3.2.4 Constant acceleration formulae

What you'll learn

How displacement, velocity and acceleration describe motion in a straight line.
How to derive the constant acceleration formulae from definitions and a velocity–time graph.
How to choose and use the correct formula for a problem.
How to handle signs, stopping, changing direction and vertical motion.

The quantities used in kinematics

Kinematics is the study of motion without considering the forces causing it. Here, motion takes place along a straight line, so one direction can be chosen as positive and the opposite direction as negative.

The five quantities used in constant-acceleration problems are often called the SUVAT variables:

$s$ is the displacement, measured in metres (m).
$u$ is the initial velocity, measured in m s⁻¹.
$v$ is the final velocity, measured in m s⁻¹.
$a$ is the constant acceleration, measured in m s⁻².
$t$ is the elapsed time, measured in seconds (s).

Definition

Displacement and velocity

Displacement is the change in position in a specified direction. Velocity is the rate of change of displacement, so it also has a direction. Distance and speed, by contrast, are scalar quantities and have no direction.

Definition

Constant acceleration

Constant acceleration means that velocity changes by equal amounts in equal time intervals. The acceleration has one fixed value throughout the period being modelled.

Choosing a positive direction

Before calculating, choose one direction as positive. Every displacement, velocity and acceleration must then use signs consistent with that choice.

For example, if upwards is positive, a falling object's velocity is negative and its acceleration is $s−2-9.8\text{ m s}^{-2}$ .

Example

Interpreting signed motion

A particle has initial velocity $s−1u=12\text{ m s}^{-1}$ and acceleration $s−2a=-3\text{ m s}^{-2}$ .

The positive initial velocity means that the particle initially moves in the chosen positive direction.
Acceleration has the opposite sign to velocity, so the particle's velocity decreases by $s−13\text{ m s}^{-1}$ each second.
After 4 seconds, its velocity reaches zero. If the same acceleration continues, the velocity then becomes negative, meaning that the particle reverses direction.

Common Mistake

Negative acceleration

Negative acceleration does not automatically mean slowing down. An object slows down only when its velocity and acceleration have opposite signs.

Deriving the first formula

Acceleration is the rate of change of velocity. For constant acceleration,

a=v−ut.a=\frac{v-u}{t}.

Rearranging gives the first constant acceleration formula:

v=u+at.\boxed{v=u+at}.

This tells you how the velocity changes after time $t$ .

Example

Finding the velocity

A train travels at $s−118\text{ m s}^{-1}$ and accelerates uniformly at $s−21.5\text{ m s}^{-2}$ for 8 seconds. Find its final velocity.

Identify $u = 18$ , $a = 1.5$ and $t = 8$ .
Use the formula containing $u$ , $v$ , $a$ and $t$ :
$v = u + a t .$
Substitute:
$v = 18 + 1.5 (8) = 30.$
The final velocity is $s−130\text{ m s}^{-1}$ .

Displacement from a velocity–time graph

On a velocity–time graph:

the gradient represents acceleration;
the signed area between the graph and the time axis represents displacement.

For constant acceleration, velocity changes linearly, so the graph is a straight line.

Velocity-time graph showing constant acceleration and displacement as the area of a rectangle and triangle

The area under the graph is a rectangle plus a triangle:

s=ut+12(v−u)t=ut+12at2.\begin{aligned} s&=ut+\frac{1}{2}(v-u)t \\ &=ut+\frac{1}{2}at^2. \end{aligned}

Therefore,

s=ut+12at2.\boxed{s=ut+\frac{1}{2}at^2}.

Alternatively, the graph is a trapezium whose parallel sides have lengths $u$ and $v$ . This gives

s=12(u+v)t.\boxed{s=\frac{1}{2}(u+v)t}.

The quantity $u+v2\frac{u+v}{2}$ is the average velocity during constant acceleration.

Key Idea

Graph meaning

For constant acceleration, the velocity–time graph is a straight line: its gradient gives acceleration and its signed area gives displacement.

Example

Finding displacement

A cyclist has initial velocity $s−14\text{ m s}^{-1}$ and accelerates uniformly at $s−22\text{ m s}^{-2}$ for 6 seconds. Find the displacement.

The known quantities are $u = 4$ , $a = 2$ and $t = 6$ , so use
$s=ut+12at2.s=ut+\frac{1}{2}at^2.$
Substitute the values:
$s=4(6)+12(2)(62).s=4(6)+\frac{1}{2}(2)(6^2).$
Calculate:
$s = 24 + 36 = 60.$
The cyclist's displacement is 60 m.

Eliminating time

Sometimes time is not given and is not required. We can eliminate $t$ from the formulae.

Starting with

v = u + a t,

we have

t=v−ua.t=\frac{v-u}{a}.

Substitute this into $s=12(u+v)ts=\frac{1}{2}(u+v)t$ :

s=12(u+v)v−ua2as=(u+v)(v−u)2as=v2−u2.\begin{aligned} s&=\frac{1}{2}(u+v)\frac{v-u}{a} \\ 2as&=(u+v)(v-u) \\ 2as&=v^2-u^2. \end{aligned}

Hence,

v2=u2+2as.\boxed{v^2=u^2+2as}.

Example

Calculating a stopping distance

A car travels at $s−124\text{ m s}^{-1}$ and brakes with constant acceleration $s−2-6\text{ m s}^{-2}$ . Find its displacement before it stops.

At the stopping instant, $v = 0$ . The known quantities are therefore $u = 24$ , $v = 0$ and $a = - 6$ .
Since time is neither known nor needed, use
$v^2=u^2+2as.$
Substitute and solve:
$02=242+2(−6)s0=576−12ss=48.\begin{aligned} 0^2&=24^2+2(-6)s \\ 0&=576-12s \\ s&=48. \end{aligned}$
The stopping displacement is 48 m.

The five formulae

The standard constant acceleration formulae are:

v=u+at,s=ut+12at2,s=vt−12at2,s=12(u+v)t,v2=u2+2as.\begin{aligned} v&=u+at, \\ s&=ut+\frac{1}{2}at^2, \\ s&=vt-\frac{1}{2}at^2, \\ s&=\frac{1}{2}(u+v)t, \\ v^2&=u^2+2as. \end{aligned}

The formula $s=vt−12at2s=vt-\frac{1}{2}at^2$ follows by writing $u = v - a t$ and substituting into $s=12(u+v)ts=\frac{1}{2}(u+v)t$ .

Choosing a formula

Each formula omits one SUVAT variable. List what you know, identify what you need, and choose the formula that excludes the remaining unwanted quantity.

$v = u + a t$ excludes $s$ .
$s=ut+12at2s=ut+\frac{1}{2}at^2$ excludes $v$ .
$s=vt−12at2s=vt-\frac{1}{2}at^2$ excludes $u$ .
$s=12(u+v)ts=\frac{1}{2}(u+v)t$ excludes $a$ .
$v^2=u^2+2as$ excludes $t$ .

Tip

Selecting efficiently

Write down the five symbols $s, u, v, a, t$ , fill in the known values with signs and circle the required quantity. The best formula contains those quantities but not the unused unknown.

Example

Finding the duration of motion

A particle travels 45 m while accelerating uniformly from $s−13\text{ m s}^{-1}$ to $s−112\text{ m s}^{-1}$ . Find the time taken.

The known quantities are $s = 45$ , $u = 3$ and $v = 12$ . Acceleration is unknown and unnecessary.
Choose the formula excluding $a$ :
$s=12(u+v)t.s=\frac{1}{2}(u+v)t.$
Substitute and solve:
$45=12(3+12)t45=7.5tt=6.\begin{aligned} 45&=\frac{1}{2}(3+12)t \\ 45&=7.5t \\ t&=6. \end{aligned}$
The motion lasts 6 seconds.

Solving quadratic equations in time

Using $s=ut+12at2s=ut+\frac{1}{2}at^2$ can produce a quadratic equation. Two positive roots may represent two genuine times at which a particle passes the same position, usually once in each direction.

Example

Returning to the starting point

A particle starts with velocity $s−114\text{ m s}^{-1}$ and constant acceleration $s−2-2\text{ m s}^{-2}$ . Find when it returns to its starting point.

Returning to the starting point means its displacement from that point is $s = 0$ . Use
$s=ut+12at2.s=ut+\frac{1}{2}at^2.$
Substitute:
$0=14t+12(−2)t2=14t−t2.0=14t+\frac{1}{2}(-2)t^2=14t-t^2.$
Factorise:
$t (14 - t) = 0,$
giving $t = 0$ or $t = 14$ .
The root $t = 0$ is the starting instant. Therefore, the particle returns after 14 seconds.

Common Mistake

Discarding roots too quickly

Do not automatically reject one root of a quadratic. Check what each root means physically, although negative time is normally outside the model.

Vertical motion under gravity

Near Earth's surface, an object in free flight is modelled with constant downward acceleration of magnitude $s−2g=9.8\text{ m s}^{-2}$ .

If upwards is positive, use $s−2a=-9.8\text{ m s}^{-2}$ . At the highest point, the instantaneous velocity is zero, but the acceleration is still $s−2-9.8\text{ m s}^{-2}$ .

Example

Finding maximum height

A ball is projected vertically upwards at $s−119.6\text{ m s}^{-1}$ . Find the maximum height above its launch point.

Choose upwards as positive. Then $u = 19.6$ , $a = - 9.8$ and, at maximum height, $v = 0$ .
Time is not required, so use
$v^2=u^2+2as.$
Substitute:
$0^2=19.6^2+2(-9.8)s.$
Solve:
$s=19.6219.6=19.6.s=\frac{19.6^2}{19.6}=19.6.$
The maximum height above the launch point is 19.6 m.

Common Mistake

When SUVAT does not apply

These formulae require constant acceleration throughout the chosen interval. Do not apply one SUVAT equation across stages with different accelerations; model each stage separately.

Exam technique

In the exam

Choose a positive direction and attach signs to $s$ , $u$ , $v$ and $a$ before substituting.
List the known and required SUVAT variables, then select the formula that omits the unused unknown.
State special conditions explicitly, such as $v = 0$ at a stopping point or maximum height, and check that roots and units make physical sense.
Keep exact values during your working and round only at the end if the question requests a decimal.

Self review

Check yourself

Can you derive $s=ut+12at2s=ut+\frac{1}{2}at^2$ from the area under a velocity–time graph?
Which formula would you choose if time were unknown and not required?
If upwards is positive, what signs should velocity and acceleration have while an object is falling?

What you'll learn

How displacement, velocity and acceleration describe motion in a straight line.
How to derive the constant acceleration formulae from definitions and a velocity–time graph.
How to choose and use the correct formula for a problem.
How to handle signs, stopping, changing direction and vertical motion.

The quantities used in kinematics

The five quantities used in constant-acceleration problems are often called the SUVAT variables:

$s$ is the displacement, measured in metres (m).
$u$ is the initial velocity, measured in m s⁻¹.
$v$ is the final velocity, measured in m s⁻¹.
$a$ is the constant acceleration, measured in m s⁻².
$t$ is the elapsed time, measured in seconds (s).

Definition

Displacement and velocity

Definition

Constant acceleration

Constant acceleration means that velocity changes by equal amounts in equal time intervals. The acceleration has one fixed value throughout the period being modelled.

Choosing a positive direction

Before calculating, choose one direction as positive. Every displacement, velocity and acceleration must then use signs consistent with that choice.

For example, if upwards is positive, a falling object's velocity is negative and its acceleration is $s−2-9.8\text{ m s}^{-2}$ .

Example

Interpreting signed motion

A particle has initial velocity $s−1u=12\text{ m s}^{-1}$ and acceleration $s−2a=-3\text{ m s}^{-2}$ .

The positive initial velocity means that the particle initially moves in the chosen positive direction.
Acceleration has the opposite sign to velocity, so the particle's velocity decreases by $s−13\text{ m s}^{-1}$ each second.
After 4 seconds, its velocity reaches zero. If the same acceleration continues, the velocity then becomes negative, meaning that the particle reverses direction.

Common Mistake

Negative acceleration

Negative acceleration does not automatically mean slowing down. An object slows down only when its velocity and acceleration have opposite signs.

Deriving the first formula

Acceleration is the rate of change of velocity. For constant acceleration,

a=v−ut.a=\frac{v-u}{t}.

Rearranging gives the first constant acceleration formula:

v=u+at.\boxed{v=u+at}.

This tells you how the velocity changes after time $t$ .

Example

Finding the velocity

A train travels at $s−118\text{ m s}^{-1}$ and accelerates uniformly at $s−21.5\text{ m s}^{-2}$ for 8 seconds. Find its final velocity.

Identify $u = 18$ , $a = 1.5$ and $t = 8$ .
Use the formula containing $u$ , $v$ , $a$ and $t$ :
$v = u + a t .$
Substitute:
$v = 18 + 1.5 (8) = 30.$
The final velocity is $s−130\text{ m s}^{-1}$ .

Displacement from a velocity–time graph

On a velocity–time graph:

the gradient represents acceleration;
the signed area between the graph and the time axis represents displacement.

For constant acceleration, velocity changes linearly, so the graph is a straight line.

Velocity-time graph showing constant acceleration and displacement as the area of a rectangle and triangle

The area under the graph is a rectangle plus a triangle:

s=ut+12(v−u)t=ut+12at2.\begin{aligned} s&=ut+\frac{1}{2}(v-u)t \\ &=ut+\frac{1}{2}at^2. \end{aligned}

Therefore,

s=ut+12at2.\boxed{s=ut+\frac{1}{2}at^2}.

Alternatively, the graph is a trapezium whose parallel sides have lengths $u$ and $v$ . This gives

s=12(u+v)t.\boxed{s=\frac{1}{2}(u+v)t}.

The quantity $u+v2\frac{u+v}{2}$ is the average velocity during constant acceleration.

Key Idea

Graph meaning

For constant acceleration, the velocity–time graph is a straight line: its gradient gives acceleration and its signed area gives displacement.

Example

Finding displacement

A cyclist has initial velocity $s−14\text{ m s}^{-1}$ and accelerates uniformly at $s−22\text{ m s}^{-2}$ for 6 seconds. Find the displacement.

The known quantities are $u = 4$ , $a = 2$ and $t = 6$ , so use
$s=ut+12at2.s=ut+\frac{1}{2}at^2.$
Substitute the values:
$s=4(6)+12(2)(62).s=4(6)+\frac{1}{2}(2)(6^2).$
Calculate:
$s = 24 + 36 = 60.$
The cyclist's displacement is 60 m.

Eliminating time

Sometimes time is not given and is not required. We can eliminate $t$ from the formulae.

Starting with

v = u + a t,

we have

t=v−ua.t=\frac{v-u}{a}.

Substitute this into $s=12(u+v)ts=\frac{1}{2}(u+v)t$ :

s=12(u+v)v−ua2as=(u+v)(v−u)2as=v2−u2.\begin{aligned} s&=\frac{1}{2}(u+v)\frac{v-u}{a} \\ 2as&=(u+v)(v-u) \\ 2as&=v^2-u^2. \end{aligned}

Hence,

v2=u2+2as.\boxed{v^2=u^2+2as}.

Example

Calculating a stopping distance

A car travels at $s−124\text{ m s}^{-1}$ and brakes with constant acceleration $s−2-6\text{ m s}^{-2}$ . Find its displacement before it stops.

At the stopping instant, $v = 0$ . The known quantities are therefore $u = 24$ , $v = 0$ and $a = - 6$ .
Since time is neither known nor needed, use
$v^2=u^2+2as.$
Substitute and solve:
$02=242+2(−6)s0=576−12ss=48.\begin{aligned} 0^2&=24^2+2(-6)s \\ 0&=576-12s \\ s&=48. \end{aligned}$
The stopping displacement is 48 m.

The five formulae

The standard constant acceleration formulae are:

v=u+at,s=ut+12at2,s=vt−12at2,s=12(u+v)t,v2=u2+2as.\begin{aligned} v&=u+at, \\ s&=ut+\frac{1}{2}at^2, \\ s&=vt-\frac{1}{2}at^2, \\ s&=\frac{1}{2}(u+v)t, \\ v^2&=u^2+2as. \end{aligned}

The formula $s=vt−12at2s=vt-\frac{1}{2}at^2$ follows by writing $u = v - a t$ and substituting into $s=12(u+v)ts=\frac{1}{2}(u+v)t$ .

Choosing a formula

Each formula omits one SUVAT variable. List what you know, identify what you need, and choose the formula that excludes the remaining unwanted quantity.

$v = u + a t$ excludes $s$ .
$s=ut+12at2s=ut+\frac{1}{2}at^2$ excludes $v$ .
$s=vt−12at2s=vt-\frac{1}{2}at^2$ excludes $u$ .
$s=12(u+v)ts=\frac{1}{2}(u+v)t$ excludes $a$ .
$v^2=u^2+2as$ excludes $t$ .

Tip

Selecting efficiently

Write down the five symbols $s, u, v, a, t$ , fill in the known values with signs and circle the required quantity. The best formula contains those quantities but not the unused unknown.

Example

Finding the duration of motion

A particle travels 45 m while accelerating uniformly from $s−13\text{ m s}^{-1}$ to $s−112\text{ m s}^{-1}$ . Find the time taken.

The known quantities are $s = 45$ , $u = 3$ and $v = 12$ . Acceleration is unknown and unnecessary.
Choose the formula excluding $a$ :
$s=12(u+v)t.s=\frac{1}{2}(u+v)t.$
Substitute and solve:
$45=12(3+12)t45=7.5tt=6.\begin{aligned} 45&=\frac{1}{2}(3+12)t \\ 45&=7.5t \\ t&=6. \end{aligned}$
The motion lasts 6 seconds.

Solving quadratic equations in time

Example

Returning to the starting point

A particle starts with velocity $s−114\text{ m s}^{-1}$ and constant acceleration $s−2-2\text{ m s}^{-2}$ . Find when it returns to its starting point.

Returning to the starting point means its displacement from that point is $s = 0$ . Use
$s=ut+12at2.s=ut+\frac{1}{2}at^2.$
Substitute:
$0=14t+12(−2)t2=14t−t2.0=14t+\frac{1}{2}(-2)t^2=14t-t^2.$
Factorise:
$t (14 - t) = 0,$
giving $t = 0$ or $t = 14$ .
The root $t = 0$ is the starting instant. Therefore, the particle returns after 14 seconds.

Common Mistake

Discarding roots too quickly

Do not automatically reject one root of a quadratic. Check what each root means physically, although negative time is normally outside the model.

Vertical motion under gravity

Near Earth's surface, an object in free flight is modelled with constant downward acceleration of magnitude $s−2g=9.8\text{ m s}^{-2}$ .

If upwards is positive, use $s−2a=-9.8\text{ m s}^{-2}$ . At the highest point, the instantaneous velocity is zero, but the acceleration is still $s−2-9.8\text{ m s}^{-2}$ .

Example

Finding maximum height

A ball is projected vertically upwards at $s−119.6\text{ m s}^{-1}$ . Find the maximum height above its launch point.

Choose upwards as positive. Then $u = 19.6$ , $a = - 9.8$ and, at maximum height, $v = 0$ .
Time is not required, so use
$v^2=u^2+2as.$
Substitute:
$0^2=19.6^2+2(-9.8)s.$
Solve:
$s=19.6219.6=19.6.s=\frac{19.6^2}{19.6}=19.6.$
The maximum height above the launch point is 19.6 m.

Common Mistake

When SUVAT does not apply

These formulae require constant acceleration throughout the chosen interval. Do not apply one SUVAT equation across stages with different accelerations; model each stage separately.

Exam technique

In the exam

Choose a positive direction and attach signs to $s$ , $u$ , $v$ and $a$ before substituting.
List the known and required SUVAT variables, then select the formula that omits the unused unknown.
State special conditions explicitly, such as $v = 0$ at a stopping point or maximum height, and check that roots and units make physical sense.
Keep exact values during your working and round only at the end if the question requests a decimal.

Self review

Check yourself

Can you derive $s=ut+12at2s=ut+\frac{1}{2}at^2$ from the area under a velocity–time graph?
Which formula would you choose if time were unknown and not required?
If upwards is positive, what signs should velocity and acceleration have while an object is falling?

What you'll learn

The quantities used in kinematics

Choosing a positive direction

Deriving the first formula

Displacement from a velocity–time graph

Eliminating time

The five formulae

Choosing a formula

Solving quadratic equations in time

Vertical motion under gravity

Review 3.2.4 Constant acceleration formulae by teaching Genie

3.2.4 Constant acceleration formulae

What you'll learn

The quantities used in kinematics

Choosing a positive direction

Deriving the first formula

Displacement from a velocity–time graph

Eliminating time

The five formulae

Choosing a formula

Solving quadratic equations in time

Vertical motion under gravity

Review 3.2.4 Constant acceleration formulae by teaching Genie

3.2.4 Constant acceleration formulae

What you'll learn

The quantities used in kinematics

Choosing a positive direction

Deriving the first formula

Displacement from a velocity–time graph

Eliminating time

The five formulae

Choosing a formula

Solving quadratic equations in time

Vertical motion under gravity

Review 3.2.4 Constant acceleration formulae by teaching Genie

3.2.4 Constant acceleration formulae Revision Guide

3.2.4 Constant acceleration formulae

What you'll learn

The quantities used in kinematics

Choosing a positive direction

Deriving the first formula

Displacement from a velocity–time graph

Eliminating time

The five formulae

Choosing a formula

Solving quadratic equations in time

Vertical motion under gravity

Review 3.2.4 Constant acceleration formulae by teaching Genie

3.2.4 Constant acceleration formulae Revision Guide