National 5PhysicsSQARevision NotesDynamicsVectors & ScalarsMotion Relationships

Motion Relationships (SQA National 5 Physics): Revision Note

Exam code: X857 75

Katie M

Written by: Katie M

Reviewed by: Leander Oates

Updated on

Motion relationships

Speed

  • The speed of an object is defined as:

The distance travelled per unit time

  • Speed can be calculated using the following relationship:

speed space equals space distance over time

v space equals space d over t

  • Where:

    • v = speed, measured in metres per second (m s-1)

    • d = distance travelled, measured in metres (m)

    • t = time, measured in seconds (s)

  • Distance and speed are both scalar quantities with magnitude only

    • The direction of the object's motion is not described by speed

Examiner Tips and Tricks

Relationship triangles are really useful for knowing how to rearrange physics equations, such as the relationship between speed, distance, and time.

To use them:

  • Cover up the quantity to be calculated; this is known as the 'subject' of the equation

  • If the other two quantities are on the same line, this means they are multiplied

  • If one quantity is above the other, this means they are divided - make sure to keep the order of which is on the top and bottom of the fraction!

Formula Triangle, downloadable AS & A Level Physics revision notes

However, they are a tool to use until you feel confident in rearranging equations. They are not meant to be used in your exam. If you give an equation in triangle form in your exam, you will not be awarded the mark.

Average and instantaneous speed

  • The speedometer of a car gives a measure of instantaneous speed

  • For example, if the speedometer reads 50 mph (miles per hour), which is equivalent to 22 m s-1, it means that the car is travelling 22 metres every second that this speed is maintained

Close-up of a car's speedometer, showing speeds from 0 to 260 km/h, with glowing red indicators and diesel fuel gauge at half.
The speedometer on a vehicle displays instantaneous speed

Image: Creative Commons licence from pxhere.com

  • As the speed of the car changes, the speedometer will give a new value for the new instantaneous speed

    • The instantaneous speed is the speed at which an object travels in each instant of time

  • Average speed considers the total distance travelled and the total time taken

    • For example, the car will not travel at 22 m s-1 for its entire journey; it will speed up and slow down accordingly

  • Average speed can be calculated using the following relationship:

average space speed space equals space fraction numerator total space distance over denominator total space time end fraction

v with bar on top space equals space d over t

  • Where:

    • v with bar on top = average speed, measured in metres per second (m s-1)

    • d = total distance travelled, measured in metres (m)

    • t = total time taken, measured in seconds (s)

  • If the car travelled a total distance of 2000 metres and the whole journey took 3 minutes, or 180 seconds, then the average speed of the car is:

average space speed space equals space 2000 over 180 space equals space 11 space straight m space straight s to the power of negative 1 end exponent

Map showing the distance travelled by a car

A hand-drawn map of a town with irregular grid roads and buildings. A red line shows a winding route labelled "Distance travelled." A north arrow is in the top right.
The red line shows the distance traveled by the car over its 3 minute journey

Examiner Tips and Tricks

A bar over the symbol of a quantity is the standard mathematical notation for the average value of that quantity, e.g. speed is represented by v, so the average speed is represented by v with bar on top.

Velocity

  • The velocity of an object is defined as:

The displacement per unit time

  • Velocity can be calculated using the following relationship:

velocity space equals space displacement over time

v space equals space s over t

  • Where:

    • v = velocity, measured in metres per second (m s-1)

    • s = displacement, measured in metres (m)

    • t = time, measured in seconds (s)

  • Velocity is a vector quantity with both magnitude and direction

    • If an object travels with a constant speed but changes direction, then its velocity is changing

    • Therefore, it is possible for an object to travel at a constant speed without a constant velocity, but it is not possible for an object to travel at a constant velocity without a constant speed

  • The magnitude of an object's velocity is its speed

    • However, the magnitude of an object's average velocity is not its average speed

Comparing speed and velocity

Two cars moving with speed 20 m/s; left car heading east, right car heading west.
The cars in the diagram above have the same speed (a scalar quantity) but different velocities (a vector quantity)

Worked Example

A car travels north on a motorway at a speed of 33 m s-1.

Determine the displacement of the car in 2.0 minutes.

Answer:

Step 1: List the known quantities

  • Velocity, v = 33 m s-1 north

  • Time, t = 2.0 minutes

Step 2: Convert the values to SI units

  • 1 minute = 60 s

  • Therefore, 2.0 minutes = 2.0 × 60 = 120 s

Step 3: Write out the correct relationship

s space equals space v t

Step 4: Substitute in the known values to calculate displacement

s space equals space 33 space cross times space 120

s space equals space 3960 space straight m space north

Step 5: Round the final answer to an appropriate number of significant figures

  • The least precise input value is 2 s.f.

  • Therefore, the answer can only be given to the same precision

s space equals space 4000 space straight m space north space open parentheses 2 space straight s. straight f. close parentheses

Examiner Tips and Tricks

Always include the direction when referring to velocity or displacement, but not when referring to speed or distance. The exception to this is if a question asks for the 'magnitude', which means you only need to give the numerical value.

Average velocity

  • Average velocity also describes the whole journey of an object

    • It considers the total displacement rather than the distance

  • Average velocity can be calculated using:

average space velocity space equals space fraction numerator total space displacement over denominator total space time end fraction

v with bar on top space equals space s over t

  • Where:

    • v with bar on top = average velocity, measured in metres per second (m s-1)

    • d = total displacement, measured in metres (m)

    • t = total time taken, measured in seconds (s)

  • If the same car had a displacement of 1500 metres over its 3 minute journey, then the magnitude of the average velocity would be:

v with bar on top space equals space 1500 over 180 space equals space 8.3 space straight m space straight s to the power of negative 1 end exponent

  • Notice how the average speed and average velocity have different magnitudes

  • The direction of the average velocity is in the same direction as the displacement

    • This could be found by drawing a straight line from start to finish and measuring the angle with a protractor, or calculating it using trigonometry if the scale of the map is known

Map showing the displacement of a car

Map showing a red winding path for distance travelled and a straight blue line for displacement. Includes a key and data: distance is 2000 metres, displacement is 1500 metres east of north.
Map showing distance travelled in red and displacement in blue. Since these values are different, the average speed and the average velocities are also different

Positive and negative velocity

  • Since velocity is a vector quantity, it can have a positive or negative value

  • A positive velocity is typically assigned to the object's initial direction of motion

  • A negative velocity is typically assigned to the opposite direction to the object's initial velocity

  • If there is no initial direction of motion, then positive velocity is generally either:

    • forwards

    • to the right

    • upwards

  • However, mathematically, any direction can be assigned as positive as long as the opposing direction is negative

Examiner Tips and Tricks

Which ever direction you assign as positive, make sure you are consistent throughout your calculation.

Worked Example

The map below shows a teacher's route through the city on their way to work. The teacher travels 1.8 km north, and then 450 m east before parking their car. The journey takes 4 minutes.

City street map with a red arrow pointing upwards along a major vertical road heading north, then turning a corner onto another street heading east. A compass indicating north is in the upper-left corner.

Which row in the table shows the teacher's average speed and average velocity for the journey?

Average speed

Average velocity

A

7.7 m s-1

7.7 m s-1 at 014

B

7.7 m s-1 at 014

7.7 m s-1

C

9.4 m s-1

7.7 m s-1 at 014

D

9.4 m s-1 at 014

9.4 m s-1

E

9.4 m s-1

9.4 m s-1 at 014

The correct answer is C

Answer:

Step 1: List the known quantities

  • 1 km = 1000 m

  • Displacement to the north = 1.8 km = 1.8 × 1000 = 1800 m

  • Displacement to the east = 450 m

  • 1 minute = 60 s

  • Total time = 4 minutes = 4 × 60 = 240 s

Step 2: Calculate the average speed

  • Determine the total distance travelled

total space distance space equals space 1800 space plus space 450 space equals space 2250 space straight m

  • Determine the average speed

average space speed space equals space fraction numerator total space distance over denominator total space time end fraction

average space speed space equals space 2250 over 240 space equals space 9.4 space straight m space straight s to the power of negative 1 end exponent

  • Therefore, the average speed is 9.4 m s-1, and it doesn't have a direction, so options A, B and D can be eliminated

Step 3: Determine the total displacement

  • The resultant displacement (blue arrow) is the vector sum of the two displacement vectors (red arrows)

The same city map shows the resultant displacement vector forming the hypotenuse of a triangle with the north and east-bound vectors.

  • The magnitude of the resultant displacement can be found using Pythagoras' theorem:

c squared space equals space a squared space plus space b squared

s squared space equals space 1800 squared space plus space 450 squared

s space equals space square root of 1800 squared space plus space 450 squared end root space equals space 1855 space straight m

  • The direction of the resultant displacement can be found using trigonometry:

tan space theta space equals space opposite over adjacent space equals space 450 over 1800

theta space equals space tan to the power of negative 1 end exponent space open parentheses 450 over 1800 close parentheses space equals space 14 degree space straight E space of space straight N

North has a bearing of 000, therefore, bearing space equals space 0 space plus space 14 space equals space 014

Step 4: Calculate the average velocity

  • The magnitude of the average velocity is

average space velocity space equals space fraction numerator total space displacement over denominator total space time end fraction

average space velocity space equals space 1855 over 240 space equals space 7.7 space straight m space straight s to the power of negative 1 end exponent

  • The direction of the average velocity is in the same direction as the resultant displacement

  • Therefore, the average velocity is 7.7 m s-1 at 014

  • This is option C

Examiner Tips and Tricks

It is best practice to convert units before you input the value into your calculation. You are less likely to forget and less likely to make mistakes this way.

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    Author
    Reviewer
    Katie M

    Author: Katie M

    Expertise: Physics Content Creator

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

    Leander Oates

    Reviewer: Leander Oates

    Expertise: Physics Content Creator

    Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.