Question

A typical value for the coefficent of quadratic air resistance on a cyclist is around c=0.20 N/m(m/s)2. Assuming thath the total mass (cyclist plust cycle) is m = 80 kg and that at t = 0 the cyclist has an intial speed v=20m/s (abouit 45 mi/h) and starts to coast to a stop under the influence of air resistance, find the characteristic time τ=m/cvo. How long will it take him to slow to 15 m/s? What about 10 m/s? And 5 m/s? (Below about 5 m/s, it is certainly not reasonable to ignore friction, so there is no point pursuing this calculation to lower speeds.)

Answer 1

Answer:

Characteristic time τ=20 s

When V = 15 m/s

t = 6.66 s

When V = 10 m/s

t = 20 s

When V = 5 m/s

t = 60 s

Explanation:

Given that

c=0.20 N/m(m/s)2

m = 80 kg

Vo=20m/s

Characteristic time τ=m/cvo.

$\tau =\dfrac{m}{c.V_o}$

$\tau =\dfrac{80}{0.2\times 20}$

Characteristic time τ=20 s

We know that velocity after t time given as follows

$V =\dfrac{V_o}{1+\dfrac{\tau}{t}}$

When V = 15 m/s

$15 =\dfrac{20}{1+\dfrac{20}{t}}$

t = 6.66 s

When V = 10 m/s

$10 =\dfrac{20}{1+\dfrac{20}{t}}$

t = 20 s

When V = 5 m/s

$5 =\dfrac{20}{1+\dfrac{20}{t}}$

t = 60 s