Speed
Every question with a fully worked solution.
- Speed - Video - Finding the average speed of a car in km per hour Watch
Theory
Speed is a rate: the distance travelled in each unit of time. \(S = \dfrac{D}{T}\), \(D = S \times T\) and \(T = \dfrac{D}{S}\). A car that goes \(270\) km in \(3\) h and \(160\) km in \(2\) h averages \(430 \div 5 = 86\) km/h. This Year 8 maths page shows how.
Speed is the distance travelled per unit of time. Its units name both, such as km/h or m/s.
One formula, three ways: \(S = \dfrac{D}{T}\), \(D = S \times T\), \(T = \dfrac{D}{S}\). Write the formula, substitute, then write the units.
The units must match. km/h needs kilometres and hours; m/s needs metres and seconds. Change units before substituting. m/s \(\times 3.6 =\) km/h and km/h \(\div 3.6 =\) m/s.
Minutes to hours: divide by \(60\): \(15\) min \(= 0.25\) h; \(20\) min \(= \dfrac{1}{3}\) h (keep the fraction). Decimal hours to minutes: multiply the decimal part by \(60\): \(4.2\) h \(= 4\) h \(12\) min.
Average speed \(=\) total distance \(\div\) total time, counting every stop. It is not the mean of the separate speeds.
Harder problems. Time to catch up \(=\) the gap \(\div\) the distance gained each hour (travellers moving towards each other close the gap at the sum of their speeds). Speed needed \(=\) distance left \(\div\) time left.
Minutes as part of an hour:
| Minutes | Of an hour | Decimal |
|---|---|---|
| 6 | \(\dfrac{1}{10}\) | 0.1 |
| 12 | \(\dfrac{1}{5}\) | 0.2 |
| 15 | \(\dfrac{1}{4}\) | 0.25 |
| 20 | \(\dfrac{1}{3}\) | \(0.333\ldots\) |
| 30 | \(\dfrac{1}{2}\) | 0.5 |
| 45 | \(\dfrac{3}{4}\) | 0.75 |
Solving a speed problem
- Write the formula you need.
- Match the units (change minutes to hours, or km to m).
- Substitute and calculate.
- Answer with units, changing decimal hours to hours and minutes if needed.
| \(3 \times 90\) | \(=\) | \(270 \text{ km}\) |
| \(270 + 160\) | \(=\) | \(430 \text{ km}\) |
| \(3 + 2\) | \(=\) | \(5 \text{ h}\) |
| \(430 \div 5\) | \(=\) | \(86\) |
\(86\) km/h
| \(\dfrac{30 \times 1000}{60 \times 60}\) | \(=\) | \(\dfrac{300}{36} = 8\dfrac{1}{3}\) |
About \(8\dfrac{1}{3}\) m/s (or \(30 \div 3.6\)).
| \(36 \text{ min}\) | \(=\) | \(\dfrac{3}{5} \text{ h}\) |
| \(45 \times \dfrac{3}{5}\) | \(=\) | \(27\) |
\(27\) km
(i) Find its speed in km/h.
(ii) Write it in m/s.
| \(10.8 \times 4\) | \(=\) | \(43.2\) |
| \(43.2 \div 3.6\) | \(=\) | \(12\) |
(i) \(43.2\) km/h (\(15\) min \(= \dfrac{1}{4}\) h)
(ii) \(12\) m/s (check: \(10\,800\) m \(\div\) \(900\) s \(= 12\))
| \(148 \div 40\) | \(=\) | \(3.7 \text{ h}\) |
| \(0.7 \times 60\) | \(=\) | \(42\) |
\(3\) h \(42\) min after 06:15: it arrives at 09:57.
(i) Find its speed on the first part.
(ii) Find its average speed for the whole trip.
| \(30 \div \dfrac{25}{60}\) | \(=\) | \(72\) |
| \(30 + 21\) | \(=\) | \(51\) |
| \(25 + 5 + 30\) | \(=\) | \(60 \text{ min}\) |
(i) \(72\) km/h
(ii) \(51\) km in \(1\) h: \(51\) km/h
(ii) A coach must cover \(240\) km in \(3\) h. The first \(100\) km take \(1.25\) h. What speed does it need for the rest?
(i) At 8:20 the cyclist is \(18 \times \dfrac{20}{60} = 6\) km ahead. The car gains \(54 - 18 = 36\) km each hour.
| \(6 \div 36 \times 60\) | \(=\) | \(10 \text{ min}\) |
| \(140 \div 1.75\) | \(=\) | \(80\) |
(i) At 8:30 am
(ii) \(140\) km in \(1.75\) h: \(80\) km/h
Common pitfalls
Frequently asked questions
What is the formula for speed?
Speed = distance ÷ time. Rearranged, distance = speed × time and time = distance ÷ speed.
How do I change km/h to m/s?
Divide by 3.6. To change m/s to km/h, multiply by 3.6.
How do I change minutes to hours?
Divide by 60. For example, 45 minutes is 0.75 hours.
How do I find average speed?
Divide the total distance by the total time, including any stops.
How do I change 3.6 hours to hours and minutes?
Multiply the decimal part by 60: 0.6 × 60 = 36, so 3 h 36 min.