Question

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

Short answer

Characteristic time is 20 s; 6.67 s to 15 m/s, 20 s to 10 m/s, and 60 s to 5 m/s.

Step-by-step solution

Calculating Characteristic Time

The characteristic time \( \tau \) is given by the formula \( \tau = \frac{m}{c v_{o}} \). Here, \( m = 80 \text{ kg} \), \( c = 0.20 \) \( \text{N}/(\text{m/s})^2 \), and \( v_{o} = 20 \text{ m/s} \). Substitute these values into the formula to find:\[\tau = \frac{80 \text{ kg}}{0.20 \text{ N/(m/s)}^2 \times 20 \text{ m/s}} = \frac{80}{4} = 20 \text{ seconds}.\]The characteristic time \( \tau \) is 20 seconds.

Finding Time to Slow to 15 m/s

The velocity of the cyclist as a function of time under the influence of quadratic air resistance can be expressed as \( v(t) = \frac{v_{o}}{1 + \frac{cv_{o}t}{m}} \). We need to solve this equation for \( t \) when \( v(t) = 15 \text{ m/s} \).Set \( v(t) = 15 \) and solve for \( t \):\[15 = \frac{20}{1 + \frac{0.20 \times 20 \times t}{80}}.\]Solving for \( t \):\[15(1 + \frac{t}{20}) = 20 \15 + \frac{15t}{20} = 20 \\frac{15t}{20} = 5 \t = \frac{20 \times 5}{15} = \frac{100}{15} \approx 6.67 \text{ seconds}.\]It takes approximately 6.67 seconds to slow down to \( 15 \text{ m/s} \).

Finding Time to Slow to 10 m/s

Next, we apply the same method to find the time \( t \) when the velocity \( v(t) = 10 \text{ m/s} \).Set \( v(t) = 10 \) and solve for \( t \):\[10 = \frac{20}{1 + \frac{0.20 \times 20 \times t}{80}}.\]Solving for \( t \):\[10(1 + \frac{t}{20}) = 20 \10 + \frac{10t}{20} = 20 \\frac{10t}{20} = 10 \t = \frac{20 \times 10}{10} = 20 \text{ seconds}.\]It takes 20 seconds to slow down to \( 10 \text{ m/s} \).

Finding Time to Slow to 5 m/s

Finally, we find the time \( t \) when the velocity \( v(t) = 5 \text{ m/s} \) using the equation:Set \( v(t) = 5 \) and solve for \( t \):\[5 = \frac{20}{1 + \frac{0.20 \times 20 \times t}{80}}.\]Solving for \( t \):\[5(1 + \frac{t}{20}) = 20 \5 + \frac{5t}{20} = 20 \\frac{5t}{20} = 15 \t = \frac{20 \times 15}{5} = 60 \text{ seconds}.\]It takes 60 seconds to slow down to \( 5 \text{ m/s} \).

Key concepts

Quadratic Air Resistance

When analyzing motion through a fluid like air, a force often comes into play that opposes the movement. This is known as air resistance. It acts against the direction of velocity and depends on several factors, including the shape and speed of the object. In the study of mechanics, air resistance is often modeled in two main ways: linear and quadratic. Quadratic air resistance is particularly important at higher speeds.

• **Quadratic dependency**: As the name suggests, this type involves the force being proportional to the square of the velocity. The formula for the force due to quadratic air resistance is given by \( F_d = c v^2 \), where \( c \) is the drag coefficient and \( v \) is the velocity.


• **Relevance**: This model is highly relevant for high-speed motions, such as cycling at speeds we deal with in this exercise. The faster the cyclist moves, the greater the opposing force they will experience due to air resistance.

This quadratic relationship makes it crucial to use the correct equations when calculating how long it takes to slow down from different speeds, as it is done in the exercise.

Characteristic Time

Characteristic time, represented by the symbol \( \tau \), is a concept that helps understand how quickly a system responds to changes. For a cyclist affected by quadratic air resistance, it serves as a critical parameter to predict the time scale over which the velocity significantly changes.

• **Formula**: The characteristic time for an object experiencing quadratic air resistance can be calculated using the formula \( \tau = \frac{m}{c v_o} \). Here, \( m \) is the mass, \( c \) is the drag coefficient, and \( v_o \) is the initial velocity.


• **Purpose**: Knowing the characteristic time helps in estimating how quickly the cyclist's speed will decrease due to air resistance alone.
• **Example**: In the exercise, we calculated the characteristic time, \( \tau \), to be 20 seconds with the given parameters. This value provides a baseline, indicating how time scales linearly with changes in air resistance.

Velocity Calculation

Calculating the velocity under the influence of quadratic air resistance involves solving differential equations that describe the motion of the cyclist. The velocity as a function of time \( v(t) \) is calculated as:

• The formula: \( v(t) = \frac{v_o}{1 + \frac{cv_o t}{m}} \) is derived from integrating Newton's second law of motion, considering the quadratic air resistance.


• Solving for specific velocities: In the exercise, this formula is used to determine when the cyclist reaches speeds of 15 m/s, 10 m/s, and 5 m/s.


• Interpretation: As time increases, the denominator grows, causing the velocity to decrease, showing how quadratic air resistance affects the slowing of the cyclist.

By understanding this calculation, one can grasp how air resistance influences the speed over time, explaining the gradual decrease to lower velocities.

Cyclist Mechanics Problem

The problem at hand involves understanding the temporal evolution of a cyclist's velocity under the force of quadratic air resistance. Here, the examination elucidates the mechanical principles involved in such a scenario.

• **Setup**: Initially, a cyclist with a mass of 80 kg and an initial speed of 20 m/s coasts to a stop, influenced by air resistance with a coefficient \( c = 0.20 \text{ N}/(\text{m/s})^2 \).


• **Physics involved**: Air resistance acts as a decelerating force, and is proportional to the square of the velocity. This makes it crucial for the cyclist to slow down from high speeds.


• **Practical Implications**: As calculated, the times to decelerate to certain lower speeds (15 m/s, 10 m/s, and 5 m/s) illustrate the application of theoretical physics in practical scenarios. These calculations help in designing safety measures and improving performance using aerodynamic insights.

This practical mechanics problem underscores the importance of physics in everyday activities, offering a fundamental understanding of how cyclists can be affected by factors such as air resistance.