Further Kinematics

What you'll learn

Use $i\mathbf{i}$ and $j\mathbf{j}$ components to describe motion in a plane.
Connect position, velocity and acceleration using vector equations.
Handle directions, bearings, minimum speed and relative motion.
Differentiate and integrate vector functions of time.

1. Vectors, speed and constant velocity

In Further Kinematics, you usually model motion in two perpendicular directions. The unit vector $i\mathbf{i}$ is often due east, and $j\mathbf{j}$ is often due north.

A vector such as $3i−4j3\mathbf{i}-4\mathbf{j}$ means 3 units in the $i\mathbf{i}$ direction and 4 units in the negative $j\mathbf{j}$ direction.

$A component diagram showing how 3\mathbf{i}-4\mathbf{j} is built from horizontal and vertical components.$

Definition

Position vector and displacement

A position vector gives the position of a particle relative to a fixed origin $O$ .
A displacement is the change in position, so $position\text{displacement}=\text{final position}-\text{initial position}$ .
The speed is the magnitude of the velocity vector.

For constant velocity,

r=r0+vt\mathbf{r}=\mathbf{r}_0+\mathbf{v}t

where $r0\mathbf{r}_0$ is the initial position vector, $v\mathbf{v}$ is the constant velocity, and $t$ is the time.

Key Idea

Work component by component

Treat the $i\mathbf{i}$ and $j\mathbf{j}$ parts like two separate one-dimensional problems, then combine them at the end for speed, distance or direction.

Example

Finding velocity, speed, bearing and position

A boat is at position $4i−3j4\mathbf{i}-3\mathbf{j}$ km at noon. At 3 p.m. it is at position $−8i+12j-8\mathbf{i}+12\mathbf{j}$ km. Find its velocity, speed, bearing, and position vector $t$ hours after noon.

The boat’s displacement is north-west, so its bearing is measured clockwise from north to a direction close to 360°.

Find the displacement over the 3 hours:
$(−8i+12j)−(4i−3j)=−12i+15j(-8\mathbf{i}+12\mathbf{j})-(4\mathbf{i}-3\mathbf{j}) = -12\mathbf{i}+15\mathbf{j}$
Divide by the time to get the velocity:
$v=−12i+15j3=−4i+5j\mathbf{v} = \frac{-12\mathbf{i}+15\mathbf{j}}{3} = -4\mathbf{i}+5\mathbf{j}$
The speed is the magnitude of the velocity:
$∣v∣=(−4)2+52=41|\mathbf{v}|=\sqrt{(-4)^2+5^2}=\sqrt{41}$
For the bearing, the boat moves west and north. The angle west of north is:
$θ=tan⁡−1(45)≈38.7∘\theta=\tan^{-1}\left(\frac{4}{5}\right)\approx 38.7^\circ$
Bearings are measured clockwise from north, so the bearing is approximately 321°.
Write the position vector after $t$ hours:
$r=(4i−3j)+t(−4i+5j)=(4−4t)i+(−3+5t)j\mathbf{r}=(4\mathbf{i}-3\mathbf{j})+t(-4\mathbf{i}+5\mathbf{j}) = (4-4t)\mathbf{i}+(-3+5t)\mathbf{j}$

Tip

Bearings

Always sketch the quadrant first. A vector with negative $i\mathbf{i}$ and positive $j\mathbf{j}$ components points north-west, so its bearing is close to 360°, not close to 0°.

2. Constant acceleration in vector form

Acceleration is the rate of change of velocity. With constant acceleration, the familiar SUVAT equations work as vector equations:

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

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

Here $u\mathbf{u}$ is the initial velocity, $v\mathbf{v}$ is the velocity at time $t$ , and $a\mathbf{a}$ is the acceleration.

If you are told a particle moves “in the direction” of a vector, that vector gives the ratio of the components, not necessarily the velocity itself.

A direction vector fixes the ratio of components; multiplying its unit vector by the speed gives the actual velocity.

Example

Using a direction vector and constant acceleration

A particle starts at the origin. Initially it moves in the direction $3i+4j3\mathbf{i}+4\mathbf{j}$ with speed 20 m s^-1. Five seconds later its velocity is $−5i+18j-5\mathbf{i}+18\mathbf{j}$ m s^-1. Find its acceleration and its position after 5 seconds.

Convert the initial direction into a unit vector, then multiply by the speed:
$u=20(35i+45j)=12i+16j\mathbf{u} = 20\left(\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}\right) = 12\mathbf{i}+16\mathbf{j}$
Use $v=u+at\mathbf{v}=\mathbf{u}+\mathbf{a}t$ , rearranged to find $a\mathbf{a}$ :
$a=(−5i+18j)−(12i+16j)5=−175i+25j\mathbf{a} = \frac{(-5\mathbf{i}+18\mathbf{j})-(12\mathbf{i}+16\mathbf{j})}{5} = -\frac{17}{5}\mathbf{i}+\frac{2}{5}\mathbf{j}$
Use the position formula with $r0=0\mathbf{r}_0=\mathbf{0}$ and $t = 5$ :
$r=(12i+16j)(5)+12(−175i+25j)(25)=352i+85j\mathbf{r} = (12\mathbf{i}+16\mathbf{j})(5) + \frac{1}{2}\left(-\frac{17}{5}\mathbf{i}+\frac{2}{5}\mathbf{j}\right)(25) = \frac{35}{2}\mathbf{i}+85\mathbf{j}$
So after 5 seconds, the position vector is $352i+85j\frac{35}{2}\mathbf{i}+85\mathbf{j}$ m.

Common Mistake

Direction is not velocity

Do not use a direction vector as a velocity until you have scaled it to the correct speed.

3. Direction tests and minimum speed

A velocity is parallel to another vector if it is a scalar multiple of it. For example, moving in the direction $i+2j\mathbf{i}+2\mathbf{j}$ means the $j\mathbf{j}$ component is twice the $i\mathbf{i}$ component, with the same overall direction.

A vector is perpendicular to $i\mathbf{i}$ if its $i\mathbf{i}$ component is zero.

For minimum speed, avoid square roots. Minimise the square of the speed instead.

Minimising speed is equivalent to minimising the positive quadratic V^2(t).

Example

Direction and minimum speed

A particle has velocity

v=(4t−8)i+(3t+1)j\mathbf{v}=(4t-8)\mathbf{i}+(3t+1)\mathbf{j}

where $t≥0t\ge 0$ . Find when it moves in the direction $i+2j\mathbf{i}+2\mathbf{j}$ , and when its speed is minimum.

Moving in the direction $i+2j\mathbf{i}+2\mathbf{j}$ means the $j\mathbf{j}$ component is twice the $i\mathbf{i}$ component:
$3 t + 1 = 2 (4 t - 8)$
Solve for $t$ :
$\Rightarrow t=\frac{17}{5}$
At $t=175t=\frac{17}{5}$ , the $i\mathbf{i}$ component is positive, so the particle is moving in the required direction.
Let $V$ be the speed. Minimise $V^2$ :
$V^2=(4t-8)^2+(3t+1)^2=25t^2-58t+65$
Differentiate and set equal to zero:
$ddt(V2)=50t−58=0⇒t=2925\frac{d}{dt}(V^2)=50t-58=0 \Rightarrow t=\frac{29}{25}$
Since $V^2$ is a positive quadratic, the speed is minimum when $t=2925t=\frac{29}{25}$ .

Tip

Minimum speed shortcut

For constant acceleration, minimum speed occurs when the velocity is perpendicular to the acceleration, provided that time lies in the allowed interval.

With constant acceleration, the speed is smallest at the instant when the velocity vector is perpendicular to the acceleration vector.

4. Relative motion and collisions

Relative motion describes one object as seen from another object.

Definition

Relative position

The position of $B$ relative to $A$ is $rB/A=rB−rA\mathbf{r}_{B/A}=\mathbf{r}_B-\mathbf{r}_A$ . If $rB/A=0\mathbf{r}_{B/A}=\mathbf{0}$ , the two objects are at the same point.

Relative position is the vector from object A to object B, found by subtracting their position vectors.

Example

Checking whether two boats collide

At noon, boat $A$ is at $6i6\mathbf{i}$ km and moves with velocity $−5i+4j-5\mathbf{i}+4\mathbf{j}$ km h^-1. Boat $B$ is at $9j9\mathbf{j}$ km and moves with velocity $−3i+j-3\mathbf{i}+\mathbf{j}$ km h^-1. Find the relative position of $B$ from $A$ , and decide whether they collide.

The two boats collide because their position paths meet at the same point at the same time.

Write position vectors after $t$ hours:
$rA=(6−5t)i+4tj,rB=−3ti+(9+t)j\mathbf{r}_A=(6-5t)\mathbf{i}+4t\mathbf{j}, \qquad \mathbf{r}_B=-3t\mathbf{i}+(9+t)\mathbf{j}$
Subtract to find the relative position:
$rB/A=rB−rA=(2t−6)i+(9−3t)j\mathbf{r}_{B/A} = \mathbf{r}_B-\mathbf{r}_A = (2t-6)\mathbf{i}+(9-3t)\mathbf{j}$
For a collision, both components must be zero:
$2t−6=0⇒t=3,9−3t=0⇒t=32t-6=0\Rightarrow t=3, \qquad 9-3t=0\Rightarrow t=3$
Both components give the same time, so the boats collide 3 hours after noon.
The collision position can be found using either boat’s position vector:
$rA(3)=−9i+12j\mathbf{r}_A(3)=-9\mathbf{i}+12\mathbf{j}$

Common Mistake

Only checking one component

A collision needs the same $i\mathbf{i}$ coordinate and the same $j\mathbf{j}$ coordinate at the same time.

5. Variable kinematics with calculus

When position, velocity or acceleration depends on $t$ , use calculus component by component.

Differentiation moves from position to velocity to acceleration, while integration reverses the process and introduces constants.

v=drdt\mathbf{v}=\frac{d\mathbf{r}}{dt}

a=dvdt\mathbf{a}=\frac{d\mathbf{v}}{dt}

Going backwards, integrate. Remember that integrating introduces constants, which you find using given conditions.

Example

Differentiating and integrating vector functions

Particle $P$ has position vector

rP=(2t2+3t−1)i+(t2+9t+4)j\mathbf{r}_P=(2t^2+3t-1)\mathbf{i}+(t^2+9t+4)\mathbf{j}

Particle $Q$ has acceleration $aQ=(6t−2)i+4tj\mathbf{a}_Q=(6t-2)\mathbf{i}+4t\mathbf{j}$ . When $t = 2$ , its velocity is $8i+10j8\mathbf{i}+10\mathbf{j}$ , and initially its position vector is $i−3j\mathbf{i}-3\mathbf{j}$ . Find $vP\mathbf{v}_P$ , $aP\mathbf{a}_P$ , and the position vector of $Q$ .

Differentiate $rP\mathbf{r}_P$ to find velocity:
$vP=(4t+3)i+(2t+9)j\mathbf{v}_P=(4t+3)\mathbf{i}+(2t+9)\mathbf{j}$
Differentiate again to find acceleration:
$aP=4i+2j\mathbf{a}_P=4\mathbf{i}+2\mathbf{j}$
Integrate $aQ\mathbf{a}_Q$ to find $vQ\mathbf{v}_Q$ :
$vQ=(3t2−2t+C)i+(2t2+D)j\mathbf{v}_Q=(3t^2-2t+C)\mathbf{i}+(2t^2+D)\mathbf{j}$
Use the velocity condition at $t = 2$ :
$8+C=8,8+D=10⇒C=0,D=28+C=8,\qquad 8+D=10 \Rightarrow C=0,\quad D=2$
Integrate again and use the initial position:
$rQ=(t3−t2+1)i+(23t3+2t−3)j\mathbf{r}_Q=(t^3-t^2+1)\mathbf{i}+\left(\frac{2}{3}t^3+2t-3\right)\mathbf{j}$

Tip

Distance between two positions

If you are given $v(t)\mathbf{v}(t)$ and asked for the distance between two points, integrate velocity to get displacement first, then take the magnitude of that displacement.

Exam technique

In the exam

Write down whether you are using position, velocity or acceleration before substituting numbers.
Keep units of time consistent, especially when questions use clock times such as noon, 1430 or 6 a.m.
For collisions, solve both components and check that they give the same value of $t$ .
For directions and bearings, sketch the vector so your angle is in the correct quadrant.

Self review

Check yourself

Can you convert “speed 15 in the direction $4i+3j4\mathbf{i}+3\mathbf{j}$ ” into a velocity vector?
If $rB/A=(3t−6)i+(12−6t)j\mathbf{r}_{B/A}=(3t-6)\mathbf{i}+(12-6t)\mathbf{j}$ , how would you test for a collision?
When should you differentiate, and when should you integrate, in a vector kinematics question?

Further Kinematics

What you'll learn

Use $i\mathbf{i}$ and $j\mathbf{j}$ components to describe motion in a plane.
Connect position, velocity and acceleration using vector equations.
Handle directions, bearings, minimum speed and relative motion.
Differentiate and integrate vector functions of time.

1. Vectors, speed and constant velocity

In Further Kinematics, you usually model motion in two perpendicular directions. The unit vector $i\mathbf{i}$ is often due east, and $j\mathbf{j}$ is often due north.

A vector such as $3i−4j3\mathbf{i}-4\mathbf{j}$ means 3 units in the $i\mathbf{i}$ direction and 4 units in the negative $j\mathbf{j}$ direction.

$A component diagram showing how 3\mathbf{i}-4\mathbf{j} is built from horizontal and vertical components.$

Definition

Position vector and displacement

A position vector gives the position of a particle relative to a fixed origin $O$ .
A displacement is the change in position, so $position\text{displacement}=\text{final position}-\text{initial position}$ .
The speed is the magnitude of the velocity vector.

For constant velocity,

r=r0+vt\mathbf{r}=\mathbf{r}_0+\mathbf{v}t

where $r0\mathbf{r}_0$ is the initial position vector, $v\mathbf{v}$ is the constant velocity, and $t$ is the time.

Key Idea

Work component by component

Treat the $i\mathbf{i}$ and $j\mathbf{j}$ parts like two separate one-dimensional problems, then combine them at the end for speed, distance or direction.

Example

Finding velocity, speed, bearing and position

The boat’s displacement is north-west, so its bearing is measured clockwise from north to a direction close to 360°.

Find the displacement over the 3 hours:
$(−8i+12j)−(4i−3j)=−12i+15j(-8\mathbf{i}+12\mathbf{j})-(4\mathbf{i}-3\mathbf{j}) = -12\mathbf{i}+15\mathbf{j}$
Divide by the time to get the velocity:
$v=−12i+15j3=−4i+5j\mathbf{v} = \frac{-12\mathbf{i}+15\mathbf{j}}{3} = -4\mathbf{i}+5\mathbf{j}$
The speed is the magnitude of the velocity:
$∣v∣=(−4)2+52=41|\mathbf{v}|=\sqrt{(-4)^2+5^2}=\sqrt{41}$
For the bearing, the boat moves west and north. The angle west of north is:
$θ=tan⁡−1(45)≈38.7∘\theta=\tan^{-1}\left(\frac{4}{5}\right)\approx 38.7^\circ$
Bearings are measured clockwise from north, so the bearing is approximately 321°.
Write the position vector after $t$ hours:
$r=(4i−3j)+t(−4i+5j)=(4−4t)i+(−3+5t)j\mathbf{r}=(4\mathbf{i}-3\mathbf{j})+t(-4\mathbf{i}+5\mathbf{j}) = (4-4t)\mathbf{i}+(-3+5t)\mathbf{j}$

Tip

Bearings

Always sketch the quadrant first. A vector with negative $i\mathbf{i}$ and positive $j\mathbf{j}$ components points north-west, so its bearing is close to 360°, not close to 0°.

2. Constant acceleration in vector form

Acceleration is the rate of change of velocity. With constant acceleration, the familiar SUVAT equations work as vector equations:

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

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

Here $u\mathbf{u}$ is the initial velocity, $v\mathbf{v}$ is the velocity at time $t$ , and $a\mathbf{a}$ is the acceleration.

If you are told a particle moves “in the direction” of a vector, that vector gives the ratio of the components, not necessarily the velocity itself.

A direction vector fixes the ratio of components; multiplying its unit vector by the speed gives the actual velocity.

Example

Using a direction vector and constant acceleration

Convert the initial direction into a unit vector, then multiply by the speed:
$u=20(35i+45j)=12i+16j\mathbf{u} = 20\left(\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}\right) = 12\mathbf{i}+16\mathbf{j}$
Use $v=u+at\mathbf{v}=\mathbf{u}+\mathbf{a}t$ , rearranged to find $a\mathbf{a}$ :
$a=(−5i+18j)−(12i+16j)5=−175i+25j\mathbf{a} = \frac{(-5\mathbf{i}+18\mathbf{j})-(12\mathbf{i}+16\mathbf{j})}{5} = -\frac{17}{5}\mathbf{i}+\frac{2}{5}\mathbf{j}$
Use the position formula with $r0=0\mathbf{r}_0=\mathbf{0}$ and $t = 5$ :
$r=(12i+16j)(5)+12(−175i+25j)(25)=352i+85j\mathbf{r} = (12\mathbf{i}+16\mathbf{j})(5) + \frac{1}{2}\left(-\frac{17}{5}\mathbf{i}+\frac{2}{5}\mathbf{j}\right)(25) = \frac{35}{2}\mathbf{i}+85\mathbf{j}$
So after 5 seconds, the position vector is $352i+85j\frac{35}{2}\mathbf{i}+85\mathbf{j}$ m.

Common Mistake

Direction is not velocity

Do not use a direction vector as a velocity until you have scaled it to the correct speed.

3. Direction tests and minimum speed

A vector is perpendicular to $i\mathbf{i}$ if its $i\mathbf{i}$ component is zero.

For minimum speed, avoid square roots. Minimise the square of the speed instead.

Minimising speed is equivalent to minimising the positive quadratic V^2(t).

Example

Direction and minimum speed

A particle has velocity

v=(4t−8)i+(3t+1)j\mathbf{v}=(4t-8)\mathbf{i}+(3t+1)\mathbf{j}

where $t≥0t\ge 0$ . Find when it moves in the direction $i+2j\mathbf{i}+2\mathbf{j}$ , and when its speed is minimum.

Moving in the direction $i+2j\mathbf{i}+2\mathbf{j}$ means the $j\mathbf{j}$ component is twice the $i\mathbf{i}$ component:
$3 t + 1 = 2 (4 t - 8)$
Solve for $t$ :
$\Rightarrow t=\frac{17}{5}$
At $t=175t=\frac{17}{5}$ , the $i\mathbf{i}$ component is positive, so the particle is moving in the required direction.
Let $V$ be the speed. Minimise $V^2$ :
$V^2=(4t-8)^2+(3t+1)^2=25t^2-58t+65$
Differentiate and set equal to zero:
$ddt(V2)=50t−58=0⇒t=2925\frac{d}{dt}(V^2)=50t-58=0 \Rightarrow t=\frac{29}{25}$
Since $V^2$ is a positive quadratic, the speed is minimum when $t=2925t=\frac{29}{25}$ .

Tip

Minimum speed shortcut

For constant acceleration, minimum speed occurs when the velocity is perpendicular to the acceleration, provided that time lies in the allowed interval.

With constant acceleration, the speed is smallest at the instant when the velocity vector is perpendicular to the acceleration vector.

4. Relative motion and collisions

Relative motion describes one object as seen from another object.

Definition

Relative position

The position of $B$ relative to $A$ is $rB/A=rB−rA\mathbf{r}_{B/A}=\mathbf{r}_B-\mathbf{r}_A$ . If $rB/A=0\mathbf{r}_{B/A}=\mathbf{0}$ , the two objects are at the same point.

Relative position is the vector from object A to object B, found by subtracting their position vectors.

Example

Checking whether two boats collide

The two boats collide because their position paths meet at the same point at the same time.

Write position vectors after $t$ hours:
$rA=(6−5t)i+4tj,rB=−3ti+(9+t)j\mathbf{r}_A=(6-5t)\mathbf{i}+4t\mathbf{j}, \qquad \mathbf{r}_B=-3t\mathbf{i}+(9+t)\mathbf{j}$
Subtract to find the relative position:
$rB/A=rB−rA=(2t−6)i+(9−3t)j\mathbf{r}_{B/A} = \mathbf{r}_B-\mathbf{r}_A = (2t-6)\mathbf{i}+(9-3t)\mathbf{j}$
For a collision, both components must be zero:
$2t−6=0⇒t=3,9−3t=0⇒t=32t-6=0\Rightarrow t=3, \qquad 9-3t=0\Rightarrow t=3$
Both components give the same time, so the boats collide 3 hours after noon.
The collision position can be found using either boat’s position vector:
$rA(3)=−9i+12j\mathbf{r}_A(3)=-9\mathbf{i}+12\mathbf{j}$

Common Mistake

Only checking one component

A collision needs the same $i\mathbf{i}$ coordinate and the same $j\mathbf{j}$ coordinate at the same time.

5. Variable kinematics with calculus

When position, velocity or acceleration depends on $t$ , use calculus component by component.

Differentiation moves from position to velocity to acceleration, while integration reverses the process and introduces constants.

v=drdt\mathbf{v}=\frac{d\mathbf{r}}{dt}

a=dvdt\mathbf{a}=\frac{d\mathbf{v}}{dt}

Going backwards, integrate. Remember that integrating introduces constants, which you find using given conditions.

Example

Differentiating and integrating vector functions

Particle $P$ has position vector

rP=(2t2+3t−1)i+(t2+9t+4)j\mathbf{r}_P=(2t^2+3t-1)\mathbf{i}+(t^2+9t+4)\mathbf{j}

Differentiate $rP\mathbf{r}_P$ to find velocity:
$vP=(4t+3)i+(2t+9)j\mathbf{v}_P=(4t+3)\mathbf{i}+(2t+9)\mathbf{j}$
Differentiate again to find acceleration:
$aP=4i+2j\mathbf{a}_P=4\mathbf{i}+2\mathbf{j}$
Integrate $aQ\mathbf{a}_Q$ to find $vQ\mathbf{v}_Q$ :
$vQ=(3t2−2t+C)i+(2t2+D)j\mathbf{v}_Q=(3t^2-2t+C)\mathbf{i}+(2t^2+D)\mathbf{j}$
Use the velocity condition at $t = 2$ :
$8+C=8,8+D=10⇒C=0,D=28+C=8,\qquad 8+D=10 \Rightarrow C=0,\quad D=2$
Integrate again and use the initial position:
$rQ=(t3−t2+1)i+(23t3+2t−3)j\mathbf{r}_Q=(t^3-t^2+1)\mathbf{i}+\left(\frac{2}{3}t^3+2t-3\right)\mathbf{j}$

Tip

Distance between two positions

If you are given $v(t)\mathbf{v}(t)$ and asked for the distance between two points, integrate velocity to get displacement first, then take the magnitude of that displacement.

Exam technique

In the exam

Write down whether you are using position, velocity or acceleration before substituting numbers.
Keep units of time consistent, especially when questions use clock times such as noon, 1430 or 6 a.m.
For collisions, solve both components and check that they give the same value of $t$ .
For directions and bearings, sketch the vector so your angle is in the correct quadrant.

Self review

Check yourself

Can you convert “speed 15 in the direction $4i+3j4\mathbf{i}+3\mathbf{j}$ ” into a velocity vector?
If $rB/A=(3t−6)i+(12−6t)j\mathbf{r}_{B/A}=(3t-6)\mathbf{i}+(12-6t)\mathbf{j}$ , how would you test for a collision?
When should you differentiate, and when should you integrate, in a vector kinematics question?

Further Kinematics

What you'll learn

1. Vectors, speed and constant velocity

2. Constant acceleration in vector form

3. Direction tests and minimum speed

4. Relative motion and collisions

5. Variable kinematics with calculus

Recap questions

You've reached the end

Further Kinematics

What you'll learn

1. Vectors, speed and constant velocity

2. Constant acceleration in vector form

3. Direction tests and minimum speed

4. Relative motion and collisions

5. Variable kinematics with calculus

Recap questions

You've reached the end

Further Kinematics

What you'll learn

1. Vectors, speed and constant velocity

2. Constant acceleration in vector form

3. Direction tests and minimum speed

4. Relative motion and collisions

5. Variable kinematics with calculus

Recap questions

You've reached the end

Further Kinematics Revision Guide

Further Kinematics

What you'll learn

1. Vectors, speed and constant velocity

2. Constant acceleration in vector form

3. Direction tests and minimum speed

4. Relative motion and collisions

5. Variable kinematics with calculus

Recap questions

You've reached the end

Further Kinematics Revision Guide