The key idea
IGCSE motion returns with equations instead of graph-reading — and with projectiles, the first genuinely two-dimensional problem most students meet. The whole topic is won by one discipline: choose a positive direction and never betray it.
Understand it · 01
Read motion graphs as physical statements
A displacement–time graph describes position relative to an origin. Its gradient is velocity, so a line becoming steeper means that the magnitude of velocity is increasing. A horizontal line means the object is at rest, even if its displacement is large. A velocity–time graph answers a different question: its gradient is acceleration and its signed area is displacement. A section below the time axis contributes negative displacement, while still contributing positive distance travelled.
When a graph is curved, draw a tangent at the instant requested to estimate an instantaneous gradient. Choose two well-separated points on that tangent rather than two points on the curve. To estimate an area numerically, divide it into shapes or narrow strips and preserve the sign of each contribution. Average speed is total distance divided by total time; average velocity uses net displacement, so an out-and-back journey can have zero average velocity.
Understand it · 02
Choose an equation after choosing an axis
The constant-acceleration equations connect five variables: displacement s, initial velocity u, final velocity v, acceleration a and elapsed time t. Start by selecting the positive direction and writing the known values with their signs. Then choose an equation that includes the unknown you want but omits any unavailable variable. The equations describe a period during which acceleration stays constant; they cannot be applied unchanged across a collision or a changing drag force.
For vertical motion with upward positive, acceleration is −g throughout free flight near Earth, including at the highest point. The velocity is momentarily zero there, but gravity has not switched off. A quadratic equation for time may have two physically meaningful roots if an object passes the same height on its way up and down. Interpret the roots against the actual question instead of discarding a value simply because it looks unfamiliar.
Understand it · 03
Separate horizontal and vertical motion
A projectile can be treated as two simultaneous one-dimensional motions if air resistance is negligible. Horizontal velocity is constant, while vertical velocity changes under gravity. Both components share the same elapsed time. For a horizontal launch, find the fall time from the vertical drop and then multiply it by the horizontal speed. For an angled launch, resolve the initial velocity first; neither component is equal to the whole launch speed.
The ideal path is parabolic because horizontal position is proportional to time while vertical displacement contains a time-squared term. This model becomes less suitable for a light object with large air resistance, a very high trajectory where g changes appreciably, or motion with another force. In a practical measurement of g, electronic timing reduces reaction-time uncertainty; plotting drop distance against time squared checks the model and makes half of g the expected gradient.
Key definitions
- Displacement
- The change in position from the starting point to the finishing point, with direction; its magnitude is the straight-line distance between those points.
- Velocity
- The rate of change of displacement.
- Acceleration
- The rate of change of velocity.
Equations, units & conditions
SUVAT 1
v = u + at
SUVAT 2
s = ut + ½at²
SUVAT 3
v² = u² + 2as
SUVAT 4
s = ½(u + v)t
Worked examples, step by step
Example 1
A ball is thrown horizontally at 15 m/s from a cliff 45 m high. How far from the base does it land? (g = 9.81 m/s²)
Follow the method
- Vertical (down positive): 45 = ½ × 9.81 × t², so t² = 9.17, t = 3.03 s.
- Horizontal: constant velocity, s = vt = 15 × 3.03.
s ≈ 45 m from the base
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 3 suggested marks
1. Find the stopping distance
A vehicle travelling at 18 m s⁻¹ slows uniformly at 3.0 m s⁻². Find the stopping distance.
Show solution and self-check: question 1
- Choose the direction of travel as positive: u = 18, v = 0, a = −3.0.
- Use v² = u² + 2as: 0 = 324 − 6s.
54 m.
Self-check · one mark per point
- Use a negative acceleration with the chosen positive direction.
- Substitute into v² = u² + 2as.
- Obtain 54 m.
Build · 3 suggested marks
2. Horizontal launch
A ball leaves a platform horizontally at 8.0 m s⁻¹ from a height of 20 m. Neglect drag and take g = 9.81 m s⁻². Find its horizontal range.
Show solution and self-check: question 2
- Vertical motion: 20 = ½ × 9.81 × t², so t = 2.019 s.
- Horizontal range = 8.0 × 2.019.
16 m to two significant figures.
Self-check · one mark per point
- Find the fall time from vertical motion.
- Use constant horizontal speed with the same time.
- Obtain approximately 16.2 m before rounding.
Common mistakes & how to avoid them
- Using SUVAT when acceleration is not constant. The equations are valid only for uniform acceleration — if the question mentions air resistance building up, SUVAT is dead and graphs take over.
- Sign chaos in vertical motion. Choose up as positive: then g = −9.81 m/s², a downward velocity is negative, and the algebra takes care of itself. Most projectile errors are sign errors.
- Mixing the components of projectile motion. Horizontal: constant velocity, no acceleration. Vertical: constant acceleration g. They share only the time — solve them in separate columns.
- Reading 'deceleration of 3 m/s²' and substituting +3. If your positive direction is the direction of motion, deceleration is negative.
Test yourself
10 questions · instant marking
Question 1 of 10
The SUVAT equations are valid only when:
Question 2 of 10
A ball is thrown vertically upwards. At its highest point:
Question 3 of 10
A car accelerates uniformly from 5 m/s to 25 m/s over 100 m. Using v² = u² + 2as, the acceleration is:
Question 4 of 10
In projectile motion (no air resistance), the horizontal velocity:
Question 5 of 10
The gradient of a displacement-time graph gives:
Question 6 of 10
A stone is dropped from rest. How far does it fall in 3.0 s? (g = 9.81 m/s²)
Question 7 of 10
A projectile is launched at 20 m/s at 30° above horizontal. Its initial vertical velocity is:
Question 8 of 10
An object decelerates uniformly from 30 m/s to rest in 6 s. The distance travelled is:
Question 9 of 10
Two balls are released from the same height: one dropped, one thrown horizontally. Ignoring air resistance, they land:
Question 10 of 10
A velocity-time graph shows a straight line crossing from positive to negative velocity. This means the object:
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
When can you use the SUVAT equations?
The SUVAT equations apply over an interval with constant acceleration along the chosen direction. They can describe ideal free fall near Earth, but cannot be applied unchanged when drag, thrust or another changing force makes acceleration vary.
Choose an interval before selecting an equation. Motion before and after an impact usually requires separate calculations because velocity can change abruptly during contact. A curved velocity–time graph also signals changing acceleration, so its signed area or a numerical method is more suitable than one constant-a substitution. Constant acceleration does not require constant speed or motion in the positive direction. A ball thrown upward has downward acceleration throughout its ideal flight, and the same vertical constant-a equations can describe both its ascent and descent when signs are consistent.
How do you approach projectile questions?
Resolve a projectile’s initial velocity into horizontal and vertical components and solve the two motions using the same elapsed time. With negligible drag, horizontal velocity stays constant and vertical acceleration is approximately g downward.
The useful connection between the two columns is time, not equality of displacement or speed. A horizontal launch starts with zero vertical velocity, whereas an angled launch generally does not. State a vertical sign convention and keep the launch and landing heights distinct. The familiar range formula assumes equal launch and landing levels; it should not be used unchanged for a launch from a balcony. If a time equation gives two positive roots, the projectile may pass a chosen height twice, so identify which event the question describes.
A launch above the landing level
A ball is launched horizontally at 6.0 m s⁻¹ and lands 1.25 s later. Neglect drag. Find its horizontal displacement and vertical drop; use g = 9.81 m s⁻².
- Horizontal displacement = 6.0 × 1.25 = 7.50 m.
- Vertical drop = ½ × 9.81 × 1.25² = 7.66 m.
7.50 m horizontally and 7.66 m downward.
How is g measured in the lab?
Measure the time taken for a released object to fall through a known distance, then test s = ½gt² when it starts from rest and drag is negligible. A graph of s against t² has expected gradient g/2.
Electronic release and detection reduce human start–stop timing uncertainty, but the distance must still match the points at which the timing begins and ends. A delay in release or an initial push introduces a systematic mismatch with the assumed initial conditions. Use several distances and inspect the shape of the graph rather than relying on one calculated value of g. Longer fall times can reduce percentage timing uncertainty, although the apparatus and safety limits matter. Repeated readings estimate scatter; they do not by themselves identify a consistently incorrect distance origin.
How do negative areas on a velocity–time graph affect distance and displacement?
Area below a velocity–time axis gives negative displacement, because the object moves against the chosen positive direction. Distance travelled uses the magnitudes of all areas, so positive and negative portions do not cancel.
Begin by separating the graph at every crossing of the time axis. Calculate the area of each section using a sign for displacement and no sign for distance. A horizontal segment below the axis represents constant negative velocity, not negative speed or continuing negative acceleration. If the graph has a positive gradient there, the object is slowing while moving in the negative direction. These distinctions are easier to follow when you imagine an origin and label motion to the right positive before reading any numbers.
An out-and-back graph
An object travels at +4 m s⁻¹ for 3 s, then at −2 m s⁻¹ for 5 s. Treat the velocity change as instantaneous. Find distance and displacement.
- Forward area = 4 × 3 = 12 m.
- Backward signed area = −2 × 5 = −10 m.
- Displacement = 12 − 10; distance = 12 + 10.
Displacement +2 m; distance 22 m.
Why is acceleration not zero at the highest point of a projectile?
At the highest point of an ideal projectile, its vertical velocity is zero but its vertical acceleration remains −g if upward is positive. An angled projectile still has a nonzero horizontal velocity at that instant.
The force of gravity depends on the local field and the object’s mass, not on whether the object is momentarily moving upward or downward. As a result, velocity passes smoothly through zero at the top while its rate of change remains downward. For a purely vertical throw the total speed is zero there; for an angled throw only the vertical component vanishes. Confusing these two cases can give a wrong kinetic energy at the highest point. Acceleration would be zero only if the resultant force were zero, which free-flight gravity does not satisfy.
How do I find instantaneous velocity from a curved displacement–time graph?
Instantaneous velocity is the gradient of a tangent to a displacement–time graph at the required instant. A secant joining two different points on the curve instead gives the average velocity over that interval.
Draw a tangent that follows the local slope, then choose two well-separated points on that line to calculate rise divided by run. They do not need to be measured data points, and they should not be selected from different parts of the original curve. Include axis scale factors when converting the gradient into units. A large gradient triangle reduces the relative effect of reading the two coordinates, but it cannot correct a poorly drawn tangent. If the tangent slopes downward, preserve the negative sign because it describes the direction of motion.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Define displacement, velocity and acceleration as vectors
- Use the four uniform-acceleration (SUVAT) equations
- Interpret and sketch displacement-time and velocity-time graphs
- Describe free fall; determine g experimentally
- Analyse projectile motion: independent horizontal and vertical components
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
Reference this page
Dr Desouky Physics Academy. Edexcel IAL Physics: Kinematics. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/kinematics
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