Equations of Constant Acceleration

AS · M1 · 15 min

When a particle moves in a straight line with constant acceleration, five quantities describe each stage of its motion, and five equations link them. These are the constant acceleration formulae, often called the suvat equations after the letters they use. They are the workhorse of Paper 4: they appear in kinematics questions on their own, and again in almost every Newton's law question, where you first find the acceleration from $F = ma$ and then use suvat to find a speed, a distance or a time.

The five quantities

For one stage of motion with constant acceleration, use these symbols:

Symbol	Meaning	Unit
$s$	displacement from the starting point	$\text{m}$
$u$	initial velocity (at the start of the stage)	$\text{m s}^{-1}$
$v$	final velocity (at the end of the stage)	$\text{m s}^{-1}$
$a$	acceleration (constant)	$\text{m s}^{-2}$
$t$	time taken for the stage	$\text{s}$

Every one of these except $t$ is a vector, so each has a sign. Choose a positive direction before you start and give every quantity its sign relative to it.

Each suvat equation contains four of the five quantities. So if you know any three, you can find the other two: pick the equation that contains the three you know and the one you want.

Where the equations come from

The quickest derivation uses a velocity–time graph. With constant acceleration the graph is a straight line from velocity $u$ at time $0$ to velocity $v$ at time $t$ .

y = 4 + x (6, 0) -- (6, 10) fill 0 6 y = 4 + x

The graph shows a particle whose velocity rises steadily from $u = 4$ to $v = 10$ in $t = 6$ seconds.

From the gradient. The gradient of a $v$ – $t$ graph is the acceleration, so $a = \dfrac{v - u}{t}$ , which rearranges to

v = u + at

From the area. The area under a $v$ – $t$ graph is the displacement. The shaded region is a trapezium with parallel sides $u$ and $v$ and width $t$ :

s = \tfrac12(u + v)t

In the graph above, $a = \tfrac{10 - 4}{6} = 1\ \text{m s}^{-2}$ and $s = \tfrac12(4 + 10)(6) = 42\ \text{m}$ .

Eliminating one variable at a time. Substitute $v = u + at$ into the area formula:

s = \tfrac12(u + u + at)t = ut + \tfrac12at^2

Substitute $u = v - at$ instead:

s = \tfrac12(v - at + v)t = vt - \tfrac12at^2

Finally, eliminate $t$ . From $v = u + at$ , $t = \dfrac{v - u}{a}$ , so

s = \tfrac12(u + v)\cdot\frac{v - u}{a} = \frac{v^2 - u^2}{2a} \quad\Rightarrow\quad v^2 = u^2 + 2as

You may be asked to derive one of these, so know where they come from. Learn all five by heart: you will use them in almost every question on the paper, and looking them up costs time.

Constant acceleration formulae

v = u + at \qquad\qquad s = \tfrac12(u + v)t

s = ut + \tfrac12at^2 \qquad\qquad s = vt - \tfrac12at^2

v^2 = u^2 + 2as

They are valid only when the acceleration is constant throughout the stage.

Choosing the right equation

Each equation leaves out exactly one of the five quantities. Find the quantity you neither know nor want, and use the equation without it.

Equation	Quantity it does not contain
$v = u + at$	$s$
$s = \tfrac12(u + v)t$	$a$
$s = ut + \tfrac12at^2$	$v$
$s = vt - \tfrac12at^2$	$u$
$v^2 = u^2 + 2as$	$t$

Solving a constant acceleration problem

Check the acceleration is constant during the stage. If the motion has several stages, treat each one separately.
State the positive direction (usually the direction of the initial motion).
List $s$ , $u$ , $v$ , $a$ , $t$ for the stage, with signs. Put a question mark against the one you want and a dash against the one you do not need.
Choose the equation that does not contain the unneeded quantity. Substitute and solve.
If you get a quadratic in $t$ , find both roots and decide what each means.
Answer in the form asked: speed (positive) or velocity (with direction), distance or displacement, to 3 significant figures.

Words in the question often give a value without stating it:

"starts from rest" or "is released from rest": $u = 0$ ;
"comes to rest" or "stops": $v = 0$ ;
"decelerates at $2\ \text{m s}^{-2}$ ": $a = -2$ , with the direction of motion positive;
"uniform acceleration" or "constant acceleration": suvat applies;
"moves with constant speed": $a = 0$ and $s = vt$ .

Finding an acceleration and a time

A car travelling along a straight road passes a point $A$ with speed $8\ \text{m s}^{-1}$ . It accelerates uniformly and passes a point $B$ , $84\ \text{m}$ further on, with speed $20\ \text{m s}^{-1}$ . Find the acceleration of the car and the time it takes to travel from $A$ to $B$ .

Solution

Take the direction of motion as positive: $s = 84$ , $u = 8$ , $v = 20$ , $a = ?$ , $t$ not needed for the first part.

The equation without $t$ is $v^2 = u^2 + 2as$ :

400 = 64 + 2a(84) \quad\Rightarrow\quad 168a = 336 \quad\Rightarrow\quad a = 2\ \text{m s}^{-2}

Now $v = u + at$ :

20 = 8 + 2t \quad\Rightarrow\quad t = 6\ \text{s}

Check with the equation that does not use $a$ (so it does not rely on the first answer): $s = \tfrac12(8 + 20)(6) = 84$ . Correct.

Braking to rest

A train is travelling at $30\ \text{m s}^{-1}$ when the brakes are applied. It decelerates uniformly and comes to rest in a distance of $450\ \text{m}$ .

(a) Find the deceleration and the time taken to stop.

(b) Find the distance travelled in the last $10\ \text{s}$ before the train stops.

Solution

(a) Direction of motion positive: $s = 450$ , $u = 30$ , $v = 0$ , $a = ?$ .

0 = 30^2 + 2a(450) \quad\Rightarrow\quad a = -\frac{900}{900} = -1

The deceleration is $1\ \text{m s}^{-2}$ . Then $0 = 30 + (-1)t$ , so $t = 30\ \text{s}$ .

(b) For the last $10\ \text{s}$ we know the final velocity $v = 0$ , the time $t = 10$ and $a = -1$ , but not the velocity at the start of those $10\ \text{s}$ . Use the equation without $u$ :

s = vt - \tfrac12at^2 = 0 - \tfrac12(-1)(10)^2 = 50\ \text{m}

Alternatively, find the distance in the first $20\ \text{s}$ and subtract from $450$ . The direct route is shorter.

When the particle turns back

If the acceleration is opposite to the initial velocity, the particle slows down, stops for an instant, then moves back the other way, still with the same acceleration. The suvat equations describe the whole of this motion in one go, because the acceleration never changes. The displacement $s$ is measured from the starting point, so it can decrease after the turning point and even become negative.

This is why an equation like $s = ut + \tfrac12at^2$ can give two positive times: the particle passes the same position once on the way out and once on the way back.

Two times at the same position

A particle moving in a straight line passes a point $O$ with velocity $12\ \text{m s}^{-1}$ . It has a constant acceleration of $3\ \text{m s}^{-2}$ in the direction opposite to its initial velocity.

(a) Find the times at which the particle is $18\ \text{m}$ from $O$ on the positive side.

(b) Find the total distance travelled by the particle in the first $6\ \text{s}$ .

Solution

Take the direction of the initial velocity as positive, so $u = 12$ and $a = -3$ .

(a) $s = 18$ , $u = 12$ , $a = -3$ , $t = ?$ :

18 = 12t - 1.5t^2 \quad\Rightarrow\quad 1.5t^2 - 12t + 18 = 0 \quad\Rightarrow\quad t^2 - 8t + 12 = 0

(t - 2)(t - 6) = 0 \quad\Rightarrow\quad t = 2 \text{ or } t = 6

The particle passes the point $18\ \text{m}$ from $O$ at $t = 2\ \text{s}$ on its way out and again at $t = 6\ \text{s}$ on its way back.

(b) The displacement at $t = 6$ is $18\ \text{m}$ , but that is not the distance travelled. First find when the particle turns round ( $v = 0$ ):

0 = 12 - 3t \quad\Rightarrow\quad t = 4

At $t = 4$ : $s = 12(4) - 1.5(16) = 24\ \text{m}$ . So the particle travels $24\ \text{m}$ out, then comes back from $24\ \text{m}$ to $18\ \text{m}$ .

\text{distance} = 24 + (24 - 18) = 30\ \text{m}

Watch out

Whenever a particle might change direction, distance travelled is not the value of $s$ . Find the time at which $v = 0$ , work out the displacement at that time, and add up the outward and return journeys separately.

Problems with several stages

Many questions describe a journey in stages: accelerate, cruise, brake. The acceleration is constant within each stage but changes between them, so apply suvat to one stage at a time. The link between stages is the velocity: the final velocity of one stage is the initial velocity of the next.

A sketch of the $v$ – $t$ graph is often the quickest way to organise a multi-stage problem (see Displacement–time and velocity–time graphs). You can then use areas and gradients, suvat, or a mixture, whichever is shorter.

When a question gives information about the same stage from two different starting points, it is often easiest to apply suvat from a single fixed point to each of them, and solve simultaneously.

Three points on a line

A particle moves along a straight line with constant acceleration. It passes points $A$ , $B$ and $C$ in that order. $AB = 24\ \text{m}$ and $BC = 60\ \text{m}$ . The particle takes $2\ \text{s}$ to travel from $A$ to $B$ and $3\ \text{s}$ to travel from $B$ to $C$ . Find the acceleration of the particle and its speed at $C$ .

Solution

Let the speed at $A$ be $u$ and the acceleration $a$ . Apply $s = ut + \tfrac12at^2$ from $A$ both times, because the speed at $A$ is common.

$A$ to $B$ : $s = 24$ , $t = 2$ :

24 = 2u + 2a \quad\Rightarrow\quad u + a = 12 \qquad (1)

$A$ to $C$ : $s = 24 + 60 = 84$ , $t = 2 + 3 = 5$ :

84 = 5u + 12.5a \qquad (2)

From (1), $u = 12 - a$ . Substitute into (2):

84 = 60 - 5a + 12.5a \quad\Rightarrow\quad 7.5a = 24 \quad\Rightarrow\quad a = 3.2\ \text{m s}^{-2}

So $u = 8.8\ \text{m s}^{-1}$ , and at $C$ :

v = u + at = 8.8 + 3.2(5) = 24.8\ \text{m s}^{-1}

Check $BC$ : the speed at $B$ is $8.8 + 6.4 = 15.2$ , and $15.2(3) + \tfrac12(3.2)(9) = 45.6 + 14.4 = 60$ . Correct.

Two particles

When two particles move on the same line, write the displacement of each from the same origin, in terms of the same time variable, and then use the condition in the question:

they meet (or one catches the other) when their positions are equal;
they are a given distance apart when the difference of their positions is that distance;
if they move towards each other from points $d$ apart, they meet when the sum of the distances each has travelled is $d$ .

If one particle starts later, its time is $t - T$ , where $T$ is the delay.

Particles moving towards each other (exam standard)

Points $O$ and $X$ are $120\ \text{m}$ apart on a straight line. Particle $A$ starts from rest at $O$ and moves towards $X$ with constant acceleration $1\ \text{m s}^{-2}$ . At the same instant particle $B$ passes through $X$ , moving towards $O$ with speed $8\ \text{m s}^{-1}$ and constant deceleration $0.2\ \text{m s}^{-2}$ . Find the time at which the particles meet, the distance $OA$ at that moment, and the speed of each particle as they meet.

Solution

After $t$ seconds, $A$ has travelled $\tfrac12(1)t^2 = 0.5t^2$ metres from $O$ .

For $B$ , take the direction of its motion as positive, so $u = 8$ and $a = -0.2$ . It has travelled $8t - 0.1t^2$ metres from $X$ .

They meet when the two distances add up to $120$ :

0.5t^2 + 8t - 0.1t^2 = 120 \quad\Rightarrow\quad 0.4t^2 + 8t - 120 = 0 \quad\Rightarrow\quad t^2 + 20t - 300 = 0

(t + 30)(t - 10) = 0 \quad\Rightarrow\quad t = 10\ \text{s}

(The root $t = -30$ is rejected.) $B$ comes to rest only at $t = \tfrac{8}{0.2} = 40\ \text{s}$ , so it is still moving towards $O$ at $t = 10$ and the model holds.

OA = 0.5(10)^2 = 50\ \text{m}

Speeds at $t = 10$ : $A$ has $v = 0 + 1(10) = 10\ \text{m s}^{-1}$ and $B$ has $v = 8 - 0.2(10) = 6\ \text{m s}^{-1}$ .

Distance in a particular second

A particle starts from rest and moves with constant acceleration. In the fifth second of its motion it travels $13.5\ \text{m}$ . Find its acceleration and the distance it travels in the first $8\ \text{s}$ .

Solution

"The fifth second" means the interval from $t = 4$ to $t = 5$ . With $u = 0$ , the displacement after $t$ seconds is $\tfrac12at^2$ , so

\tfrac12a(5^2) - \tfrac12a(4^2) = 13.5 \quad\Rightarrow\quad 4.5a = 13.5 \quad\Rightarrow\quad a = 3\ \text{m s}^{-2}

In the first $8\ \text{s}$ : $s = \tfrac12(3)(64) = 96\ \text{m}$ .

Common mistakes

Suvat errors

Using suvat when the acceleration is not constant. If $a$ depends on $t$ , or the question gives $v$ as a function of $t$ , use calculus instead (see Variable acceleration (note not yet published)).
Applying one equation across two stages with different accelerations. Split the motion and carry the velocity across.
Wrong sign on $a$ . A deceleration of $2\ \text{m s}^{-2}$ is $a = -2$ when the motion is positive. Write the sign once and only once.
Taking $s$ as the distance travelled when the particle has turned back.
Discarding a valid root of a quadratic in $t$ . Both positive roots usually mean something; decide which the question wants.
Assuming the speed at the midpoint of a distance is the average speed. With constant acceleration the average velocity is $\tfrac12(u + v)$ , but the particle reaches that speed at the midpoint in time, not in distance.

Exam technique

Exam tip

Write the list $s = \ldots,\ u = \ldots,\ v = \ldots,\ a = \ldots,\ t = \ldots$ before substituting. It earns no marks on its own but makes the method mark for the equation almost certain.
Quote the equation you use, then substitute. " $v^2 = u^2 + 2as$ , $0 = 30^2 + 2a(450)$ " is clear to an examiner; a bare number is not.
For "show that" questions, carry exact values or at least four significant figures until the end.
If a later part needs a value you found earlier, use the unrounded value. Rounding $a$ to 3 s.f. and then reusing it can change the third figure of the final answer.
Before using any suvat equation in a forces question, check that the forces (and so the acceleration) are constant over the stage. When a string breaks or a particle reaches a rough patch, the acceleration changes and a new stage begins.

Summary

Suvat applies only to motion in a straight line with constant acceleration, one stage at a time.
$v = u + at$ , $s = \tfrac12(u + v)t$ , $s = ut + \tfrac12at^2$ , $s = vt - \tfrac12at^2$ , $v^2 = u^2 + 2as$ .
Each equation omits one of $s, u, v, a, t$ : choose the one that omits the quantity you neither know nor want.
They come from the gradient and area of a straight-line $v$ – $t$ graph.
Fix a positive direction. Decelerations are negative accelerations.
A quadratic in $t$ often has two meaningful roots: the particle passes a point twice.
In multi-stage problems the final velocity of one stage is the initial velocity of the next.
For two particles, write each position from a common origin in terms of a common time.

Practice

Question

A cyclist accelerates uniformly from $3\ \text{m s}^{-1}$ to $9\ \text{m s}^{-1}$ in $8\ \text{s}$ . Find the acceleration and the distance travelled in this time.
A car travelling at $25\ \text{m s}^{-1}$ brakes with constant deceleration $5\ \text{m s}^{-2}$ . Find the distance it travels before stopping and the time taken.
A particle has initial velocity $4\ \text{m s}^{-1}$ and constant acceleration $1.5\ \text{m s}^{-2}$ . Find its velocity after $6\ \text{s}$ and the distance it travels in that time.
A particle starts from rest and moves with constant acceleration. It travels $14\ \text{m}$ during the fourth second of its motion. Find the acceleration and the distance travelled in the first $6\ \text{s}$ .
A particle passes through $O$ with velocity $15\ \text{m s}^{-1}$ and has constant acceleration $-2.5\ \text{m s}^{-2}$ . (a) Find the times at which it is $40\ \text{m}$ from $O$ on the positive side. (b) Find the time at which it returns to $O$ and the total distance it has then travelled.
A car passes a point $A$ and moves with constant acceleration to a point $B$ , $150\ \text{m}$ away, taking $10\ \text{s}$ . Its speed at $B$ is $20\ \text{m s}^{-1}$ . Find its speed at $A$ , its acceleration, and its speed at the midpoint of $AB$ .
A train starts from rest and accelerates uniformly at $0.5\ \text{m s}^{-2}$ for $40\ \text{s}$ . It then decelerates uniformly, coming to rest after a further $600\ \text{m}$ . Find the deceleration, the total time and the total distance.
Car $P$ starts from rest at a point $A$ and moves along a straight road with constant acceleration $0.5\ \text{m s}^{-2}$ . Ten seconds later car $Q$ starts from rest at $A$ and follows $P$ with constant acceleration $2\ \text{m s}^{-2}$ . Find the time after $P$ starts at which $Q$ draws level with $P$ , the distance from $A$ , and the speed of each car at that moment.
A particle moving with constant acceleration passes points $A$ , $B$ and $C$ in that order, where $AB = BC = 40\ \text{m}$ . It takes $4\ \text{s}$ to go from $A$ to $B$ and $2\ \text{s}$ to go from $B$ to $C$ . Find the acceleration, the speed at $A$ and the speed at $C$ .

Answers

$a = \dfrac{9 - 3}{8} = 0.75\ \text{m s}^{-2}$ . $s = \tfrac12(3 + 9)(8) = 48\ \text{m}$ .
$0 = 25^2 + 2(-5)s$ , so $s = \dfrac{625}{10} = 62.5\ \text{m}$ . $0 = 25 - 5t$ , so $t = 5\ \text{s}$ .
$v = 4 + 1.5(6) = 13\ \text{m s}^{-1}$ . $s = 4(6) + \tfrac12(1.5)(36) = 24 + 27 = 51\ \text{m}$ (the particle never reverses, so this is also the distance).
$\tfrac12a(16) - \tfrac12a(9) = 14$ , so $3.5a = 14$ and $a = 4\ \text{m s}^{-2}$ . First $6\ \text{s}$ : $\tfrac12(4)(36) = 72\ \text{m}$ .
(a) $40 = 15t - 1.25t^2$ gives $t^2 - 12t + 32 = 0$ , so $(t - 4)(t - 8) = 0$ and $t = 4\ \text{s}$ or $8\ \text{s}$ . (b) $0 = 15t - 1.25t^2$ gives $t = 12\ \text{s}$ . It turns at $t = 6$ , when $s = 90 - 45 = 45\ \text{m}$ , so the distance is $45 + 45 = 90\ \text{m}$ .
$150 = \tfrac12(u + 20)(10)$ , so $u = 10\ \text{m s}^{-1}$ . $a = \dfrac{20 - 10}{10} = 1\ \text{m s}^{-2}$ . At the midpoint ( $s = 75$ ): $v^2 = 10^2 + 2(1)(75) = 250$ , so $v = 15.8\ \text{m s}^{-1}$ , which is more than the average speed of $15\ \text{m s}^{-1}$ .
Speed after $40\ \text{s}$ : $0.5 \times 40 = 20\ \text{m s}^{-1}$ ; distance $\tfrac12(20)(40) = 400\ \text{m}$ . Braking: $0 = 400 + 2a(600)$ , so $a = -\tfrac13$ and the deceleration is $0.333\ \text{m s}^{-2}$ . Braking time: $600 = \tfrac12(20)t$ , so $t = 60\ \text{s}$ . Total time $100\ \text{s}$ ; total distance $1000\ \text{m}$ .
With $t$ the time since $P$ started: $P$ is $0.25t^2$ from $A$ and $Q$ is $(t - 10)^2$ from $A$ . Level when $(t - 10)^2 = 0.25t^2$ , so $t - 10 = 0.5t$ (taking $t > 10$ ) and $t = 20\ \text{s}$ . Distance $= 0.25(400) = 100\ \text{m}$ . Speeds: $P$ $0.5(20) = 10\ \text{m s}^{-1}$ , $Q$ $2(10) = 20\ \text{m s}^{-1}$ .
From $A$ : $40 = 4u + 8a$ and $80 = 6u + 18a$ . The first gives $u = 10 - 2a$ ; substituting, $80 = 60 + 6a$ , so $a = \tfrac{10}{3} = 3.33\ \text{m s}^{-2}$ and $u = \tfrac{10}{3} = 3.33\ \text{m s}^{-1}$ . At $C$ : $v = \tfrac{10}{3} + 6 \times \tfrac{10}{3} = \tfrac{70}{3} = 23.3\ \text{m s}^{-1}$ .

The five quantities

Where the equations come from

Choosing the right equation

When the particle turns back

Problems with several stages

Two particles

Common mistakes

Exam technique

Summary

Practice

Console