Kinematics: Complete A Level H2 Physics Cheatsheet
A Level Physics (H2), Chapter 2 · Read time: ~10 minutes
Key Formulas
Reading displacement, velocity and acceleration from graphs
On a displacement-time graph, the gradient at any point gives the instantaneous velocity. For a curved graph, this means the gradient of the tangent at that specific point, not a line joining two far-apart points.
On a velocity-time graph, the gradient gives acceleration, and the area between the line and the time-axis gives displacement. This holds even when the line curves, though for a curve the area must be estimated (e.g. counting squares) rather than calculated with a simple shape formula.
You should also be able to go the other way: given a velocity-time graph, sketch the corresponding acceleration-time graph (its gradient), and vice versa.
Trap: Constant acceleration on a v-t graph is a straight sloped line, which corresponds to a **horizontal** line on the a-t graph, not a sloped one. Students frequently sketch the a-t graph with the same shape as the v-t graph instead of correctly taking its gradient.
The SUVAT equations
The four equations of motion apply only when acceleration is constant. If acceleration changes during the motion, split the journey into stages of constant acceleration, or work from a graph instead.
Pick whichever equation excludes the variable you don't have and don't need, this avoids solving two equations simultaneously when one alone will do.
Sign convention matters: choose one direction as positive at the start of the question and apply it consistently to every velocity and acceleration, including gravity, throughout every stage.
Trick: A deceleration is simply a negative acceleration in your chosen positive direction, don't create a separate 'deceleration' equation. If an object is thrown upward and you take upward as positive, g enters every equation as −9.81 m/s² for the entire motion, both going up and coming back down.
Projectile motion
When air resistance is negligible, a projectile's horizontal and vertical motions are completely independent: horizontal velocity stays constant (no horizontal force acts), while vertical motion is uniformly accelerated by gravity.
Solve the two directions as two separate 1-D kinematics problems, linked only by the fact that they share the same time, t.
For a projectile launched at angle θ with speed u: resolve first (ux = u cosθ, uy = u sinθ), find the time of flight from the vertical motion alone, then find the horizontal range using range = ux × time of flight.
Trap: Never substitute a horizontal quantity into a vertical SUVAT equation, or vice versa. Each direction has its own initial velocity, its own acceleration (a = 0 horizontally, a = g vertically), and its own set of SUVAT variables, they are only linked through the shared value of t.
Worked Example
A ball is kicked from ground level with an initial speed of 20 m/s at an angle of 30° above the horizontal. Air resistance is negligible. Take g = 9.81 m/s².
(a) Show that the initial vertical component of the ball's velocity is 10.0 m/s. [1]
uy = u sinθ
= 20 × sin 30°
= 10.0 m/s
(b) Calculate the total time taken for the ball to return to the ground. [3]
Taking upward as positive, the ball returns to ground level when its vertical displacement is zero:
s = uy t − ½gt² = 0
t(uy − ½gt) = 0, so t = 2uy / g
= (2 × 10.0) / 9.81
= 2.04 s
(c) Calculate the horizontal distance (range) travelled by the ball. [2]
ux = u cosθ = 20 × cos 30° = 17.3 m/s
Range = ux × t
= 17.3 × 2.04
= 35.3 m
Why this question is a good test of the topic: part (b) only uses vertical-motion variables, and part (c) only uses horizontal-motion variables, joined solely by the time found in part (b). Mixing components from different directions into one equation is the single most common error on projectile-motion questions.
Frequently Asked Questions
What does the gradient of a velocity-time graph represent?
Acceleration. For a curved graph, use the gradient of the tangent at the point of interest.
Can I use the SUVAT equations if acceleration isn't constant?
No. They only apply during a phase of constant acceleration. For changing acceleration, split the motion into constant-acceleration stages or use a graph instead.
Why are horizontal and vertical motions independent in projectile motion?
Because gravity acts only vertically. With air resistance neglected, there's no horizontal force, so horizontal velocity stays constant throughout the flight.
How do I find the maximum height of a projectile?
Vertical velocity is zero at maximum height. Use v² = uy² − 2gh (taking upward as positive) and solve for h, or find the time to reach the top as t = uy / g.
Which paper is this tested in?
Paper 1 MCQs often test graph interpretation. Paper 2 and Paper 3 typically include a multi-part SUVAT or projectile-motion calculation similar to the worked example above.
SUVAT and projectile motion questions taking too long?
Small group A Level H2 Physics classes at TGC Academy's Bishan, Bukit Timah and Potong Pasir centres, taught by Andrew Seah, MOE Award-Winning Teacher and Marshall Cavendish textbook author.