Ratio, proportion & rates of change

Compound Measures

Year 10 · Year 11

✓By the end of this lesson students will be able to calculate speed, distance, or time given the other two values.
✓By the end of this lesson students will be able to calculate density, mass, or volume given the other two values.
✓By the end of this lesson students will be able to calculate pressure, force, or area given the other two values.
✓By the end of this lesson students will be able to convert between different units for compound measures (e.g., km/h to m/s, g/cm³ to kg/m³).
✓By the end of this lesson students will be able to interpret and calculate rates of change from graphs, including gradients of distance-time and velocity-time graphs (Higher).

Key concepts

Speed

Speed is a measure of how fast an object is moving. It is defined as the distance travelled per unit of time. It is a scalar quantity.

Speed = Distance / Time

Density

Density is a measure of how much mass is contained in a given volume. It describes how 'packed' the material is. Materials with higher density have more mass in the same volume.

Density = Mass / Volume

Pressure

Pressure is a measure of the force applied perpendicular to a surface per unit area. It is often measured in Pascals (Pa), where 1 Pa = 1 N/m².

Pressure = Force / Area

Rates of Change & Gradients (Higher)

A rate of change describes how one quantity changes in relation to another. In maths, this is often represented by the gradient of a graph. For a straight line, the gradient is constant. For a curve, the instantaneous rate of change is given by the gradient of the tangent at that point. Speed is the rate of change of distance with respect to time (gradient of a distance-time graph). Acceleration is the rate of change of velocity with respect to time (gradient of a velocity-time graph).

Gradient = Change in y / Change in x

Key facts to remember

1Speed = Distance / Time (S = D/T)
2Density = Mass / Volume (D = M/V)
3Pressure = Force / Area (P = F/A)
4Units must be consistent for calculations (e.g., if distance is in km, time should be in hours for speed in km/h).
5Common unit conversions: 1 km = 1000 m, 1 hour = 60 minutes = 3600 seconds, 1 kg = 1000 g.
61 N/m² is equivalent to 1 Pascal (Pa).
7On a distance-time graph, the gradient of the line represents the speed.
8On a velocity-time graph, the gradient of the line represents the acceleration (Higher Tier).

Worked examples

Example 1

A train travels 180 km in 2 hours and 20 minutes. Calculate its average speed in km/h.

IConvert the time to hours: 2 hours 20 minutes = 2 + (20/60) hours = 2 + (1/3) hours = 7/3 hours.

IIIdentify the distance: Distance = 180 km.

IIIApply the speed formula: Speed = Distance / Time.

IVSubstitute the values: Speed = 180 km / (7/3) h.

VCalculate: Speed = (180 × 3) / 7 km/h = 540 / 7 km/h.

Answer

77.1 km/h (to 3 significant figures)

Always ensure units are consistent before calculation. Converting minutes to a decimal part of an hour is crucial.

Example 2

A cylindrical metal rod has a mass of 3.14 kg. Its radius is 2 cm and its length is 25 cm. Calculate its density in g/cm³. (Use π = 3.14)

ICalculate the volume of the cylinder using the formula V = πr²h:

IIVolume = 3.14 × (2 cm)² × 25 cm = 3.14 × 4 cm² × 25 cm = 3.14 × 100 cm³ = 314 cm³.

IIIConvert the mass to grams: Mass = 3.14 kg = 3.14 × 1000 g = 3140 g.

IVApply the density formula: Density = Mass / Volume.

VSubstitute the values: Density = 3140 g / 314 cm³.

Answer

10 g/cm³

Pay attention to the required units for the final answer and perform conversions as needed.

Example 3

The distance-time graph below shows a person's journey. From t=0 to t=2 hours, the distance increases linearly from 0 km to 10 km. From t=2 to t=4 hours, the distance increases linearly from 10 km to 15 km. a) Calculate the speed during the first 2 hours. b) Calculate the average speed for the entire 4-hour journey.

Ia) For the first 2 hours:

II Change in distance (Δd) = 10 km - 0 km = 10 km.

III Change in time (Δt) = 2 h - 0 h = 2 h.

IV Speed = Gradient = Δd / Δt = 10 km / 2 h.

Vb) For the entire 4-hour journey:

VI Total distance travelled = 15 km - 0 km = 15 km.

VII Total time taken = 4 h - 0 h = 4 h.

VIII Average Speed = Total Distance / Total Time = 15 km / 4 h.

Answer

a) 5 km/h b) 3.75 km/h

On a distance-time graph, the gradient of the line represents the speed. A steeper gradient means a higher speed.

Common mistakes

✗Not converting units to be consistent before performing calculations (e.g., using km and minutes together for speed).
✗Mixing up the formulas (e.g., calculating density as volume/mass instead of mass/volume).
✗Incorrectly rearranging formulas (e.g., calculating distance as speed/time instead of speed × time).
✗Forgetting to convert time given in minutes to decimal hours (e.g., 3 hours 30 minutes is 3.5 hours, not 3.3 hours).
✗Confusing average speed (total distance / total time) with the average of different speeds during a journey.

Exam tips

★Always write down the formula you are using at the start of your working.
★Show all your working steps clearly, especially any unit conversions.
★Check that your units are consistent before performing any calculations and ensure your final answer has the correct units.
★Use 'formula triangles' (e.g., D-S-T, M-D-V, F-P-A) as a memory aid, but also understand how to rearrange the formula algebraically.

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