This physics grades 11-12 lesson on suvat equations notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on suvat equations and check yourself with the quiz.
Quick answer: The suvat equations connect displacement, initial velocity, final velocity, acceleration and time for motion in a straight line with constant acceleration. List the three known quantities, choose the equation that leaves out the unwanted one, and give every vector a sign. In free fall near the Earth, a = 9.81 m/s² downwards.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.4, IB A.1
Suvat equations notes: the key ideas
What are the suvat equations?
When an object moves in a straight line with constant acceleration, five variables describe the motion: displacement s, initial velocity u, final velocity v, acceleration a and time t. Any three of them let you find the other two using the five equations of motion, usually called the suvat equations.
Uniform acceleration means that the velocity changes by the same amount in every second. The acceleration is the gradient of the velocity-time graph and the displacement is the area under it.
v = u + at
s = ½(u + v)t
s = ut + ½at²
v² = u² + 2as
s = vt − ½at²
Symbols: s displacement (m), u initial velocity (m/s), v final velocity (m/s), a acceleration (m/s²), t time (s).
How do you choose the right equation?
Write down s, u, v, a and t, fill in the three you know, mark the one you want, and pick the equation that does not contain the fifth. Choose a positive direction first, then give every vector quantity a sign. A car braking has a negative acceleration if it moves in the positive direction.
| Known | Wanted | Use |
|---|---|---|
| u, a, t | v | v = u + at |
| u, v, t | s | s = ½(u + v)t |
| u, a, t | s | s = ut + ½at² |
| u, v, s | a | v² = u² + 2as |
A car accelerates uniformly from 12 m/s to 28 m/s in 8.0 s. Acceleration: a = (v − u) ÷ t = (28 − 12) ÷ 8.0 = 2.0 m/s². Distance: s = ½(u + v)t = ½ × 40 × 8.0 = 160 m. The car has a = 2.0 m/s² and travels 160 m.
How do you treat free fall?
Near the Earth’s surface, with air resistance ignored, every falling object has the same downward acceleration g = 9.81 m/s². Treat it as a suvat problem with a = ±9.81 m/s² depending on your positive direction. An object thrown upwards slows, stops for an instant at the top, then speeds up downwards, all with the same acceleration.
A stone is dropped from a bridge and falls 44 m. Take down as positive: u = 0, a = 9.81 m/s², s = 44 m. v² = 0 + 2 × 9.81 × 44 = 863.3, so v = 29.4 m/s. Time: t = v ÷ a = 29.4 ÷ 9.81 = 3.00 s. The stone hits at 29.4 m/s after 3.00 s.
A ball is thrown up at 14 m/s. Take up as positive: u = 14 m/s, a = −9.81 m/s². At the top v = 0, so 0 = 14² − 2 × 9.81 × h and h = 196 ÷ 19.62 = 10.0 m. Time to the top: t = 14 ÷ 9.81 = 1.43 s, so the flight lasts 2.85 s and the ball rises 10.0 m.
The acceleration at the top of a throw is not zero. Only the velocity is zero for an instant. Also remember that the suvat equations fail when the acceleration changes, for example when air resistance is large, or when a car changes gear.
What do the graphs tell you?
On a displacement-time graph the gradient is the velocity, and a curved line means acceleration. On a velocity-time graph the gradient is the acceleration and the area is the displacement. For a straight velocity-time line the area is a trapezium, which is exactly the equation s = ½(u + v)t.
Always draw a quick sketch with an arrow for the positive direction. Most lost marks in suvat problems come from a sign that was never decided, not from the algebra.
- v = u + at and s = ½(u + v)t
- s = ut + ½at² and v² = u² + 2as
- Free-fall acceleration g = 9.81 m/s² downwards
- Gradient of v-t graph = acceleration; area = displacement
- Valid only when the acceleration is constant
Equations of motion for constant acceleration: frequently asked questions
When can you use the suvat equations?
Use them only for motion in a straight line with constant acceleration. They fail if the acceleration changes, such as a falling object with growing air resistance. For a changing acceleration, use graph areas or calculus.
How do you choose which equation to use?
List s, u, v, a and t, and write which three are known and which one is wanted. Choose the equation that does not contain the remaining unused variable. If more than one works, pick the simplest.
What is the acceleration of a ball at the top of its flight?
It is still 9.81 m/s² downwards. Only the velocity is momentarily zero. The ball is changing its velocity at the same rate at every point, so the acceleration never switches off.
Why do I need a sign convention?
Velocity, displacement and acceleration are vectors, so direction matters. Choose a positive direction at the start, such as up, and use plus or minus signs for every quantity. This avoids contradictory answers.
How are suvat equations linked to graphs?
On a velocity-time graph the gradient is the acceleration and the area under the line is the displacement. For constant acceleration the line is straight, and its trapezium area gives s = ½(u + v)t.
