3.2.5 Constant acceleration in two dimensions (A-level only)

What you'll learn

How to represent displacement, velocity and acceleration as two-dimensional vectors.
How to extend the constant-acceleration formulae to vector form.
How to solve motion separately in horizontal and vertical directions.
How to model projectiles when gravity is the only acceleration.

Prerequisites: vectors and components

A vector is a quantity with both magnitude and direction. Displacement, velocity and acceleration are all vectors, so they may be written using column vectors.

For example,

s−1\mathbf{u}= \begin{pmatrix} 6\\ 8 \end{pmatrix} \text{ m s}^{-1}

means that the initial velocity has a horizontal component of 6 m s⁻¹ and a vertical component of 8 m s⁻¹.

The letters $i\mathbf{i}$ and $j\mathbf{j}$ can also represent unit vectors in the positive horizontal and vertical directions:

s−1.\mathbf{u}=6\mathbf{i}+8\mathbf{j}\text{ m s}^{-1}.

These two forms mean exactly the same thing.

Definition

Vector components

The components of a vector are its effects in chosen perpendicular directions, usually the horizontal $x$ -direction and vertical $y$ -direction.

The magnitude of a two-dimensional vector

v=(vxvy)\mathbf{v}= \begin{pmatrix} v_x\\ v_y \end{pmatrix}

is found using Pythagoras:

∣v∣=vx2+vy2.|\mathbf{v}|=\sqrt{v_x^2+v_y^2}.

For a velocity vector, this magnitude is the object's speed.

If $θ\theta$ is the direction measured anticlockwise from the positive horizontal direction, then

tan⁡θ=vyvx.\tan\theta=\frac{v_y}{v_x}.

You must use the signs of the components, or a sketch, to place the direction in the correct quadrant.

Example

Finding speed and direction from velocity

An object has velocity $s−1\mathbf{v}=5\mathbf{i}+12\mathbf{j}\text{ m s}^{-1}$ . Find its speed and direction.

Use the magnitude formula to find the speed:
$s−1.|\mathbf{v}|=\sqrt{5^2+12^2}=13\text{ m s}^{-1}.$
Since both components are positive, the velocity points up and to the right. Its direction above the horizontal satisfies
$tan⁡θ=125.\tan\theta=\frac{12}{5}.$
Therefore,
$θ=tan⁡−1(125)≈67.4∘.\theta=\tan^{-1}\left(\frac{12}{5}\right)\approx67.4^\circ.$
The velocity is 13 m s⁻¹ at approximately $67.4∘67.4^\circ$ above the horizontal.

Common Mistake

Velocity is not speed

Velocity is a vector, but speed is its scalar magnitude. Do not give a column vector when the question asks for speed.

Constant acceleration in vector form

Constant acceleration means that the acceleration vector remains unchanged in both magnitude and direction.

The familiar one-dimensional constant-acceleration formulae extend directly to vectors:

v=u+at,\mathbf{v}=\mathbf{u}+\mathbf{a}t,

s=ut+12at2,\mathbf{s}=\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2,

s=12(u+v)t,\mathbf{s}=\frac{1}{2}(\mathbf{u}+\mathbf{v})t,

and

∣v∣2=∣u∣2+2a⋅s.|\mathbf{v}|^2=|\mathbf{u}|^2+2\mathbf{a}\cdot\mathbf{s}.

Here:

$u\mathbf{u}$ is initial velocity.
$v\mathbf{v}$ is velocity after time $t$ .
$a\mathbf{a}$ is constant acceleration.
$s\mathbf{s}$ is displacement during the time interval.
$a⋅s\mathbf{a}\cdot\mathbf{s}$ is the scalar product of acceleration and displacement.

The last equation uses a scalar product because acceleration and displacement may point in different directions. In components,

a⋅s=axsx+aysy.\mathbf{a}\cdot\mathbf{s}=a_xs_x+a_ys_y.

Key Idea

One vector equation contains two scalar equations

A two-dimensional vector equation is equivalent to one equation for the horizontal components and another for the vertical components. The same time $t$ appears in both because both parts describe the same moving object.

Vector diagram showing initial position, velocity components, downward acceleration, displacement and a parabolic trajectory

Finding velocity and displacement

If the initial velocity and constant acceleration are known, use

v=u+at\mathbf{v}=\mathbf{u}+\mathbf{a}t

to find velocity and

s=ut+12at2\mathbf{s}=\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2

to find displacement from the initial position.

Example

Motion with a constant acceleration vector

A particle has initial velocity

s−1\mathbf{u}= \begin{pmatrix} 4\\ -1 \end{pmatrix} \text{ m s}^{-1}

and constant acceleration

s−2.\mathbf{a}= \begin{pmatrix} 2\\ 3 \end{pmatrix} \text{ m s}^{-2}.

Find its velocity and displacement after 4 seconds.

Substitute into $v=u+at\mathbf{v}=\mathbf{u}+\mathbf{a}t$ :
$s−1.\begin{aligned} \mathbf{v} &= \begin{pmatrix} 4\\ -1 \end{pmatrix} + 4\begin{pmatrix} 2\\ 3 \end{pmatrix}\\ &= \begin{pmatrix} 12\\ 11 \end{pmatrix} \text{ m s}^{-1}. \end{aligned}$
Substitute into $s=ut+12at2\mathbf{s}=\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2$ :
$m.\begin{aligned} \mathbf{s} &= 4\begin{pmatrix} 4\\ -1 \end{pmatrix} +\frac{1}{2}(4^2) \begin{pmatrix} 2\\ 3 \end{pmatrix}\\ &= \begin{pmatrix} 16\\ -4 \end{pmatrix} + \begin{pmatrix} 16\\ 24 \end{pmatrix}\\ &= \begin{pmatrix} 32\\ 20 \end{pmatrix} \text{ m}. \end{aligned}$
Interpret the vectors: after 4 seconds, the particle is moving right and upwards, and it has moved 32 m horizontally and 20 m vertically from its initial position.

Position vectors

A position vector gives an object's position relative to a fixed origin.

If its initial position vector is $r0\mathbf{r}_0$ and its displacement after time $t$ is $s\mathbf{s}$ , then its new position vector is

r=r0+s.\mathbf{r}=\mathbf{r}_0+\mathbf{s}.

Therefore, under constant acceleration,

r=r0+ut+12at2.\mathbf{r}=\mathbf{r}_0+\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2.

Common Mistake

Position and displacement

The expression $ut+12at2\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2$ gives displacement from the starting point. Add $r0\mathbf{r}_0$ if the question asks for position relative to an origin.

Projectiles

A projectile is an object moving freely under gravity after it has been launched. Air resistance is usually ignored.

Taking upwards as positive, its acceleration is

a=(0−g),\mathbf{a}= \begin{pmatrix} 0\\ -g \end{pmatrix},

where $s−2g=9.8\text{ m s}^{-2}$ unless stated otherwise.

This means:

Horizontal acceleration is zero, so horizontal velocity is constant.
Vertical acceleration is $- g$ , so vertical velocity changes with time.

If an object is projected with speed $U$ at angle $α\alpha$ above the horizontal, its initial velocity components are

u=(Ucos⁡αUsin⁡α).\mathbf{u}= \begin{pmatrix} U\cos\alpha\\ U\sin\alpha \end{pmatrix}.

Hence its displacement after time $t$ is

t−12gt2),\mathbf{s}= \begin{pmatrix} U\cos\alpha\,t\\ U\sin\alpha\,t-\frac{1}{2}gt^2 \end{pmatrix},

and its velocity is

v=(Ucos⁡αUsin⁡α−gt).\mathbf{v}= \begin{pmatrix} U\cos\alpha\\ U\sin\alpha-gt \end{pmatrix}.

Tip

At the highest point

At the highest point, the vertical velocity is zero, but the horizontal velocity is usually not zero. The projectile is still moving.

Example

Finding a projectile's maximum height

A ball is projected from ground level at 20 m s⁻¹ at an angle of $30∘30^\circ$ above the horizontal. Find the time taken to reach its highest point and its maximum height.

Resolve the initial velocity. The vertical component is
$s−1.u_y=20\sin30^\circ=10\text{ m s}^{-1}.$
At maximum height, $v_y=0$ . Apply $v_y=u_y-gt$ :
$0 = 10 - 9.8 t,$
so
$s.t=\frac{10}{9.8}=\frac{50}{49}\text{ s}.$
Use the vertical displacement formula:
$m.\begin{aligned} y &=10\left(\frac{50}{49}\right) -\frac{1}{2}(9.8)\left(\frac{50}{49}\right)^2\\ &=\frac{250}{49}\text{ m}. \end{aligned}$
The maximum height is $m\frac{250}{49}\text{ m}$ , approximately 5.10 m.

Forming and solving component equations

Many problems give a condition such as:

the particle reaches a particular point;
the particle returns to its initial height;
its velocity is parallel or perpendicular to a given vector;
its speed has a specified value.

Write the relevant vector equation, then compare components.

For example, if

(xy)=(3t8t−4.9t2),\begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} 3t\\ 8t-4.9t^2 \end{pmatrix},

then $x = 3 t$ and $y=8t-4.9t^2$ . A condition on height gives an equation in $t$ using the second component.

Common Mistake

Two possible times

A projectile can pass through the same height once while rising and again while falling. A quadratic equation may therefore produce two valid positive times.

Exam technique

In the exam

Define positive horizontal and vertical directions, then keep the signs consistent.
Resolve any angled velocity into components before using the constant-acceleration formulae.
Use the same time in both component equations and check whether every positive solution is physically relevant.
Distinguish carefully between displacement, position, velocity and speed.
Include vector notation and SI units, and keep exact values until a decimal answer is requested.

Self review

Check yourself

How would you find the speed and direction from a velocity column vector?
Why is the horizontal velocity of a projectile constant when air resistance is ignored?
What condition would you use to find when a projectile reaches its maximum height?

What you'll learn

How to represent displacement, velocity and acceleration as two-dimensional vectors.
How to extend the constant-acceleration formulae to vector form.
How to solve motion separately in horizontal and vertical directions.
How to model projectiles when gravity is the only acceleration.

Prerequisites: vectors and components

A vector is a quantity with both magnitude and direction. Displacement, velocity and acceleration are all vectors, so they may be written using column vectors.

For example,

s−1\mathbf{u}= \begin{pmatrix} 6\\ 8 \end{pmatrix} \text{ m s}^{-1}

means that the initial velocity has a horizontal component of 6 m s⁻¹ and a vertical component of 8 m s⁻¹.

The letters $i\mathbf{i}$ and $j\mathbf{j}$ can also represent unit vectors in the positive horizontal and vertical directions:

s−1.\mathbf{u}=6\mathbf{i}+8\mathbf{j}\text{ m s}^{-1}.

These two forms mean exactly the same thing.

Definition

Vector components

The components of a vector are its effects in chosen perpendicular directions, usually the horizontal $x$ -direction and vertical $y$ -direction.

The magnitude of a two-dimensional vector

v=(vxvy)\mathbf{v}= \begin{pmatrix} v_x\\ v_y \end{pmatrix}

is found using Pythagoras:

∣v∣=vx2+vy2.|\mathbf{v}|=\sqrt{v_x^2+v_y^2}.

For a velocity vector, this magnitude is the object's speed.

If $θ\theta$ is the direction measured anticlockwise from the positive horizontal direction, then

tan⁡θ=vyvx.\tan\theta=\frac{v_y}{v_x}.

You must use the signs of the components, or a sketch, to place the direction in the correct quadrant.

Example

Finding speed and direction from velocity

An object has velocity $s−1\mathbf{v}=5\mathbf{i}+12\mathbf{j}\text{ m s}^{-1}$ . Find its speed and direction.

Use the magnitude formula to find the speed:
$s−1.|\mathbf{v}|=\sqrt{5^2+12^2}=13\text{ m s}^{-1}.$
Since both components are positive, the velocity points up and to the right. Its direction above the horizontal satisfies
$tan⁡θ=125.\tan\theta=\frac{12}{5}.$
Therefore,
$θ=tan⁡−1(125)≈67.4∘.\theta=\tan^{-1}\left(\frac{12}{5}\right)\approx67.4^\circ.$
The velocity is 13 m s⁻¹ at approximately $67.4∘67.4^\circ$ above the horizontal.

Common Mistake

Velocity is not speed

Velocity is a vector, but speed is its scalar magnitude. Do not give a column vector when the question asks for speed.

Constant acceleration in vector form

Constant acceleration means that the acceleration vector remains unchanged in both magnitude and direction.

The familiar one-dimensional constant-acceleration formulae extend directly to vectors:

v=u+at,\mathbf{v}=\mathbf{u}+\mathbf{a}t,

s=ut+12at2,\mathbf{s}=\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2,

s=12(u+v)t,\mathbf{s}=\frac{1}{2}(\mathbf{u}+\mathbf{v})t,

and

∣v∣2=∣u∣2+2a⋅s.|\mathbf{v}|^2=|\mathbf{u}|^2+2\mathbf{a}\cdot\mathbf{s}.

Here:

$u\mathbf{u}$ is initial velocity.
$v\mathbf{v}$ is velocity after time $t$ .
$a\mathbf{a}$ is constant acceleration.
$s\mathbf{s}$ is displacement during the time interval.
$a⋅s\mathbf{a}\cdot\mathbf{s}$ is the scalar product of acceleration and displacement.

The last equation uses a scalar product because acceleration and displacement may point in different directions. In components,

a⋅s=axsx+aysy.\mathbf{a}\cdot\mathbf{s}=a_xs_x+a_ys_y.

Key Idea

One vector equation contains two scalar equations

Vector diagram showing initial position, velocity components, downward acceleration, displacement and a parabolic trajectory

Finding velocity and displacement

If the initial velocity and constant acceleration are known, use

v=u+at\mathbf{v}=\mathbf{u}+\mathbf{a}t

to find velocity and

s=ut+12at2\mathbf{s}=\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2

to find displacement from the initial position.

Example

Motion with a constant acceleration vector

A particle has initial velocity

s−1\mathbf{u}= \begin{pmatrix} 4\\ -1 \end{pmatrix} \text{ m s}^{-1}

and constant acceleration

s−2.\mathbf{a}= \begin{pmatrix} 2\\ 3 \end{pmatrix} \text{ m s}^{-2}.

Find its velocity and displacement after 4 seconds.

Substitute into $v=u+at\mathbf{v}=\mathbf{u}+\mathbf{a}t$ :
$s−1.\begin{aligned} \mathbf{v} &= \begin{pmatrix} 4\\ -1 \end{pmatrix} + 4\begin{pmatrix} 2\\ 3 \end{pmatrix}\\ &= \begin{pmatrix} 12\\ 11 \end{pmatrix} \text{ m s}^{-1}. \end{aligned}$
Substitute into $s=ut+12at2\mathbf{s}=\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2$ :
$m.\begin{aligned} \mathbf{s} &= 4\begin{pmatrix} 4\\ -1 \end{pmatrix} +\frac{1}{2}(4^2) \begin{pmatrix} 2\\ 3 \end{pmatrix}\\ &= \begin{pmatrix} 16\\ -4 \end{pmatrix} + \begin{pmatrix} 16\\ 24 \end{pmatrix}\\ &= \begin{pmatrix} 32\\ 20 \end{pmatrix} \text{ m}. \end{aligned}$
Interpret the vectors: after 4 seconds, the particle is moving right and upwards, and it has moved 32 m horizontally and 20 m vertically from its initial position.

Position vectors

A position vector gives an object's position relative to a fixed origin.

If its initial position vector is $r0\mathbf{r}_0$ and its displacement after time $t$ is $s\mathbf{s}$ , then its new position vector is

r=r0+s.\mathbf{r}=\mathbf{r}_0+\mathbf{s}.

Therefore, under constant acceleration,

r=r0+ut+12at2.\mathbf{r}=\mathbf{r}_0+\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2.

Common Mistake

Position and displacement

The expression $ut+12at2\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2$ gives displacement from the starting point. Add $r0\mathbf{r}_0$ if the question asks for position relative to an origin.

Projectiles

A projectile is an object moving freely under gravity after it has been launched. Air resistance is usually ignored.

Taking upwards as positive, its acceleration is

a=(0−g),\mathbf{a}= \begin{pmatrix} 0\\ -g \end{pmatrix},

where $s−2g=9.8\text{ m s}^{-2}$ unless stated otherwise.

This means:

Horizontal acceleration is zero, so horizontal velocity is constant.
Vertical acceleration is $- g$ , so vertical velocity changes with time.

If an object is projected with speed $U$ at angle $α\alpha$ above the horizontal, its initial velocity components are

u=(Ucos⁡αUsin⁡α).\mathbf{u}= \begin{pmatrix} U\cos\alpha\\ U\sin\alpha \end{pmatrix}.

Hence its displacement after time $t$ is

t−12gt2),\mathbf{s}= \begin{pmatrix} U\cos\alpha\,t\\ U\sin\alpha\,t-\frac{1}{2}gt^2 \end{pmatrix},

and its velocity is

v=(Ucos⁡αUsin⁡α−gt).\mathbf{v}= \begin{pmatrix} U\cos\alpha\\ U\sin\alpha-gt \end{pmatrix}.

Tip

At the highest point

At the highest point, the vertical velocity is zero, but the horizontal velocity is usually not zero. The projectile is still moving.

Example

Finding a projectile's maximum height

A ball is projected from ground level at 20 m s⁻¹ at an angle of $30∘30^\circ$ above the horizontal. Find the time taken to reach its highest point and its maximum height.

Resolve the initial velocity. The vertical component is
$s−1.u_y=20\sin30^\circ=10\text{ m s}^{-1}.$
At maximum height, $v_y=0$ . Apply $v_y=u_y-gt$ :
$0 = 10 - 9.8 t,$
so
$s.t=\frac{10}{9.8}=\frac{50}{49}\text{ s}.$
Use the vertical displacement formula:
$m.\begin{aligned} y &=10\left(\frac{50}{49}\right) -\frac{1}{2}(9.8)\left(\frac{50}{49}\right)^2\\ &=\frac{250}{49}\text{ m}. \end{aligned}$
The maximum height is $m\frac{250}{49}\text{ m}$ , approximately 5.10 m.

Forming and solving component equations

Many problems give a condition such as:

the particle reaches a particular point;
the particle returns to its initial height;
its velocity is parallel or perpendicular to a given vector;
its speed has a specified value.

Write the relevant vector equation, then compare components.

For example, if

(xy)=(3t8t−4.9t2),\begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} 3t\\ 8t-4.9t^2 \end{pmatrix},

then $x = 3 t$ and $y=8t-4.9t^2$ . A condition on height gives an equation in $t$ using the second component.

Common Mistake

Two possible times

A projectile can pass through the same height once while rising and again while falling. A quadratic equation may therefore produce two valid positive times.

Exam technique

In the exam

Define positive horizontal and vertical directions, then keep the signs consistent.
Resolve any angled velocity into components before using the constant-acceleration formulae.
Use the same time in both component equations and check whether every positive solution is physically relevant.
Distinguish carefully between displacement, position, velocity and speed.
Include vector notation and SI units, and keep exact values until a decimal answer is requested.

Self review

Check yourself

How would you find the speed and direction from a velocity column vector?
Why is the horizontal velocity of a projectile constant when air resistance is ignored?
What condition would you use to find when a projectile reaches its maximum height?

What you'll learn

Prerequisites: vectors and components

Constant acceleration in vector form

Finding velocity and displacement

Position vectors

Projectiles

Forming and solving component equations

Review 3.2.5 Constant acceleration in two dimensions (A-level only) by teaching Genie

3.2.5 Constant acceleration in two dimensions (A-level only)

What you'll learn

Prerequisites: vectors and components

Constant acceleration in vector form

Finding velocity and displacement

Position vectors

Projectiles

Forming and solving component equations

Review 3.2.5 Constant acceleration in two dimensions (A-level only) by teaching Genie

3.2.5 Constant acceleration in two dimensions (A-level only)

What you'll learn

Prerequisites: vectors and components

Constant acceleration in vector form

Finding velocity and displacement

Position vectors

Projectiles

Forming and solving component equations

Review 3.2.5 Constant acceleration in two dimensions (A-level only) by teaching Genie

3.2.5 Constant acceleration in two dimensions (A-level only) Revision Guide

3.2.5 Constant acceleration in two dimensions (A-level only)

What you'll learn

Prerequisites: vectors and components

Constant acceleration in vector form

Finding velocity and displacement

Position vectors

Projectiles

Forming and solving component equations

Review 3.2.5 Constant acceleration in two dimensions (A-level only) by teaching Genie

3.2.5 Constant acceleration in two dimensions (A-level only) Revision Guide