Time, Distance, and Speed
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Master the fundamental relationships between speed, distance, and time, and calculate average speed for multi-stage journeys.
Imagine traveling on a matatu along the Nairobi–Mombasa highway. If the matatu cruises steadily at \(80\text{ km/h}\), every single hour adds exactly \(80\text{ km}\) of progress. In \(2\text{ hours}\), it covers \(160\text{ km}\); in \(3\text{ hours}\), \(240\text{ km}\). Speed is simply the rate at which distance changes with respect to time.
(a) Concrete Scenario
Consider a bus traveling from Nairobi to Mombasa, a distance of \(480\text{ km}\). At a constant speed of \(80\text{ km/h}\), the journey duration is: \[t = \frac{480\text{ km}}{80\text{ km/h}} = 6\text{ hours}\] If the driver doubles the speed to \(160\text{ km/h}\), the travel time is cut in half (\(3\text{ hours}\)). If the speed is halved to \(40\text{ km/h}\), the time doubles (\(12\text{ hours}\)). Thus, time and speed are inversely proportional for a fixed distance.
(b) Geometric Insight
When you plot a Distance–Time graph with time on the horizontal \(x\)-axis and distance on the vertical \(y\)-axis:
- The gradient (slope) of the graph equals the speed: \(\text{Gradient} = \frac{\Delta d}{\Delta t} = v\).
- A steeper line means a higher speed.
- A horizontal line (slope = 0) means the vehicle has stopped (stationary).
(c) The Fundamental Triangle
The core relationship forms an algebraic triangle: \[d = v \times t, \quad v = \frac{d}{t}, \quad t = \frac{d}{v}\] When a journey has multiple stages or varying speeds, the average speed is strictly the total distance divided by the total elapsed time: \[v_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}}\]
Interactive Lab: Live Matatu Journey
Adjust the speed slider and press Play Journey to observe the matatu move and watch the distance–time graph plot dynamically in real time.
Time Elapsed: 0.0 hrs
Estimated Total Time: 6.0 hrs
Key Formulas
Distance: Total ground covered equals speed multiplied by time. (Units: \(\text{km}\) or \(\text{m}\))
Speed: Distance covered per unit of time. (Units: \(\text{km/h}\) or \(\text{m/s}\))
Time: Duration taken to cover a distance at a given speed. (Units: \(\text{hours}\) or \(\text{seconds}\))
Average Speed: The single steady speed that would cover the entire multi-stage trip in the exact same total time.
Unit Conversion Shortcuts:
- Convert \(\text{km/h} \rightarrow \text{m/s}\): Multiply by \(\frac{5}{18}\) (or divide by \(3.6\)).
- Convert \(\text{m/s} \rightarrow \text{km/h}\): Multiply by \(\frac{18}{5}\) (or multiply by \(3.6\)).
- Convert minutes to hours: Divide by \(60\) (e.g., \(45\text{ min} = \frac{45}{60} = 0.75\text{ h}\)).
Worked Examples
Problem: A boda-boda rider travels at a steady speed of \(45\text{ km/h}\) for \(3\text{ hours}\) from Machakos to Kitui. What total distance does the rider cover?
- Identify the knowns: \(v = 45\text{ km/h}\), \(t = 3\text{ h}\).
- Select formula: \(d = v \times t\).
- Calculate: \(d = 45 \times 3 = 135\text{ km}\).
Conclusion: The rider travels \(135\text{ km}\).
Problem: An athlete in Eldoret runs a distance of \(1500\text{ m}\) in \(5\text{ minutes}\). Calculate the average speed in metres per second (\(\text{m/s}\)).
- Convert time to standard units (seconds): \[t = 5\text{ min} \times 60\text{ s/min} = 300\text{ seconds}\]
- Identify distance: \(d = 1500\text{ m}\).
- Apply speed formula: \[v = \frac{d}{t} = \frac{1500\text{ m}}{300\text{ s}} = 5\text{ m/s}\]
Conclusion: The athlete's speed is \(5\text{ m/s}\) (equivalent to \(18\text{ km/h}\)).
Problem: A safari van travels from Nairobi to Nakuru. The first leg of \(60\text{ km}\) is through city traffic at \(30\text{ km/h}\). The remaining \(100\text{ km}\) along the highway is covered at \(50\text{ km/h}\). What is the average speed of the van for the entire trip?
- Calculate time for Leg 1: \[t_1 = \frac{d_1}{v_1} = \frac{60\text{ km}}{30\text{ km/h}} = 2.0\text{ hours}\]
- Calculate time for Leg 2: \[t_2 = \frac{d_2}{v_2} = \frac{100\text{ km}}{50\text{ km/h}} = 2.0\text{ hours}\]
- Find total distance and total time: \[d_{\text{total}} = 60 + 100 = 160\text{ km}\] \[t_{\text{total}} = 2.0 + 2.0 = 4.0\text{ hours}\]
- Calculate overall average speed: \[v_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{160\text{ km}}{4.0\text{ h}} = 40\text{ km/h}\]
Note: The arithmetic mean of the two speeds would have been \(\frac{30+50}{2} = 40\text{ km/h}\) only because the times for both legs happened to be identical (\(2\text{ h}\) each). If the times differed, the arithmetic mean would be incorrect!
Common Mistakes
Real World
Practice