Answer to Question #274460 in Mechanics | Relativity for Gyut

Question

On a 40-km bike ride, a cyclist rides the first 20 km at 20 km/h.If the cyclist rides very fast for the final 20 km, what is the maximum value his average speed could approach?

Answer 1

Explanations & Calculations

• Write a relationship for the average speed and integrate it to see the maximum.

$\qquad\qquad \begin{aligned} \small V_{avg}&=\small \frac{40\,km}{\frac{20\,km}{20\,kmh^{-1}}+\frac{20\,km}{v}}\\ &=\small \frac{40}{\Big(1+\frac{20}{v}\Big)} \end{aligned}$

• For the average speed to be maximum, the denominator needs to be minimum which is $\small 1$ .
• For any speed during the final 20 kilometres, the denominator value is beyond 1.
• Therefore, the maximum average speed he could attain is $\small 40\,kmh^{-1}$ .