE · QUESTION

Use the graph above to determine g.

SOLUTION

Step 1 Gradient from a large triangle drawn on the line: gradient = (1.275 - 0.193)/(0.26 - 0.04) = 4.92 m s⁻²

Step 2 Gradient = g/2, so g = 2 ×4.92 = 9.84 m s⁻², within 0.3% of 9.81 m s⁻².

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.

5 Work and energy

For a speed at a given height, conservation of energy gives the answer without the time: v² = u² + 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 = ½gt² by plotting h against t² is the pattern for many practical questions: choose axes that give a straight line, then read the constant from the gradient.

1. Sort into scalars and vectors: momentum, speed, weight, energy, displacement.

2. A force of 80 N acts at 30° to the vertical. Calculate its horizontal and vertical components.

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.

4. A cyclist accelerates uniformly from 3.0 m s⁻¹ to 9.0 m s⁻¹ over 48 m. Calculate the acceleration.

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?

6. On a velocity–time graph a line crosses the time axis with a constant negative gradient. Describe the motion at that instant.

1. Vectors: momentum, weight, displacement. Scalars: speed, energy.

2. The angle is from the vertical, so vertical = 80cos 30° = 69 N and horizontal = 80sin 30° = 40 N.

3. √(1.2² + 0.90²) = 1.5 m s⁻¹, at tan⁻¹(0.90/1.2) = 37° downstream from the straight-across direction: 1.5 m s⁻¹ at 37° downstream.

4. v² = u² + 2as, so a = (9.0² - 3.0²)/(2 ×48) = (72/96) = 0.75 m s⁻².

5. Vertical motion starts from zero: t = √(2 ×1.8 ÷9.81) = 0.61 s. Horizontal distance = 6.0 ×0.606 = 3.6 m.

6. 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.