The SUVAT equations — when to use each one — Physics

Before this

Acceleration — changing velocity · coming soon
Rearranging formulae · coming soon

This leads to

A formula that stops working

You are driving at 72 km/h on a dry road. A child runs out in front of you, and you brake hard. Good brakes on a dry road slow a car by about 8 m/s every second. How far does the car travel before it stops?

You already know a formula for distance: distance = speed × time. Try to use it. Which speed? At the start the car is doing 72 km/h. At the end it is doing zero. In between, the speed is different at every instant. The formula needs one speed, and this motion does not have one.

So the formula you trust is not wrong. It is just built for motion at a steady speed, and braking is not steady. We need a tool for motion where the velocity keeps changing.

For one very common kind of changing motion, that tool exists. It is a set of five short equations, known by the letters they use: SUVAT. With the right one, the braking question takes about thirty seconds. By the end of this lecture you will be able to answer it, and you will know where every one of the five equations comes from.

Five letters for one kind of motion

The SUVAT equations work for one kind of motion only: motion in a straight line with constant acceleration. Constant acceleration means the velocity changes by the same amount every second. A car braking steadily, a ball falling, a train pulling away smoothly: all of these are close to constant acceleration.

Five quantities describe this motion. Each has a fixed letter:

Letter	Quantity	Unit
$s$	displacement: how far the object ends up from where it started, in a stated direction	m
$u$	initial velocity: the velocity at the start	m/s
$v$	final velocity: the velocity at the end	m/s
$a$	acceleration: the change in velocity each second	m/s²
$t$	time taken	s

Notice the word displacement, not distance. Displacement has a direction, so it can be positive or negative. Velocity and acceleration have directions too. So at the start of every problem, choose one direction and call it positive. Say it in words: "take forwards as positive". Every sign in the problem then follows from that choice.

Two ideas are enough

Many students try to memorise the five equations as five separate facts. That is the hard way. All five come from just two ideas.

Idea 1: what acceleration means. Acceleration is the change in velocity divided by the time taken. The velocity changes from $u$ to $v$ in time $t$ , so

a = \frac{v - u}{t} which rearranges to v = u + a t

Read $v = u + a t$ in words. The final velocity is the starting velocity, plus $a$ metres per second added on in each of the $t$ seconds. That is the first equation.

Idea 2: the average velocity sits exactly halfway. Look at an object whose velocity rises steadily from 4 m/s to 16 m/s in 6 s. That is an acceleration of 2 m/s².

Velocity rising steadily from u to v. The average velocity, 10 m/s, is exactly halfway between u and v. The two darker triangles are the same size: the time spent below the average balances the time spent above it.

What is its average velocity? Pair up the moments. At 1 s after the start, the velocity is 6 m/s. At 1 s before the end, it is 14 m/s. The average of that pair is 10 m/s. At 2 s from each end, the pair is 8 m/s and 12 m/s. Again the average is 10 m/s. Every early moment has a partner late in the motion, and every pair averages to the same value. So for constant acceleration, the average velocity is exactly halfway between the start and the end:

average velocity = \frac{u + v}{2}

This only works because the velocity changes steadily. If the object sped up quickly at first and slowly later, the pairs would not balance.

Now distance = speed × time works again, because we have one honest velocity to use: the average. So the displacement is

s = \frac{( u + v )}{2} t

That is it. That is the whole idea. Two equations, from two ideas. The other three come from combining these two with a little algebra. Step through the algebra below. Each line comes with the reason it is allowed.

Step through — five equations from two ideas

Press Next to add one step. Watch which letter disappears each time.

Each step gives one line and the reason it is allowed.

Idea 1: $v = u + a t$ . Acceleration is the change in velocity per second. No $s$ in this equation.
Idea 2: $s = \frac{1}{2} (u + v) t$ . Displacement is average velocity × time. No $a$ in this equation.
Remove $v$ : $s = \frac{1}{2} (u + u + a t) t = u t + \frac{1}{2} a t^{2}$ . Replace $v$ in step 2 using step 1.
Remove $u$ : $s = \frac{1}{2} (v - a t + v) t = v t - \frac{1}{2} a t^{2}$ . Step 1 rearranged gives $u = v - a t$ . Replace $u$ in step 2.
Remove $t$ : $s = \frac{1}{2} (u + v) \times \frac{v - u}{a} = \frac{v ^{2} - u ^{2}}{2 a}$ . Step 1 rearranged gives $t = \frac{v - u}{a}$ . Replace $t$ in step 2, and use $(v + u) (v - u) = v^{2} - u^{2}$ .
Rearrange: $v^{2} = u^{2} + 2 a s$ . Multiply both sides by $2 a$ , then add $u^{2}$ .

The SUVAT equations (constant acceleration only)

$v = u + a t$ · $s = u t + \frac{1}{2} a t^{2}$ · $s = v t - \frac{1}{2} a t^{2}$

$v^{2} = u^{2} + 2 a s$ · $s = \frac{1}{2} (u + v) t$

The missing letter chooses the equation

Look at the five equations again. Each one contains exactly four of the five letters. Each one leaves out a different letter.

Letter left out	Equation to use
$s$	$v = u + a t$
$v$	$s = u t + \frac{1}{2} a t^{2}$
$u$	$s = v t - \frac{1}{2} a t^{2}$
$t$	$v^{2} = u^{2} + 2 a s$
$a$	$s = \frac{1}{2} (u + v) t$

A typical problem gives you three of the letters and asks for a fourth. The fifth letter is neither given nor wanted. That letter does not matter to you, so choose the equation that does not contain it. This gives a method that works every time:

1. Write the five letters in a column: $s, u, v, a, t$ .
2. State your positive direction. Fill in every value you know, with its sign. Watch for hidden values: "from rest" means $u = 0$ , and "comes to a stop" means $v = 0$ .
3. Mark the letter you want with a question mark.
4. The letter left blank tells you the equation. Rearrange, substitute, and calculate.
5. Check that the answer is sensible: its size, its sign and its unit.

One more habit before the examples. Every value must be in metres, seconds, m/s and m/s². If a speed is given in km/h, convert it first. Divide by 3.6, because 1 km/h is 1000 m in 3600 s.

Stop and think

A car brakes steadily from 20 m/s to a stop. When its speed has dropped to 10 m/s, has it covered half of its stopping distance?

Show the answer

No. It has covered three-quarters. Use $v^{2} = u^{2} + 2 a s$ , which shows that distance depends on the square of the speed. With a deceleration of 8 m/s², the full stop takes $\frac{2 0 ^{2}}{2 \times 8} = 25$ m. Slowing from 20 m/s to 10 m/s takes $\frac{2 0 ^{2} - 1 0 ^{2}}{2 \times 8} = 18.75$ m. That is 75% of the total. The car spends most of its braking distance at high speed. This is why a small increase in speed makes a big difference to stopping distance.

Worked examples

Worked Example 1 — Pulling away from traffic lights

A car accelerates from rest at 3 m/s² for 8 s. How far does it travel?

Take forwards as positive. $s = ?$ , $u = 0$ (from rest), $v$ not given, $a = 3$ m/s², $t = 8$ s.

The letter left out is $v$ . So use $s = u t + \frac{1}{2} a t^{2}$ :

s = 0 \times 8 + \frac{1}{2} \times 3 \times 8^{2} = \frac{1}{2} \times 3 \times 64 = 96 m

Check: the final velocity is $v = 0 + 3 \times 8 = 24$ m/s, so the average velocity is 12 m/s. Then $12 \times 8 = 96$ m. Both routes agree.

Worked Example 2 — A ball rolling to a stop

A ball rolling at 12 m/s slows down steadily and stops after 36 m. Find its acceleration.

Take the direction of rolling as positive. $s = 36$ m, $u = 12$ m/s, $v = 0$ (it stops), $a = ?$ , $t$ not given.

The letter left out is $t$ . So use $v^{2} = u^{2} + 2 a s$ :

0^{2} = 1 2^{2} + 2 \times a \times 36 \Rightarrow 0 = 144 + 72 a \Rightarrow a = - 2 m/s^{2}

The answer is negative. That is not a mistake. Forwards was positive, and the acceleration points backwards, against the motion. That is what slowing down means. You can describe it in words as "a deceleration of 2 m/s²", but in the equation, $a = - 2$ m/s².

Worked Example 3 — The braking question

Back to the opening question. A car at 72 km/h brakes at 8 m/s² until it stops. Find the stopping distance and the stopping time.

Convert first: $72 \div 3.6 = 20$ m/s. Take forwards as positive. So $u = 20$ m/s, $v = 0$ , $a = - 8$ m/s².

For the distance, $t$ is not needed, so use $v^{2} = u^{2} + 2 a s$ :

0 = 2 0^{2} + 2 \times (- 8) \times s \Rightarrow 16 s = 400 \Rightarrow s = 25 m

For the time, $s$ is not needed, so use $v = u + a t$ :

0 = 20 + (- 8) t \Rightarrow t = 2.5 s

Check with the average-velocity equation: $s = \frac{1}{2} (20 + 0) \times 2.5 = 25$ m. The car needs 25 m to stop, and that is after the driver has reacted. Real stopping distances also include the distance travelled during the reaction time.

Now change the braking yourself. A harder brake means a bigger deceleration. Watch how the stopping distance changes.

Simulate — braking from 20 m/s

Choose the deceleration. Press Play, or step through 0.5 s at a time. The dots mark where the car was every 0.5 s.

Deceleration: 8 m/s²

For a car braking from $u = 20$ m/s, the stopping distance is $s = \frac{u ^{2}}{2 a}$ and the stopping time is $t = \frac{u}{a}$ :

Deceleration (m/s²)	Stopping distance (m)	Stopping time (s)
2	100	10
4	50	5
5	40	4
8	25	2.5
10	20	2

Doubling the deceleration halves both the distance and the time.

The table in the simulation shows a pattern. Doubling the deceleration halves the stopping distance. The equation explains why. With $v = 0$ , the equation $v^{2} = u^{2} + 2 a s$ gives $s = \frac{u ^{2}}{2 a}$ , so the distance is inversely proportional to $a$ .

Worked Example 4 — A stone dropped into a well

A stone is dropped from rest into a well. It hits the water 1.5 s later. How deep is the water surface below the top? Ignore air resistance.

Falling is constant acceleration too: $g = 9.8$ m/s² downwards. Here the motion is downwards, so take down as positive. Then $u = 0$ , $a = 9.8$ m/s², $t = 1.5$ s, $s = ?$ , and $v$ is not needed.

s = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times 9.8 \times 1. 5^{2} = 11.025 m \approx 11 m

The SUVAT equations do not care whether the motion is along a road or straight down. Constant acceleration is the only condition.

Worked Example 5 — When the starting velocity is unknown

A cyclist accelerates at 1 m/s² for 6 s and reaches 8 m/s. How far does she travel in those 6 s?

Take forwards as positive. $s = ?$ , $u$ not given, $v = 8$ m/s, $a = 1$ m/s², $t = 6$ s. The letter left out is $u$ , so use $s = v t - \frac{1}{2} a t^{2}$ :

s = 8 \times 6 - \frac{1}{2} \times 1 \times 6^{2} = 48 - 18 = 30 m

Check: $u = v - a t = 8 - 6 = 2$ m/s. Then $s = \frac{1}{2} (2 + 8) \times 6 = 30$ m. The two routes agree.

Three classic mistakes

Trap 1 — Forgetting to convert km/h

If you put 72 into the equations as if it were in m/s, every answer is wrong, even when the method is perfect. Convert first: divide km/h by 3.6 to get m/s.

Trap 2 — Using SUVAT when the acceleration is not constant

The equations depend on Idea 2, the halfway average, and that is only true when the velocity changes steadily. A car whose acceleration changes, or a skydiver once air resistance builds up, cannot be solved with one set of SUVAT values. Check the word "constant" or "uniform" in the question.

Trap 3 — Losing the minus sign of a deceleration

"A deceleration of 2 m/s²" goes into the equations as $a = - 2$ m/s² when forwards is positive. If you put in +2, the equations describe an object speeding up, and $v^{2} = u^{2} + 2 a s$ may give you no answer at all.

Try it yourself

Work each one on paper first. Write the five letters every time. Open a hint only when you are stuck.

A sprinter starts from rest and accelerates at 4 m/s² for 2.5 s. Find her speed at the end and the distance she covers.

Hint 1
"From rest" gives you one of the five letters. Write all five down.

Hint 2
For the speed, $s$ is not needed. For the distance, $v$ is not needed.

Full answer
$u = 0$ , $a = 4$ m/s², $t = 2.5$ s. Speed: $v = u + a t = 0 + 4 \times 2.5 = 10$ m/s. Distance: $s = u t + \frac{1}{2} a t^{2} = \frac{1}{2} \times 4 \times 2. 5^{2} = 12.5$ m. Check: average velocity 5 m/s × 2.5 s = 12.5 m.
A car travelling at 54 km/h brakes steadily and stops in 3.0 s. Find its acceleration and its stopping distance.

Hint 1
Convert 54 km/h to m/s first.

Hint 2
For the acceleration, use the equation without $s$ . For the distance, the equation without $a$ avoids using your own answer.

Full answer
$54 \div 3.6 = 15$ m/s. Take forwards as positive: $u = 15$ m/s, $v = 0$ , $t = 3.0$ s. Acceleration: $0 = 15 + 3 a$ , so $a = - 5$ m/s², a deceleration of 5 m/s². Distance: $s = \frac{1}{2} (u + v) t = \frac{1}{2} \times 15 \times 3.0 = 22.5$ m.
A ball is thrown straight up at 14 m/s. Using $g = 9.8$ m/s², find the greatest height it reaches and the time it takes to get there. Ignore air resistance.

Hint 1
At the very top, the ball stops for an instant. Which letter does that give you?

Hint 2
Take up as positive. Then gravity gives $a = - 9.8$ m/s².

Full answer
Take up as positive: $u = 14$ m/s, $v = 0$ at the top, $a = - 9.8$ m/s². Height (no $t$ ): $0 = 1 4^{2} + 2 (- 9.8) s$ , so $s = \frac{196}{19.6} = 10$ m. Time (no $s$ ): $0 = 14 - 9.8 t$ , so $t = \frac{14}{9.8} \approx 1.43$ s.
Two cars brake with the same deceleration, 6 m/s². One is going at 15 m/s, the other at 30 m/s. Find both stopping distances. What happens to the stopping distance when the speed doubles, and why?

Hint 1
With $v = 0$ , rearrange $v^{2} = u^{2} + 2 a s$ to make $s$ the subject. Remember the sign of $a$ .

Hint 2
Look at where $u$ appears in your formula for $s$ . What happens to $u^{2}$ when $u$ doubles?

Full answer
With $v = 0$ and $a = - 6$ m/s², $0 = u^{2} - 12 s$ , so $s = \frac{u ^{2}}{12}$ . At 15 m/s: $s = \frac{225}{12} = 18.75$ m. At 30 m/s: $s = \frac{900}{12} = 75$ m. Doubling the speed makes the braking distance four times as long, because the distance depends on $u^{2}$ , and $2^{2} = 4$ . This is why speed limits near schools are so low.

The bigger idea

Almost 400 years ago, Galileo rolled balls down gentle slopes and measured how far they went in equal times. He found that the distances in the first, second, third and fourth intervals were in the ratio 1 : 3 : 5 : 7, the odd numbers. That is exactly what $s = \frac{1}{2} a t^{2}$ predicts. For a stone falling from rest, it covers 4.9 m in the first second, 14.7 m in the next, then 24.5 m, then 34.3 m.

The five equations are not five facts. They are one idea about steady change, seen through five windows. When the acceleration is not constant, the same two ideas still hold for each tiny moment. Adding up those moments is what calculus does, and you will meet it later.

Key concepts in this lesson

The SUVAT equations describe straight-line motion with constant acceleration, using $s$ , $u$ , $v$ , $a$ and $t$ .
All five come from two ideas: $v = u + a t$ (the meaning of acceleration) and $s = \frac{1}{2} (u + v) t$ (the average velocity is halfway between $u$ and $v$ ).
Each equation leaves out one letter, so the letter that is neither given nor wanted tells you which equation to use.
Choose a positive direction first, and then every displacement, velocity and acceleration carries a sign, so a deceleration is a negative $a$ .
All values must be in m, s, m/s and m/s², so speeds in km/h are divided by 3.6 first.
With $v = 0$ , stopping distance is $\frac{u ^{2}}{2 a}$ , so doubling the speed makes the braking distance four times as long.