A.1 Kinematics

Kinematics describes motion using position, velocity and acceleration. These notes emphasise constant-acceleration relations, graphical intuition and projectile decomposition. Each practice problem includes a hidden solution you can reveal.

Core ideas

Displacement — vector from initial to final position.
Velocity — rate of change of displacement; slope of position–time graph.
Acceleration — rate of change of velocity; slope of velocity–time graph.
Projectile motion — treat horizontal and vertical components independently when air resistance is negligible.

Important formulas

v = u + at

Final velocity after time t under constant acceleration.

s = ut + \tfrac{1}{2}at^{2}

Displacement under constant acceleration.

s = \tfrac{1}{2}(u+v)t

Average velocity × time for constant acceleration.

v^{2} = u^{2} + 2as

Relates velocities, acceleration and displacement.

a = \dfrac{\Delta v}{\Delta t}

Acceleration = change in velocity ÷ time.

Graphical intuition

Position–time: slope = velocity; curvature indicates acceleration.
Velocity–time: slope = acceleration; area under curve = displacement.
Use areas and slopes to move between graphs rather than memorising steps.

Practice problems

Uniform acceleration — basic

A scooter starts from rest and accelerates uniformly at

1.8\ \mathrm{m\,s^{-2}}

for

5\ \mathrm{s}.

Find its final speed and the distance travelled.

Projectile components

A projectile is launched at

20\ \mathrm{m\,s^{-1}}

at an angle of

40^{\circ}

above the horizontal. Neglect air resistance. Find the time to reach maximum height and the maximum height (use

g = 9.8\ \mathrm{m\,s^{-2}}

Velocity–time graph

A velocity–time graph is a straight line from

v = 12\ \mathrm{m\,s^{-1}}

t = 0

v = -4\ \mathrm{m\,s^{-1}}

t = 4\ \mathrm{s}.

Find the acceleration and the displacement over the 4 s interval.

Clarity tip: Always list knowns and unknowns, choose the equation that avoids the unknown you cannot find directly, and check units at the end.