Equations of motion: the three (or five) SUVAT equations for constant acceleration (v = u + at, s = ut + ½at², v² = u² + 2as, s = ½(u + v)t, s = vt − ½at²), where each comes from on the velocity-time graph, which one to use for which question, the sign rule, free fall with g, ten worked examples, the mistakes that lose marks, and a printable sheet

"Equations of motion" is the first set of formulas in every physics course, and the first place students lose marks, not because the algebra is hard but because five letters, five equations and a sign rule have to be managed at once. This page gives the equations in both the SUVAT form (UK, India, Australia) and the US kinematic form, shows where each comes from on the velocity-time graph so they do not have to be memorised blind, gives the table that chooses the right equation from the missing variable, and works ten examples of the kind set at grade 11, GCSE, A level, AP Physics 1 and CBSE Class 11. The printable has the equations, the chooser and practice on two pages.

Download the equations of motion sheet (PDF) (the five equations in both notations, the velocity-time graph derivations, the chooser table, the sign rule, free fall, the ten worked examples and ten practice questions with answers; two pages, free).

The equations

For motion in a straight line with constant acceleration, with s displacement, u initial velocity, v final velocity, a acceleration and t time:

The first three are "the three equations of motion" in CBSE and most textbooks; the fourth and fifth are combinations of them that save a step. Units: metres, metres per second, metres per second squared, seconds. Each equation contains four of the five letters, which is the key to choosing one.

Where they come from: the velocity-time graph

Constant acceleration draws a straight line on a velocity-time graph, starting at u and ending at v after time t.

1. The gradient of the line is the acceleration: a = (v − u) ÷ t, which rearranges to v = u + at.
2. The area under the line is the displacement: a rectangle of height u and width t (area ut) plus a triangle of height (v − u) and width t (area ½(v − u)t = ½at²). So s = ut + ½at².
3. The same area is also a trapezium: average height ½(u + v) times width t, so s = ½(u + v)t.
4. Put t = (v − u) ÷ a into equation 4: s = ½(u + v)(v − u) ÷ a = (v² − u²) ÷ 2a, so v² = u² + 2as.
5. Swap the roles of u and v in equation 2, counting back from the end: s = vt − ½at².

A student who can draw the graph can rebuild all five in a minute, which is worth more than memorising them.

Which equation to use

Write down the three quantities the question gives and the one it asks for. The letter you neither have nor want chooses the equation.

If two quantities are unknown, use two equations (usually find v or t first).

The sign rule

Choose a positive direction (usually the direction of the initial motion, or upwards for throws) and keep it for every quantity:

• A velocity in the positive direction is positive; the opposite way, negative.
• Slowing down in the positive direction means a is negative (a deceleration of 3 m/s² is a = −3).
• For objects thrown upwards with up positive, a = −9.8 m/s² throughout, on the way up and the way down.
• A negative displacement means the object ended up behind its start.

Free fall

Near the Earth's surface, without air resistance, everything accelerates downwards at g = 9.8 m/s² (9.81 in A level and some US courses; 10 in rough work; the sign depends on which way is positive).

• Dropped from rest (u = 0, down positive): v = gt, s = ½gt², v² = 2gs. From 45 m: t = √(2 × 45 ÷ 9.8) = 3.0 s, v = 29.7 m/s.
• Thrown upwards at u (up positive, a = −9.8): highest point when v = 0, so time up = u ÷ 9.8 and maximum height = u² ÷ (2 × 9.8). Thrown at 14 m/s: 1.43 s up, 10 m high, and it returns to the hand at 14 m/s downwards after 2.86 s.
• Thrown downwards at u: use the dropped equations with u included.

Ten worked examples

The mistakes that lose marks

• Using the equations when acceleration is not constant (a rocket, a car whose driver changes the pedal). Split the motion into stages.
• Forgetting the sign: a braking car has negative a; a ball on its way down has negative v when up is positive.
• Mixing units: km/h must become m/s (divide by 3.6) before the equations are used.
• Using average speed = (u + v) ÷ 2 when acceleration is not constant; it holds only for the straight-line graph.
• Reading s as distance travelled when it is displacement: a ball thrown up and caught has s = 0, not twice the height.
• Dropping the ½ in ½at², or squaring t but not the ½.

Where it sits in school

The three equations are CBSE Class 9 (Motion, with the graphical derivation) and Class 11 (Kinematics); GCSE Physics (v² = u² + 2as is on the equation sheet, the others are expected) and A level Mechanics with all five; Year 11 Physics in Australia; and grade 11 and 12 Physics and AP Physics 1 (unit 1, Kinematics) in the United States, where they are called the kinematic equations and written with v₀ and Δx. Two-dimensional projectiles use the same equations in each direction separately, with a = −g vertically and a = 0 horizontally.

Practice: ten questions

1. A car accelerates from 5 m/s to 25 m/s in 8 s. Find a and the distance.
2. A bus at 12 m/s decelerates at 1.5 m/s². How far does it go before stopping?
3. A runner accelerates from rest at 3 m/s² over 18 m. Final speed?
4. A stone is dropped down a well and hits the water after 2.5 s. How deep is the well?
5. A ball is thrown up at 25 m/s. How high does it go, and how long until it returns?
6. A train travels 500 m while slowing from 40 m/s to 20 m/s. Find a and the time taken.
7. A rocket sled accelerates at 30 m/s² for 4 s from rest. Speed and distance?
8. A ball is thrown down from a window at 5 m/s and hits the ground at 20 m/s. Height of the window?
9. A car at 72 km/h brakes to a stop in 40 m. Deceleration?
10. A stone thrown up at 10 m/s: where is it and how fast is it moving after 1.5 s?

Answers: 1) a = 2.5 m/s²; s = ½(5 + 25) × 8 = 120 m. 2) 0 = 144 − 3s, s = 48 m. 3) v² = 2 × 3 × 18 = 108, v = 10.4 m/s. 4) s = 4.9 × 6.25 = 30.6 m. 5) h = 625 ÷ 19.6 = 31.9 m; total time 2 × 25 ÷ 9.8 = 5.1 s. 6) 400 = 1,600 + 1,000a, a = −1.2 m/s²; t = 20 ÷ 1.2 = 16.7 s. 7) v = 120 m/s; s = 240 m. 8) 400 = 25 + 19.6s, s = 19.1 m. 9) 72 km/h = 20 m/s; 0 = 400 + 80a, a = −5 m/s². 10) s = 15 − 11.0 = 4.0 m above the hand; v = 10 − 14.7 = −4.7 m/s, moving down.

Download the equations of motion sheet (PDF)

Where to get help

A 1-on-1 online physics class with a tutor who holds a master's degree in science, for CBSE Class 9 and 11, GCSE and A level, AP Physics 1 and the Australian Year 11 course: the velocity-time graph drawn first for every problem on the shared screen, the chooser table until it is automatic, then the sign rule on thrown objects where most marks go. The recording and notes after every class, the same tutor every week, US$10 to $13 an hour. The physics page has the courses; the free 30-minute first class is a real lesson, usually on the homework the student brings.

Questions parents ask

See how we teach this, 1-on-1 online →