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Constant acceleration: the four formulae · Fathom

Constant acceleration: the four formulae

8 minAS Level4.2.4

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can choose and use the four constant-acceleration formulae, set up a sign convention, and handle a problem that needs two stages.

Key to the symbols

Symbol	Meaning	Unit
$u$	initial (starting) velocity	$\text{m s}^{-1}$
$v$	final velocity	$\text{m s}^{-1}$
$a$	acceleration (constant)	$\text{m s}^{-2}$
$s$	displacement	m
$t$	time taken	s
$s_1$ , $s_2$	displacements in the first and second stages of a journey	m

People call these the suvat formulae, after the five letters.

Five quantities, four equations

If acceleration is constant, the whole motion is fixed by five quantities:

$u$ — initial velocity
$v$ — final velocity
$a$ — acceleration
$s$ — displacement
$t$ — time

Each formula links four of them and leaves one out.

That's the point: pick the formula that leaves out the quantity you don't know and don't need.

$v = u + at$

$s = ut + \dfrac{1}{2}at^2$

$v^2 = u^2 + 2as$

$s = \dfrac{1}{2}(u + v)t$

The last one is the area of the velocity–time trapezium, which is where these formulae come from. $v = u + at$ is the gradient fact $a = \frac{v-u}{t}$ rearranged, and substituting it into $s = \tfrac12(u+v)t$ gives the other two.

Definition

Constant-acceleration formulae

For a particle moving in a straight line with constant acceleration: $v = u + at, \qquad s = ut + \tfrac{1}{2}at^2, \qquad v^2 = u^2 + 2as, \qquad s = \tfrac{1}{2}(u+v)t.$ They may only be used while the acceleration stays constant.

Set the sign convention first

Before substituting anything, decide which direction is positive and write it down. Then:

"starts from rest" means $u = 0$ ;
"comes to rest" means $v = 0$ ;
for vertical motion with upwards positive, $a = -10$ (gravity acts downwards), whether the particle is moving up or down;
a deceleration of $4\ \text{m s}^{-2}$ means slowing down at that rate, so $a = -4$ when the motion is in the positive direction;
a displacement in the negative direction is negative.

Signs are where most marks are lost. A negative answer can be perfectly correct: it means the displacement or velocity is in the negative direction.

For vertical motion in 9709, use $g = 10\ \text{m s}^{-2}$ , and say whether up or down is positive.

Worked example: Braking to a stop

A car travelling at $20\ \text{m s}^{-1}$ brakes with a constant deceleration of $4\ \text{m s}^{-2}$ and stops.

Find (a) the distance it takes to stop and (b) the time taken.

Have a go first?

Worked example: Two stages in one journey

A cyclist starts from rest and accelerates at $3\ \text{m s}^{-2}$ for $4\ \text{s}$ . She then travels at the velocity she has reached for a further $5\ \text{s}$ .

Find the total distance travelled.

Have a go first?

Which formula should I use?

Ask which quantity the question does not mention:

Missing quantity	Use
$v$	$s = ut + \tfrac{1}{2}at^2$
$s$	$v = u + at$
$t$	$v^2 = u^2 + 2as$
$a$	$s = \tfrac{1}{2}(u+v)t$
$u$	$s = vt - \tfrac{1}{2}at^2$ (a fifth form)

Then substitute and solve. If a question asks for two unknowns, find one first and use it in a second equation — that is the "more than one equation" case the syllabus mentions.

For a particle thrown upwards, the highest point has $v = 0$ ; the time to return to the starting level is twice the time to the highest point, because the motion is symmetric.

Common mistakes

The formulae work for any motion.

They require constant acceleration. If the acceleration changes, split the motion into stages or use calculus.

Deceleration means using $a = -4$ always.

The sign depends on your convention. If the car moves in the negative direction, a deceleration is a positive acceleration.

A negative displacement must be a mistake.

In the exam

Write down the five quantities with their signs before choosing a formula. Examiners award method marks for a correct substitution even if the arithmetic slips.

If the motion has two stages, label them and solve each separately. Carry the final velocity of one stage into the next as its initial velocity.

Finish with units, and check that the size of the answer is plausible: a car stopping from $20\ \text{m s}^{-1}$ in $50\ \text{m}$ is reasonable; $5000\ \text{m}$ is not.

Check yourself

CalculateQuestion 1 · Recall

CalculateQuestion 2 · Apply

Answer the 2 questions above to continue.