Kinematics — Topic Summary

1 Scalars & Vectors — Position & Displacement

Scalars vs Vectors

A scalar has magnitude only (a number and a unit). A vector has both magnitude and direction. This distinction matters enormously in physics.

Type	Quantity	Example
Scalar	Distance	You drove 14 km
Scalar	Speed	Moving at 60 km/h
Scalar	Time	Elapsed time = 3 s
Vector	Displacement	14 km north
Vector	Velocity	60 km/h eastward
Vector	Acceleration	9.8 m/s² downward

Position vs Displacement

Position (d) is where an object is relative to a reference point. Displacement (Δd) is the change in position — it is a vector.

Displacement

Δd = d_f − d_i

💡 A runner completing 4 laps of a 400 m track travels a distance of 1600 m, but their displacement is 0 m — they end where they started.

Defining a Positive Direction

Always choose and state a positive direction before solving. Common conventions: right is +, up is +, north is +. Negative values mean the opposite direction.

✏️

Worked example 1: A car starts at position d_i = 50 m east. It moves to d_f = 200 m east. Find displacement.
Δd = 200 − 50 = 150 m east (positive)

✏️

Worked example 2: A ball is thrown upward from d_i = 0 m and lands back at d_f = 0 m. Distance travelled = 2 × height. Displacement = 0 m.

2 Velocity & Speed

Definitions

Speed is the rate of change of distance — a scalar. Velocity is the rate of change of displacement — a vector (includes direction).

Average velocity

v_avg = Δd / Δt = (d_f − d_i) / (t_f − t_i)

Average speed

speed = total distance / total time

🔑 Instantaneous velocity is the velocity at a single moment in time — the limit of Δd/Δt as Δt approaches zero. On a d-t graph, it equals the slope of the tangent at that point.

Units

SI unit for velocity: metres per second (m/s). Convert km/h to m/s by dividing by 3.6 (i.e., × 1000/3600).

✏️

Worked example 1: A cyclist travels 300 m north in 12 s.
v = Δd/Δt = 300/12 = 25 m/s north

✏️

Worked example 2: A car drives 400 m east, then turns and drives 300 m west, total time 35 s.
Displacement = 400 − 300 = 100 m east. Velocity = 100/35 ≈ 2.86 m/s east.
Total distance = 700 m. Speed = 700/35 = 20 m/s.

3 Acceleration

Definition

Acceleration is the rate of change of velocity. It is a vector. An object accelerates whenever its velocity changes — in magnitude, direction, or both.

Average acceleration

a = Δv / Δt = (v_f − v_i) / t

💡 Slowing down is still acceleration! A car braking from 30 m/s to 0 m/s has a negative acceleration (if forward is positive). The term deceleration just means acceleration in the opposite direction of motion.

Uniform vs Non-Uniform

Type	Description	On a v-t graph
Uniform acceleration	Acceleration is constant	Straight line (constant slope)
Non-uniform acceleration	Acceleration changes over time	Curved line (changing slope)
Zero acceleration	Constant velocity	Horizontal line (slope = 0)

✏️

Worked example 1: A car accelerates from rest to 27 m/s in 9 s.
a = (27 − 0) / 9 = 3 m/s²

✏️

Worked example 2: A car brakes from 30 m/s to 6 m/s in 4 s.
a = (6 − 30) / 4 = −24/4 = −6 m/s² (deceleration)

4 UAM Equations (Uniform Acceleration)

These five equations apply only when acceleration is constant. They connect five variables: d (displacement), v₀ (initial velocity), v (final velocity), a (acceleration), t (time). Choose the equation that contains the four variables you know plus the one you want to find.

Equation 1

v = v₀ + at

Equation 2

d = v₀t + ½at²

Equation 3

v² = v₀² + 2ad

Equation 4

d = ((v₀ + v) / 2) × t

Equation 5

d = vt − ½at²

Variable Reference

Variable	Meaning	Unit	Note
d	Displacement	m	Vector — can be negative
v₀	Initial velocity	m/s	State at the start of the interval
v	Final velocity	m/s	State at the end of the interval
a	Acceleration	m/s²	Must be constant to use these equations
t	Time	s	Always positive

✏️

Worked example 1: v₀ = 0, a = 2.5 m/s², t = 8 s. Find displacement.
Use Eq. 2: d = 0(8) + ½(2.5)(8²) = 0 + ½(2.5)(64) = 80 m

✏️

Worked example 2: A car brakes from 25 m/s to rest over 62.5 m. Find acceleration.
Use Eq. 3: 0² = 25² + 2a(62.5) → 0 = 625 + 125a → a = −625/125 = −5 m/s²

⚠️ Free fall: When an object falls under gravity only, a = g = 9.8 m/s² downward. If upward is your positive direction, use a = −9.8 m/s².

5 Motion Graphs

Position-Time (d-t) Graph

The slope of a d-t graph equals velocity. A straight line means constant velocity. A curve means changing velocity (acceleration). Slope = Δd/Δt = v.

Velocity-Time (v-t) Graph

The slope of a v-t graph equals acceleration. The area under a v-t graph equals displacement.

Slope upward: positive acceleration (speeding up if also moving in + direction)
Slope downward: negative acceleration
Horizontal line: constant velocity (zero acceleration)
Area above x-axis: positive displacement; area below: negative displacement

Acceleration-Time (a-t) Graph

The area under an a-t graph equals the change in velocity (Δv). A horizontal line indicates uniform (constant) acceleration.

Graph	Slope means	Area means
d-t graph	Velocity	— (not directly used)
v-t graph	Acceleration	Displacement
a-t graph	Rate of change of acceleration	Change in velocity (Δv)

✏️

Worked example 1: A v-t graph shows constant v = 15 m/s for 6 s.
Area = 15 × 6 = 90 m (displacement)

✏️

Worked example 2: A v-t graph shows a triangle: v rises from 0 to 20 m/s in 5 s, stays at 20 m/s for 5 s, then drops to 0 in 4 s.
Area = ½(5)(20) + (5)(20) + ½(4)(20) = 50 + 100 + 40 = 190 m

6 Projectile Motion

A projectile is any object launched into the air and moving under gravity alone (no thrust). The key insight: horizontal and vertical motions are completely independent. Time is the only link between them.

Horizontal Component

No horizontal force acts on the projectile, so horizontal acceleration is zero. Horizontal velocity is constant throughout the flight.

Horizontal accel.

aₓ = 0

Horizontal velocity

vₓ = constant = v₀ cos θ

Horizontal range

x = vₓ × t

Vertical Component

Gravity acts downward at 9.8 m/s². The vertical motion follows the UAM equations with a = −9.8 m/s² (if upward is positive).

Vertical accel.

aᵧ = −9.8 m/s² (if up is +)

Initial vert. velocity

vᵧ₀ = v₀ sin θ

Vertical velocity

vᵧ = vᵧ₀ − g·t

🔑

Strategy for projectile problems:
1. Split initial velocity into vₓ = v₀ cos θ and vᵧ₀ = v₀ sin θ
2. Find the flight time from the vertical equations
3. Use t in the horizontal equation to find range

✏️

Worked example 1: Ball launched horizontally at 12 m/s from a 20 m cliff.
Vertical: 20 = ½(9.8)t² → t² = 4.08 → t ≈ 2.02 s
Horizontal: x = 12 × 2.02 ≈ 24.2 m

✏️

Worked example 2: Ball kicked at 25 m/s at 37° above horizontal.
vₓ = 25 cos 37° ≈ 20 m/s · vᵧ₀ = 25 sin 37° ≈ 15 m/s
Time to peak: t_up = 15/9.8 ≈ 1.53 s · Total time ≈ 3.06 s
Range: x = 20 × 3.06 ≈ 61.2 m
Max height: vᵧ² = vᵧ₀² − 2(9.8)h → h = 15²/(2 × 9.8) ≈ 11.5 m

7 Common Mistakes to Avoid

Mistake	What to do instead
Not defining a positive direction	Always state your positive direction first. Stick to it throughout the problem.
Confusing distance and displacement	Distance is total path length (scalar). Displacement is net change in position (vector). Use Δd = d_f − d_i.
Forgetting initial values (v₀ ≠ 0)	Read the problem carefully. "Starts from rest" means v₀ = 0. Otherwise always list v₀.
Using UAM when a is not constant	The five UAM equations require constant acceleration. Check that this holds before applying them.
Using g = 9.8 m/s² with wrong sign	If upward is +, then a = −9.8 m/s². If downward is +, then a = +9.8 m/s².
Mixing horizontal and vertical in projectile	Treat x and y as completely separate. Only time is shared between them.
Using total distance instead of displacement for velocity	Average velocity uses displacement (Δd), not total distance. Average speed uses total distance.