Average speed for a there-and-back journey
Average speed for a there-and-back journey
IP 1
7 marks
A cyclist rides \(d\text{ km}\) up a hill at a constant \(u\text{ km/h}\) and returns down the same road at a constant \(v\text{ km/h}\).
- For \(d=15\), \(u=12\) and \(v=20\), find the average speed for the whole journey.[2]
- Show that, for every distance \(d\), the average speed is \(\dfrac{2uv}{u+v}\text{ km/h}\).[3]
- Show that \(\dfrac{u+v}{2}-\dfrac{2uv}{u+v}=\dfrac{(u-v)^2}{2(u+v)}\). Hence explain why the average speed is less than the average of the two speeds unless \(u=v\).[2]
Solution:
Solution locked
680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.
Create a free account to practise similar questions.
Answer:(a) \(15\text{ km/h}\) (b) \(\dfrac{2d}{\dfrac du+\dfrac dv}=\dfrac{2uv}{u+v}\) (c) the difference is \(\dfrac{(u-v)^2}{2(u+v)}>0\) for \(u\ne v\)
Need help? Join our IP Math tuition programme.
Learn more