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

The characteristic time, tau, is 20 seconds. It will take the cyclist approximately 7.22 seconds to slow down to 15 m/s, 10.82 seconds to slow down to 10 m/s, and 14.42 seconds to slow down to 5 m/s.

Step-by-step explanation:

To find the characteristic time, τ, we can use the formula τ = m / (c * vo). Given that the mass, m, of the cyclist plus the cycle is 80 kg and the coefficient of quadratic air resistance, c, is 0.20 N/m(m/s)2, and the initial speed, vo, is 20 m/s:

τ = 80 kg / (0.20 N/m(m/s)2 * 20 m/s) = 20 s

So the characteristic time is 20 seconds.

To find how long it will take the cyclist to slow down to a certain speed, we can use the formula t = τ * ln(vf/vo), where vf is the final speed. Plugging in the values for vf and vo:

t = 20 s * ln(15 m/s / 20 m/s) = 7.22 s

t = 20 s * ln(10 m/s / 20 m/s) = 10.82 s

t = 20 s * ln(5 m/s / 20 m/s) = 14.42 s

Answer 2

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

Step-by-step explanation:

Given that

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

m = 80 kg

Vo=20m/s

Characteristic time τ=m/cvo.

`\tau =(m)/(c.V_o)`

`\tau =(80)/(0.2* 20)`

Characteristic time τ=20 s

We know that velocity after t time given as follows

`V =(V_o)/(1+(\tau)/(t))`

When V = 15 m/s

`15 =(20)/(1+(20)/(t))`

t = 6.66 s

When V = 10 m/s

`10 =(20)/(1+(20)/(t))`

t = 20 s

When V = 5 m/s

`5 =(20)/(1+(20)/(t))`

t = 60 s