Constant acceleration and SUVAT equations

The SUVAT equations connect displacement, velocity, acceleration and time when acceleration is constant. Their power comes from choosing the equation that contains exactly the quantities you know and the one you need.

This lesson considers motion along a straight line. You must choose a positive direction and keep that choice throughout each stage of the motion.

Prerequisites

You should be able to:

• distinguish distance from displacement, and speed from velocity, using kinematics language;
• work confidently with negative numbers, powers and square roots;
• solve linear and quadratic equations;
• convert common units using quantities and units in mechanics.

The five quantities

The name SUVAT comes from five symbols:

Symbol	Meaning	SI unit
$s$	displacement during the time interval	$\mathrm m$
$u$	velocity at the start of the interval	$\mathrm{m\,s^{-1}}$
$v$	velocity at the end of the interval	$\mathrm{m\,s^{-1}}$
$a$	constant acceleration	$\mathrm{m\,s^{-2}}$
$t$	elapsed time	$\mathrm s$

The words start and end refer to the chosen time interval. If a journey is split into stages, the final velocity of one stage may become the initial velocity of the next.

Displacement, velocity and acceleration are directed quantities, so they can be negative. Time is normally positive. Distance and speed are magnitudes, so they are non-negative.

The model condition

SUVAT equations are valid only while acceleration is constant. They cannot be applied across an interval in which acceleration changes. If the motion has several constant-acceleration stages, treat each stage separately.

The constant acceleration equations

For constant acceleration,

$$\boxed{v=u+at}$$ $$\boxed{s=ut+\frac12at^2}$$ $$\boxed{s=vt-\frac12at^2}$$ $$\boxed{s=\frac{u+v}{2}t}$$ $$\boxed{v^2=u^2+2as}$$

You do not need all five equations in every problem. Each equation omits one SUVAT quantity:

Equation	Quantity omitted
$v=u+at$	$s$
$s=ut+\frac12at^2$	$v$
$s=vt-\frac12at^2$	$u$
$s=\frac{u+v}{2}t$	$a$
$v^2=u^2+2as$	$t$

This gives a reliable selection method:

1. Write down the quantities you know, with signs and units.
2. Identify the quantity you need.
3. Choose an equation containing those quantities but no unwanted unknown.
4. Substitute only after choosing the equation.
5. Solve, then interpret the sign and check whether the answer is physically possible.

Why the equations work

Constant acceleration means velocity changes at a constant rate, so

$$a=\frac{v-u}{t}.$$

Rearranging gives $v=u+at$.

On a velocity-time graph, constant acceleration produces a straight line. Displacement is the signed area under that line, which is average velocity multiplied by time:

$$s=\frac{u+v}{2}t.$$

Substituting $v=u+at$ gives

$$s=\frac{u+(u+at)}{2}t =ut+\frac12at^2.$$

The other equations follow by similar substitution. This connection is explored further in kinematics graphs and calculus in kinematics.

Direct use of one equation

Worked example 1: accelerating from rest

A cyclist starts from rest and accelerates uniformly at $1.8\ \mathrm{m\,s^{-2}}$ for $7$ seconds. Find the cyclist’s speed and displacement after $7$ seconds.

Choose the direction of motion as positive. Then

$$u=0,\qquad a=1.8,\qquad t=7.$$

For the final velocity, use $v=u+at$:

$$v=0+1.8(7)=12.6\ \mathrm{m\,s^{-1}}.$$

For displacement, use $s=ut+\frac12at^2$:

$$\begin{aligned} s&=0(7)+\frac12(1.8)(7^2)\\ &=44.1\ \mathrm m. \end{aligned}$$

The cyclist travels $\boxed{44.1\ \mathrm m}$ and reaches a speed of $\boxed{12.6\ \mathrm{m\,s^{-1}}}$.

Self-check 1

A train increases its speed uniformly from $8\ \mathrm{m\,s^{-1}}$ to $20\ \mathrm{m\,s^{-1}}$ in $15$ seconds. Find its acceleration and displacement.

Answer

Using $v=u+at$,

$$a=\frac{v-u}{t}=\frac{20-8}{15}=0.8\ \mathrm{m\,s^{-2}}.$$

Using average velocity,

$$s=\frac{u+v}{2}t=\frac{8+20}{2}(15)=210\ \mathrm m.$$

Signs, slowing down and stopping

Acceleration is not automatically positive. If positive is chosen in the direction of the initial motion, slowing down is represented by $a<0$.

The precise rule is:

• velocity and acceleration with the same sign means speed is increasing;
• velocity and acceleration with opposite signs means speed is decreasing.

Worked example 2: braking distance

A car is travelling at $24\ \mathrm{m\,s^{-1}}$. The brakes produce a constant acceleration of magnitude $6\ \mathrm{m\,s^{-2}}$. Find the time and distance required for the car to stop.

Take the car’s initial direction as positive. Braking acceleration acts in the opposite direction, so

$$u=24,\qquad v=0,\qquad a=-6.$$

For time, use $v=u+at$:

$$0=24-6t,$$

so

$$t=4\ \mathrm s.$$

For displacement, the equation omitting time is efficient:

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

Therefore

$$0^2=24^2+2(-6)s,$$

so

$$12s=576 \quad\Longrightarrow\quad s=48\ \mathrm m.$$

The car stops after $\boxed{4\ \mathrm s}$ and travels $\boxed{48\ \mathrm m}$ while braking.

What happens after $v=0$?

If the same acceleration continued beyond $4$ seconds in this mathematical model, $v$ would become negative and the car would reverse. Real brakes do not make a stationary car reverse, so the constant-acceleration model ends when the car stops.

Self-check 2

A particle has initial velocity $18\ \mathrm{m\,s^{-1}}$ and constant acceleration $-3\ \mathrm{m\,s^{-2}}$. Find when it first comes to rest and its displacement by then.

Answer

At rest, $v=0$. Hence

$$0=18-3t \quad\Longrightarrow\quad t=6\ \mathrm s.$$

Then

$$s=18(6)+\frac12(-3)(6^2)=54\ \mathrm m.$$

Unknown time and quadratic equations

When displacement, initial velocity and acceleration are known, using

$$s=ut+\frac12at^2$$

often produces a quadratic in $t$. A quadratic may have two valid positive roots because a particle can pass the same position twice.

Worked example 3: passing a point twice

A particle moves along a line with initial velocity $14\ \mathrm{m\,s^{-1}}$ and constant acceleration $-2\ \mathrm{m\,s^{-2}}$. Find the times at which its displacement from its starting point is $40$ m.

Here

$$u=14,\qquad a=-2,\qquad s=40.$$

Use $s=ut+\frac12at^2$:

$$40=14t+\frac12(-2)t^2.$$

Rearrange:

$$t^2-14t+40=0.$$

Factorise:

$$(t-4)(t-10)=0.$$

Therefore

$$\boxed{t=4\ \mathrm s\quad\text{or}\quad t=10\ \mathrm s}.$$

These are not duplicate mathematical answers. At $t=4$,

$$v=14-2(4)=6\ \mathrm{m\,s^{-1}},$$

so the particle passes the point in the positive direction. It reaches its greatest displacement when $v=0$, at $t=7$ seconds, then reverses. At $t=10$,

$$v=14-2(10)=-6\ \mathrm{m\,s^{-1}},$$

so it passes the same point on its return.

Do not discard a root without interpreting it

Reject a time only when it lies outside the stated interval or conflicts with the model. Two positive roots can both be meaningful.

Vertical motion under gravity

Near Earth’s surface, a freely moving particle is modelled as having constant downward acceleration of magnitude $g$. Unless a question specifies otherwise, use the given value of $g$, commonly

$$g=9.8\ \mathrm{m\,s^{-2}}.$$

Air resistance is neglected. The sign of $a$ depends on the chosen positive direction:

• positive upwards gives $a=-g$;
• positive downwards gives $a=+g$.

The acceleration remains downward even while an object is travelling upwards. At the highest point, velocity is instantaneously zero, but acceleration is still $-g$ if upwards is positive.

Worked example 4: thrown vertically upwards

A ball is projected vertically upwards from a point $1.5$ m above the ground with speed $19.6\ \mathrm{m\,s^{-1}}$. Ignore air resistance and take $g=9.8\ \mathrm{m\,s^{-2}}$. Find:

1. the time taken to reach its highest point;
2. the greatest height above the ground;
3. the speed just before it reaches the ground.

Choose upwards as positive. During the whole flight,

$$u=19.6,\qquad a=-9.8.$$

At the highest point, $v=0$. Therefore

$$0=19.6-9.8t,$$

giving

$$t=2\ \mathrm s.$$

Let $s$ be displacement from the launch point. At the top,

$$\begin{aligned} 0^2&=19.6^2+2(-9.8)s,\\ s&=19.6\ \mathrm m. \end{aligned}$$

The launch point is $1.5$ m above the ground, so the greatest height is

$$19.6+1.5=\boxed{21.1\ \mathrm m}.$$

At the ground, displacement from the launch point is $s=-1.5$ m. Hence

$$\begin{aligned} v^2&=19.6^2+2(-9.8)(-1.5)\\ &=413.56. \end{aligned}$$

Thus $v=\pm\sqrt{413.56}$. The ball is moving downwards at impact, so its velocity is negative:

$$v\approx-20.3\ \mathrm{m\,s^{-1}}.$$

The question asks for speed, which is the magnitude. The speed is therefore

$$\boxed{20.3\ \mathrm{m\,s^{-1}}}$$

to $3$ significant figures.

Self-check 3

A stone is released from rest $45$ m above the ground. Taking $g=9.8\ \mathrm{m\,s^{-2}}$, find the time it takes to reach the ground and its impact speed.

Answer

Choose downwards as positive. Then $u=0$, $a=9.8$ and $s=45$.

$$45=\frac12(9.8)t^2 \quad\Longrightarrow\quad t=\sqrt{\frac{90}{9.8}}\approx3.03\ \mathrm s.$$

Only the positive time is physically relevant. Also,

$$v^2=0^2+2(9.8)(45),$$

so

$$v=\sqrt{882}=\boxed{29.7\ \mathrm{m\,s^{-1}}}$$

to $3$ significant figures.

Motion in stages

A single SUVAT calculation cannot cross a point where acceleration changes. Split the motion into stages and transfer shared information between them.

Worked example 5: accelerate, then brake

A train leaves a station from rest and accelerates uniformly at $0.6\ \mathrm{m\,s^{-2}}$ for $30$ seconds. It then travels at constant speed for $50$ seconds before braking uniformly to rest in $20$ seconds. Find the total distance travelled.

Stage 1: accelerating

$$u=0,\qquad a=0.6,\qquad t=30.$$

The speed after this stage is

$$v=0+0.6(30)=18\ \mathrm{m\,s^{-1}}.$$

The distance is

$$s_1=0(30)+\frac12(0.6)(30^2)=270\ \mathrm m.$$

Stage 2: constant speed

Here $a=0$, so

$$s_2=vt=18(50)=900\ \mathrm m.$$

Stage 3: braking

The speed changes uniformly from $18$ to $0\ \mathrm{m\,s^{-1}}$. Using average velocity,

$$s_3=\frac{18+0}{2}(20)=180\ \mathrm m.$$

Therefore the total distance is

$$s_1+s_2+s_3=270+900+180=\boxed{1350\ \mathrm m}.$$

Notice that the end speed of Stage 1 supplies the speed for Stages 2 and 3.

Common misconceptions

Treating speed as velocity

A speed of $12\ \mathrm{m\,s^{-1}}$ gives a magnitude but no direction. After choosing positive, the corresponding velocity may be $12$ or $-12\ \mathrm{m\,s^{-1}}$.

Making deceleration positive automatically

The word decelerates means speed decreases. It does not determine the sign of acceleration until a positive direction is chosen.

Using distance where the formula needs displacement

In SUVAT, $s$ is signed displacement. If a particle goes out and returns to its start, then $s=0$ although the distance travelled is positive.

Mixing stages

An average acceleration over a whole journey does not make SUVAT valid over that journey. Use separate calculations whenever acceleration changes.

Assuming $v=0$ means $a=0$

At the top of vertical motion, $v=0$ for an instant while $a=-g$ remains nonzero.

Taking the positive square root without checking

From $v^2=k$, algebra gives $v=\pm\sqrt{k}$. The direction of motion selects the appropriate velocity. If the question asks for speed, report $|v|$.

Mixed self-check

A lift is moving upwards at $2\ \mathrm{m\,s^{-1}}$ when it begins to accelerate downwards at a constant $0.5\ \mathrm{m\,s^{-2}}$.

1. How long does it take to stop momentarily?
2. How far above its initial position is it then?
3. Find its displacement from the initial position after $8$ seconds.
4. Find the total distance travelled during those $8$ seconds.

Answer

Choose upwards as positive, so $u=2$ and $a=-0.5$.

At the turning point, $v=0$:

$$0=2-0.5t \quad\Longrightarrow\quad t=4\ \mathrm s.$$

Its displacement then is

$$s=2(4)+\frac12(-0.5)(4^2)=4\ \mathrm m.$$

After $8$ seconds,

$$s=2(8)+\frac12(-0.5)(8^2)=0.$$

The lift has returned to its initial position, so its displacement is $0$ m. It travelled $4$ m upwards and then $4$ m downwards, so the total distance is

$$\boxed{8\ \mathrm m}.$$

Problem-solving checklist

Before accepting an answer, ask:

• Is acceleration constant over the interval?
• Have I chosen and stated a positive direction?
• Are all quantities in compatible units?
• Is $s$ displacement rather than distance?
• Does my equation contain only one unknown?
• Have I interpreted every root and sign in context?
• Is the size of the result plausible?

Next steps

• Use gradient and area to interpret the same equations in kinematics graphs.
• Handle velocity and acceleration that vary with time in calculus in kinematics.
• Apply constant vertical acceleration alongside constant horizontal velocity in projectiles.
• Connect acceleration to its physical cause using Newton’s laws.