PHYSICS · LESSON 02 OF 9

Vectors and motion

Describe motion with position, velocity and acceleration as vectors; with constant acceleration, four equations predict everything, and horizontal and vertical motions are independent.

What this lesson explains

Kinematics — the description of motion — is the language of all mechanics. Before asking why something moves (forces), you need to describe how it moves: where it is, how fast, in which direction, and how that changes. Braking distances, falling objects, thrown or launched objects and conveyor systems are all kinematics.

Before you begin

Right-triangle trigonometry

For a vector of magnitude A at angle θ above the +x axis: A_x = A cos θ, A_y = A sin θ, and A = √A_x² + A_y², tan θ = A_y/A_x.

Quadratic formula

at² + bt + c = 0 has t = −b ± √b² − 4ac / 2a. In kinematics, keep the root that makes physical sense (usually t > 0).

Derivatives as rates

v = dx/dt and a = dv/dt (see Mathematics: “Derivative as a rate”).

The idea, made visible

Many physical quantities have a direction as well as a size. Vectors (displacement, velocity, acceleration, force) have magnitude and direction; scalars (distance, speed, time, mass, energy) have magnitude only. In one dimension the direction is shown by a sign after choosing a positive direction: +5 m/s means 5 m/s in the positive direction, −5 m/s means 5 m/s the other way.

Displacement Δx = x_f − x_i is the change in position, not the distance travelled: walk 3 m east and 3 m back and your displacement is 0 though you walked 6 m. Velocity is the rate of change of position, v = dx/dt; acceleration is the rate of change of velocity, a = dv/dt. Acceleration is not “speed”: a car braking while moving forward has negative acceleration (with forward positive), and a ball thrown straight up has zero velocity at the top of its flight but acceleration −9.8 m/s².

The horizontal velocity (gold) stays 17.3 m/s throughout. The vertical velocity (red) starts at +10 m/s, is zero at the top and is −10 m/s on landing. Arrows are not to the same scale as the axes.

On graphs: the slope of a position–time graph is velocity; the slope of a velocity–time graph is acceleration; the area under a velocity–time graph is displacement.

When acceleration is constant, the four “kinematic equations” below follow from these definitions. Choose the one that contains the unknown and three known quantities. Free fall near the Earth’s surface, neglecting air resistance, is constant acceleration with a = −g = −9.8 m/s² (taking up as positive).

In two dimensions, resolve vectors into perpendicular components. For a projectile with no air resistance, the horizontal and vertical motions are independent: horizontally a_x = 0, so v_x is constant; vertically a_y = −g. The same time t links them. At the top of the path v_y = 0, but v_x and the acceleration g are unchanged.

The projectile (20 m/s at 30°, g = 9.8 m/s²) at selected times
t (s)	x (m)	y (m)	v_x (m/s)	v_y (m/s)
0	0	0	17.3	10.0
0.5	8.66	3.78	17.3	5.10
1.02	17.7	5.10	17.3	0
1.5	26.0	3.98	17.3	−4.70
2.04	35.3	0	17.3	−10.0

v_x never changes; v_y decreases by 9.8 m/s every second. The motion is symmetric about the top.

Key terms

Position and displacement: x is location relative to a chosen origin; displacement Δx = x_f − x_i (vector).
Average velocity: v_avg = Δx / Δt; average speed = total distance/time (scalar, never negative).
Instantaneous velocity: v = dx / dt, the slope of the x–t graph.
Acceleration: a = dv / dt, the slope of the v–t graph. Units m/s².
Free fall: Motion under gravity alone: acceleration g = 9.8 m/s² downward, independent of mass (air resistance neglected).
Projectile: An object moving in two dimensions under gravity alone: constant horizontal velocity, constant downward acceleration g.

The formulas and what they mean

Symbols, meanings and units
Symbol	Meaning	Unit
x, y	position coordinates	m
v₀, v	initial and final velocity (signed)	m/s
a	acceleration (signed, constant here)	m/s²
t	time elapsed	s
g	magnitude of free-fall acceleration, 9.8 m/s² (9.81 in some texts)	m/s²
θ	launch angle above the horizontal	° or rad

Constant-acceleration equations

v = v₀ + at; Δx = v₀t + 1 / 2at²; v² = v₀² + 2aΔx; Δx = v₀ + v / 2t

Conditions and limits: Only when a is constant. All quantities signed relative to one chosen positive direction.

Vector components

A_x = A cos θ, A_y = A sin θ; A = √A_x² + A_y²

Conditions and limits: θ measured from the +x axis. If θ is measured from another direction, sin and cos swap — draw the triangle.

Projectile motion

x = v₀ cos θ · t; y = v₀ sin θ · t − 1 / 2gt²; v_y = v₀ sin θ − gt

Conditions and limits: No air resistance; up positive; launch from the origin.

Level-ground results

time of flight T = 2v₀ sin θ / g; range R = v₀² sin 2θ / g; max height H = (v₀ sin θ)² / 2g

Conditions and limits: Only when launch and landing heights are equal. Otherwise solve the y-equation for t.

Relative velocity

v_{A relative to C} = v_{A relative to B} + v_{B relative to C}

Conditions and limits: Vector addition (Galilean relativity, speeds much less than light).

Why it works: Where Δx = v₀t + ½at² comes from

For constant a, the velocity–time graph is a straight line from v₀ to v = v₀ + at.

Displacement = area under the v–t graph.
Area of (velocity × time) has units of metres.
The area is a rectangle v₀t plus a triangle 1 / 2t(v − v₀) = 1 / 2t(at).
Split the trapezium.
Δx = v₀t + 1 / 2at².
Add the two parts.
Eliminating t with t = (v − v₀)/a gives v² = v₀² + 2aΔx.
Useful when time is not given.

In calculus terms, the equations come from integrating a = dv/dt and v = dx/dt with a constant.

A first worked example

Kinematics problem-solving

Sketch the motion. Choose an origin and a positive direction (for 2D: x horizontal, y up).
List knowns and the unknown with signs: v₀, v, a, Δx, t. In free fall a = −9.8 m/s² (up positive).
Check that the acceleration is constant during the stage you analyse; split the motion into stages if it changes.
Choose the equation that contains the unknown and only known quantities.
For 2D: resolve the initial velocity, treat x and y separately, and connect them through t.
Check sign, units and size of the answer.

Displacement versus distance

Problem. A runner goes 40 m east and then 10 m west in 10 s. Find the displacement, the distance, the average velocity and the average speed.

Displacement = +40 + (−10) = +30 m (30 m east).
Signed sum of the moves, east positive.
Distance = 40 + 10 = 50 m.
Total path length, no signs.
Average velocity = 30 m / 10 s = 3 m/s east; average speed = 50 m / 10 s = 5 m/s.
Velocity uses displacement; speed uses distance.

Result: Displacement 30 m east, distance 50 m, average velocity 3 m/s east, average speed 5 m/s.

What it means: After a round trip, average velocity is zero but average speed is not.

A different case

Adding two vectors by components

Problem. Forces of 6 N due west and 8 N due north act on the same object. Find the resultant force (magnitude and direction).

Take east as +x and north as +y. Components: (−6, 0) N and (0, 8) N.
West is the negative x direction.
Add components: R = (−6 + 0, 0 + 8) = (−6, 8) N.
Vectors add component by component.
Magnitude: √6² + 8² = 10 N.
Pythagoras on the perpendicular components.
Direction: tan θ = 8/6, θ = 53.1° north of west.
The signs (−, +) put the resultant in the north-west quadrant.

Result: 10 N at 53.1° north of west.

What it means: Adding the magnitudes (14 N) would be wrong: it is only correct when the vectors point the same way.

More worked cases

Each case below uses a different skill. Every step and result is shown.

Speed after constant acceleration

Problem. A cart starts from rest and accelerates at 3 m/s² for 4 s. Find its final speed and the distance travelled.

v₀ = 0, a = 3 m/s², t = 4 s.
List the knowns.
v = v₀ + at = 0 + 3 × 4 = 12 m/s.
Equation containing v, v₀, a, t.
Δx = v₀t + 1 / 2at² = 0 + 1 / 2(3)(16) = 24 m.
Equation containing Δx.

Result: v = 12 m/s; Δx = 24 m.

What it means: Check with the average velocity: (0 + 12)/2 × 4 = 24 m.

Braking distance

Problem. A car travelling at 25 m/s brakes with constant deceleration 5 m/s². How far does it travel before stopping?

Take forward positive: v₀ = 25 m/s, v = 0, a = −5 m/s².
Braking acceleration opposes the motion, so it is negative.
v² = v₀² + 2aΔx ⇒ 0 = 625 + 2(−5)Δx.
Time is not needed, so use the equation without t.
Δx = 625 / 10 = 62.5 m.
Solve.

Result: 62.5 m.

What it means: Braking distance is proportional to v²: at double the speed (50 m/s) it would be four times as long, 250 m.

Displacement from a velocity–time graph with a reversal

Problem. An object moves at +4 m/s for 3 s, then at −2 m/s for 2 s. Find its displacement and the distance travelled.

First stretch: area = 4 × 3 = +12 m (above the time axis).
Area under a v–t graph is displacement.
Second stretch: area = (−2) × 2 = −4 m (below the axis).
Area below the axis counts as negative displacement.
Displacement = 12 − 4 = 8 m; distance = 12 + 4 = 16 m.
Signed areas give displacement; their sizes add to distance.

Result: Displacement +8 m; distance 16 m.

What it means: This is the graph version of the runner example at the start of the lesson: reversing direction makes distance larger than displacement.

Vertical throw

Problem. A ball is thrown straight up at 15 m/s. Find its maximum height and the time to reach it (g = 9.8 m/s²).

Up positive: v₀ = +15 m/s, a = −9.8 m/s²; at the top v = 0.
The ball stops momentarily at the top.
Height: 0 = 15² + 2(−9.8)Δy ⇒ Δy = 225 / 19.6 = 11.5 m.
v² = v₀² + 2aΔy.
Time: 0 = 15 − 9.8t ⇒ t = 1.53 s.
v = v₀ + at.

Result: Maximum height ≈ 11.5 m, reached after ≈ 1.53 s.

What it means: At the top the velocity is 0 but the acceleration is still −9.8 m/s²; otherwise the ball would stay there.

Projectile on level ground

Problem. A ball is launched at 20 m/s, 30° above the horizontal, over level ground. Find the time of flight, the range and the maximum height.

Components: v_x = 20 cos 30° = 17.32 m/s, v_y0 = 20 sin 30° = 10.0 m/s.
Resolve the launch velocity.
Landing: y = 0 = 10t − 4.9t² ⇒ t = 10 / 4.9 = 2.04 s (t = 0 is the launch).
Vertical motion alone decides the time.
Range: x = 17.32 × 2.041 = 35.3 m.
Constant horizontal velocity times the same time.
Maximum height: v_y = 0 ⇒ H = 10² / 2 × 9.8 = 5.10 m.
v_y² = v_y0² − 2gH.

Result: T ≈ 2.04 s, R ≈ 35.3 m, H ≈ 5.10 m.

What it means: Check with R = v₀² sin 2θ/g = 400 × sin 60°/9.8 = 35.3 m. A 60° launch would give the same range but a higher path.

A horizontal launch from a height

Problem. A stone is thrown horizontally at 8 m/s from a cliff 20 m high. Where does it land?

Vertical: v_y0 = 0, Δy = −20 m: −20 = −4.9t² ⇒ t = √20 / 4.9 = 2.02 s.
Horizontal launch means zero initial vertical velocity.
Horizontal: x = 8 × 2.02 = 16.2 m.
v_x is constant.

Result: It lands about 16.2 m from the foot of the cliff after 2.02 s.

What it means: The level-ground range formula does not apply here, because launch and landing heights differ.

Common misunderstandings

Misunderstanding: Treating distance and displacement as the same.
Correct idea: Displacement is a signed vector change in position; distance is the total path length.
Misunderstanding: Deceleration means negative acceleration.
Correct idea: An object slows down when a and v have opposite signs. Moving in the negative direction and speeding up also gives negative a.
Misunderstanding: “At the top, acceleration is zero.”
Correct idea: For a vertical throw, velocity is zero at the top. For a projectile with horizontal motion, only the vertical velocity component is zero. Acceleration is still g downward in both cases.
Misunderstanding: Using constant-acceleration equations when a changes.
Correct idea: Split the motion into stages of constant a, or use calculus.
Misunderstanding: Using the range formula when launch and landing heights differ.
Correct idea: Solve the vertical equation for t instead.
Misunderstanding: Mixing sin and cos in components.
Correct idea: cos goes with the side adjacent to the angle. Draw the triangle.

Keep in mind

v = v₀ + at; Δx = v₀t + 1 / 2at²; v² = v₀² + 2aΔx; Δx = v₀ + v / 2tConstant-acceleration equations
A_x = A cos θ, A_y = A sin θ; A = √A_x² + A_y²Vector components
x = v₀ cos θ · t; y = v₀ sin θ · t − 1 / 2gt²; v_y = v₀ sin θ − gtProjectile motion
time of flight T = 2v₀ sin θ / g; range R = v₀² sin 2θ / g; max height H = (v₀ sin θ)² / 2gLevel-ground results
v_{A relative to C} = v_{A relative to B} + v_{B relative to C}Relative velocity

Scope of this lesson

Air resistance is neglected throughout; terminal speed is an optional section of Friction and inclined surfaces.
g is taken as 9.8 m/s²; use your course’s value (9.81 m/s²) if required — answers change slightly.

Next: Relative motion and reference frames. Every velocity is measured relative to an observer. Name the frame, then add velocities as vectors: v_A/C = v_A/B + v_B/C.

An older bookmark may point here. That section is now its own lesson: Relative motion and reference frames.

Original study text. Sources and credits.

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