AS Physics: Kinematics and Equations of Motion — Revision Notes

Condensed for the final weeks. For the full explanation, use the Kinematics and Equations of Motion study guide.

Assessed on Paper 1 (multiple choice, 1 h 15, 40 marks) and Paper 2 (structured, 1 h 15, 60 marks), both testing AO1 and AO2 on AS content, so both recall of the equations and their application to unfamiliar situations can be examined.

The four equations of motion

Valid only for uniform acceleration — check this condition before reaching for any of the four.

v = u + at
s = ut + 1/2 a t^2
v^2 = u^2 + 2as
s = (u + v) t / 2

Choosing the right one: list u, v, a, s, t; identify the three you know and the one you want; pick the equation containing exactly those four.

“Derive” is an explicit syllabus command word here, not just “recall”: v = u + at follows directly from the definition of acceleration (a = Δv/Δt, rearranged), and the others follow by combining that with the definition of velocity — know the derivation, not just the finished equations.

Determining g experimentally: drop an object through a known, measured height and time the fall electronically (a light gate or a timer released by an electromagnet), then rearrange s = ½gt² (since u = 0) to g = 2s/t². Repeating the drop and averaging t reduces the effect of random timing error.

Scalars and vectors

Scalar	Vector
distance, speed, mass, time, energy, work, power	displacement, velocity, acceleration, force, momentum, weight

Vectors combine by the parallelogram rule or by resolving into perpendicular components, whichever suits the quantities given:

horizontal:  F cos(theta)
vertical:    F sin(theta)

Motion graphs

Graph	Gradient	Area under
Displacement–time	Velocity	—
Velocity–time	Acceleration	Displacement
Acceleration–time	—	Change in velocity

For a curve, take the gradient of the tangent at the point, drawn as a straight line touching the curve at exactly that instant rather than joining two separate points on it.

Note “displacement”, not “distance”: area below the axis on a velocity–time graph is negative displacement, since the object is moving in the opposite direction to whichever was defined as positive.

Projectile motion — the whole method

Treat horizontal and vertical independently, applying the equations of motion separately in each direction. They share only the time.

HORIZONTAL   a = 0        so    s = u_x t          (constant velocity)
VERTICAL     a = -g       so    use suvat with u_y

At maximum height:   v_y = 0
Time of flight:      solve the vertical equation for s = 0
Range:               s = u_x x (time of flight)
Max height:          use v^2 = u_y^2 + 2as with v_y = 0 at the top

For a projectile launched at angle θ with speed u: u_x = u cos θ, u_y = u sin θ. See the Kinematics and Equations of Motion study guide for the full derivations and worked reasoning behind every method above.

Air resistance

Without it, the trajectory is a symmetrical parabola. With air resistance the path becomes asymmetric: reduced range and maximum height, a steeper descent than ascent, and the impact speed is less than the launch speed, since a resistive force continuously removes kinetic energy from the projectile throughout its flight.

Exam traps

• Applying suvat when acceleration is not uniform.
• Mixing horizontal and vertical quantities in one equation.
• Sign errors: choose a positive direction and hold it. Taking up as positive makes g = −9.81.
• Confusing distance with displacement, or speed with velocity.
• Forgetting that at maximum height the vertical velocity is zero but the horizontal is not.
• Using the gradient of a chord where a tangent is required.
• Reciting the suvat equations without being able to derive v = u + at from the definition of acceleration, when a question explicitly asks for a derivation.
• Timing a single drop when determining g experimentally, rather than repeating and averaging to reduce random timing error.

Self-test

1. State the four equations of motion.
2. A ball is thrown at 20 m/s at 30° above the horizontal. Find its initial horizontal and vertical velocity components.
3. What does the area under a velocity–time graph give?
4. At the top of a projectile’s path, which velocity component is zero?
5. Give two effects of air resistance on a projectile’s path.
6. Derive v = u + at from the definition of acceleration.
7. Describe a method for determining g experimentally, including the equation used.

Answers: 1. v = u + at; s = ut + ½at²; v² = u² + 2as; s = (u+v)t/2. 2. u_x = 20 cos 30° = 17.3 m/s; u_y = 20 sin 30° = 10 m/s. 3. Displacement (not distance — area below the axis counts as negative). 4. The vertical component; the horizontal component is unchanged. 5. Any two: reduced range, reduced maximum height, asymmetric trajectory with a steeper descent, impact speed lower than launch speed. 6. Acceleration a = Δv/Δt = (v − u)/t; rearranging gives at = v − u, so v = u + at. 7. Drop an object through a known height, timing the fall with a light gate or electromagnet-released timer; rearrange s = ½gt² to g = 2s/t², repeating and averaging t to reduce random error.