SUVAT Equations: A Complete Guide for IB Physics Students

What does SUVAT stand for?

Five quantities, and the word is just their initials:

• s — displacement, in metres (m)
• u — initial velocity, in metres per second (m s⁻¹)
• v — final velocity (m s⁻¹)
• a — acceleration (m s⁻²)
• t — time (s)

Note that s is displacement, not distance. Displacement has a direction. If a ball is thrown straight up and lands back in your hand, the distance it travelled is not zero, but its displacement is.

The one condition

These equations only work when acceleration is constant. That is not a small print detail, it is the whole basis of them. If the acceleration changes during the motion, every SUVAT equation gives you a wrong answer and the working earns nothing.

Constant acceleration covers most of what you meet at this level: an object falling freely near the Earth, a car braking evenly, a trolley running down a slope. It does not cover an object falling with a lot of air resistance, because the acceleration drops as the object speeds up.

If a question describes a journey in stages — speeding up, then steady, then braking — the acceleration is constant inside each stage but not across the whole journey. Treat each stage as its own SUVAT problem. There is a worked example of exactly this further down.

About the word itself

“SUVAT” is a British teaching habit rather than an official term. IB subject guides and your data booklet call them the equations of motion or the kinematic equations. They are the same four equations whatever you call them, so if your teacher says “equations of motion”, nothing has changed.

The equations, and what each one leaves out

There are five in common use. Four are printed in the IB Physics data booklet. The fifth is not, which matters and is covered below.

The third column is the one to learn. Students memorise the first column and then stare at a question wondering which to use. The first column is not the useful part.

Written out properly, the four in your data booklet are:

v = u + at

s = ut + ½at²

v² = u² + 2as

s = u + v2 t

Notice that the last one is just average velocity multiplied by time. When acceleration is constant, the average velocity sits exactly halfway between the starting and finishing velocity, which is why (u + v) ÷ 2 works. It does not work when the acceleration varies.

Where the equations come from

You do not have to memorise four unrelated formulas. They all fall out of one picture: a velocity–time graph for constant acceleration. Because the acceleration is constant, the line is straight.

Two readings give you everything.

The gradient of the line is how fast the velocity changes, which is the acceleration. So a = (v − u) ÷ t. Rearrange that and you have v = u + at.

The area under the line is the displacement. That area is a rectangle with a triangle sitting on top of it. The rectangle is u × t. The triangle is ½ × base × height, which is ½ × t × (v − u), and since v − u = at, that becomes ½at². Add the two together and you have s = ut + ½at².

Take the whole shape as a trapezium instead — average height times width — and you get s = ½(u + v)t. Three down from one picture.

The fourth one, step by step

v² = u² + 2as is the one that looks like it came from nowhere. It does not. It is the other two with the time thrown away, and it takes four lines. Worth doing once so it stops feeling like magic.

1. Start with v = u + at and make t the subject. t = v − ua
2. Put that into s = ½(u + v)t, so the t disappears. s = ½(u + v) × v − ua
3. Multiply out the top. (u + v)(v − u) is the difference of two squares, which is v² − u². s = v² − u²2a
4. Multiply both sides by 2a and move the u² across. v² = u² + 2as

That is why this equation has no t in it. The t was deliberately cancelled away, which is exactly what makes it the one to reach for when a question never mentions time.

Choosing the right equation in ten seconds

This is a method, not a knack, and it works every time.

1. Write the five letters down the side of your page: s, u, v, a, t.
2. Fill in every value the question gives you, including the ones hidden in words. “Starts from rest” means u = 0. “Comes to a stop” means v = 0. “Dropped” means u = 0. “Falls freely” means a = 9.81 m s⁻² downwards.
3. Put a ring around the letter you are asked for.
4. One letter will be left blank — not given, not asked for. That is the one you do not care about.
5. Use the equation that leaves that letter out. Check the table above.

Try it. A car starts from rest and accelerates at 3.0 m s⁻² for 8.0 s. How far does it travel?

If two letters come out blank, you are missing information. Usually that means the question has more than one stage, or it wants you to find something another way first — often an acceleration from a force.

Four worked examples

Signs, g, and the mistakes examiners see most

Pick a positive direction and write it down. Up or down, left or right — it does not matter which you choose, but it has to be decided before you start and it has to stay the same all the way through. Most lost marks in vertical motion questions are sign errors, and nearly all of them come from never having chosen.

If you take upwards as positive, then a = −9.81 m s⁻² for anything falling freely, a ball thrown upwards has a positive u, and a position below the starting point has a negative s.

• g is 9.81 m s⁻² in IB Physics. That is the value printed in your data booklet. Some textbooks round it to 9.8 or 10. Use what the booklet gives unless the question says otherwise.
• Free fall does not depend on mass. A heavy ball and a light ball dropped together land together, ignoring air resistance. This comes up in multiple choice every year.
• At the top of its flight, a thrown object has v = 0 but a is still 9.81 m s⁻². The velocity is zero for an instant. The acceleration never stops. Students write a = 0 at the top and lose the question.
• Show every step. Method marks are awarded for the working, not only the final number. Do three lines in your head, write one number, and a single slipped sign leaves the examiner nothing to credit.
• Units and significant figures. A velocity with no units, or an answer to eight decimal places when the data had three significant figures, is a mark the scheme openly offers and you quietly decline.

Projectile motion: SUVAT twice

Once you can use SUVAT in a straight line, projectiles are the same thing done twice. One rule makes them easy:

For a projectile with no air resistance:

• Horizontally, there is no force, so a = 0. The horizontal velocity never changes, and the horizontal distance is simply u_x × t.
• Vertically, a = 9.81 m s⁻² downwards the whole time, including at the very top.

A launch at an angle θ splits into u_x = u cos θ going across and u_y = u sin θ going up. Everything after that is two SUVAT lists, and the time connects them. Work out t from whichever direction hands it to you, then carry that same t across to the other.

Worked example: a ball kicked at an angle

The three things that go wrong

• Putting 9.81 into the horizontal list. Gravity pulls downwards, not sideways, so the horizontal acceleration of a projectile is zero.
• Using the launch speed as the vertical u. Only the u sin θ part goes up. Using the whole 15.0 would give a height more than four times too big.
• Forgetting that the time is shared. It is the one number both lists have in common, and it is almost always the thing to find first.

One more case worth having ready: something launched horizontally, like a ball rolling off a table. Then u_y = 0, so the vertical list starts from rest and the fall takes exactly as long as it would if you had simply dropped it. The forward speed changes where it lands, not when.

How IB Physics actually asks it

This is the part a general revision page cannot tell you, because it is specific to your course and your exam papers.

Your data booklet gives you four equations, not five

It prints v = u + at, s = ut + ½at², v² = u² + 2as and s = ½(u + v)t. The fifth form, s = vt − ½at², is not there. You will rarely need it, and when you do you can reach it by substituting u = v − at into the second equation. Do not walk into an exam relying on a formula that will not be in front of you.

The booklet also gives g = 9.81 m s⁻². Use that value, not 9.8 and not 10.

You have the booklet in both written papers, so the thing worth practising is finding your way around it quickly. Students lose real minutes hunting for the kinematics line under pressure. Keep it open beside you for every practice question until the page it is on is automatic.

The values hidden inside the wording

Half of a kinematics question is reading. Examiners give you numbers in words, and if you do not translate them you will think information is missing. These phrases come up again and again:

Where it turns up

Uniformly accelerated motion sits near the start of the course, but it does not stay there. It comes back as:

• Force questions. You find the acceleration from Newton’s second law, a = F ÷ m, and then feed it straight into SUVAT. If a question gives you a force and asks for a distance, that is the route.
• Energy questions. A velocity from SUVAT becomes a kinetic energy, ½mv². Many problems can be done either by energy or by SUVAT, and the energy route is often shorter when no time is involved.
• Graph questions. You may be asked to read a gradient or an area off a velocity–time graph rather than use a formula. It is the same physics as the picture earlier on this page.
• Projectile questions, in the two-list form above.
• Data analysis. A free-fall or trolley-on-a-ramp experiment that measures g is a standard way to examine this topic, and a standard internal assessment too. There you are usually plotting something so that the gradient gives you g, rather than plugging numbers into one equation.

Because it turns up everywhere, it is worth being genuinely fast at rather than merely able to do. The ten-second method above is what makes the difference in a timed paper.

If you also take IB Maths

In Maths AA, and in parts of AI, you meet the same motion again through calculus: differentiate displacement to get velocity, differentiate again for acceleration, integrate to go back the other way. That approach is more powerful, because it copes with acceleration that changes, which SUVAT cannot.

SUVAT is what you get when you do that calculus once, for the special case where the acceleration stays the same. It is worth knowing that the two subjects are describing the same thing, because it stops you learning kinematics twice.

Practice questions, with full answers

Reading worked examples feels like learning and mostly is not. These are in rough order of difficulty. Write the five letters down the side of your page, do each one properly, then open the answer.

Use g = 9.81 m s⁻² and ignore air resistance throughout.

1. A train starts from rest and accelerates uniformly at 0.50 m s⁻² for 40 s. How far does it travel in that time?

   Show the answer

   u = 0, a = 0.50, t = 40, s is wanted, v is blank. So use s = ut + ½at².

   s = 0 + ½(0.50)(40)² = 400 m

   A common slip here is squaring only the 40 in your head and forgetting the half.

2. A car travelling at 25 m s⁻¹ brakes uniformly and stops in 60 m. Find its acceleration.

   Show the answer

   u = 25, v = 0, s = 60, a is wanted, t is blank. So use v² = u² + 2as.

   0 = (25)² + 2a(60)

   a = −625 ÷ 120 = −5.2 m s⁻²

   The minus sign is part of the answer, and it is not optional. It says the acceleration points backwards along the motion. Writing “5.2 m s⁻²” alone is a deceleration of 5.2, which you should say in words if you drop the sign.

3. A stone is dropped from the top of a cliff and hits the sea 3.0 s later. How high is the cliff, and how fast is the stone moving when it hits?

   Show the answer

   Take downwards as positive to keep every number positive. u = 0, a = 9.81, t = 3.0.

   s = 0 + ½(9.81)(3.0)² = 44 m

   v = 0 + (9.81)(3.0) = 29 m s⁻¹

   Two significant figures, because 3.0 s has two. The calculator says 44.145 and 29.43; the data does not support that much precision.

4. An aeroplane must reach 65 m s⁻¹ to take off and can accelerate at 2.6 m s⁻². What is the shortest runway it can use, starting from rest?

   Show the answer

   u = 0, v = 65, a = 2.6, s is wanted, t is blank. So v² = u² + 2as again.

   (65)² = 0 + 2(2.6)s

   s = 4225 ÷ 5.2 = 812.5, so 810 m to two significant figures

5. A ball is thrown vertically upwards at 8.0 m s⁻¹. At what time is it 2.0 m above the point it was thrown from?

   Show the answer

   Upwards positive: u = +8.0, a = −9.81, s = +2.0, t is wanted, v is blank. Use s = ut + ½at².

   2.0 = 8.0t − 4.905t²

   Rearranged into a quadratic, 4.905t² − 8.0t + 2.0 = 0, which gives

   t = 0.31 s  or  1.32 s

   Both are real answers and both are right. The ball passes 2.0 m on the way up at 0.31 s and again on the way down at 1.32 s. If a question wants one of them it will say “on the way up” or “first”. Throwing away the second root without a reason is how students lose a mark on a question they had correct.

6. A ball rolls horizontally off a table 1.2 m high at 3.5 m s⁻¹. How far from the foot of the table does it land?

   Show the answer

   Two lists. Vertically it starts from rest, so u_y = 0, and taking downwards as positive, a = 9.81 and s = 1.2.

   1.2 = ½(9.81)t², so t = 0.49 s

   Carry that time across to the horizontal list, where a = 0.

   s = 3.5 × 0.4946 = 1.7 m

   The 3.5 m s⁻¹ plays no part in working out the time. The ball falls in exactly the time it would have taken if you had simply dropped it off the edge.

If you got the method right and the number wrong, that is an arithmetic problem and it is fixable in an afternoon. If you were not sure which equation to reach for, go back to the list method and do these again tomorrow. That is the part that has to become automatic.

The whole thing on one card

Photograph this and revise from the photograph.

Frequently asked questions

One last thing

If you take one habit from this page, take the list. Five letters down the side, fill in what you have, ring what you want, find the blank one. It turns a question you stare at into a question you start.

And when you practise, mark your own work against the mark scheme line by line instead of just checking the final number. Most marks lost on kinematics are not lost because the student did not know the physics. They go on a missing minus sign, a step done in the head, or a unit left off.