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You observe a 1350 -kg sports car rolling along flat pavement in a straight line. The only horizontal forces acting on it are a constant rolling friction and air resistance (proportional to thesquare of its speed). You take the following data during a time interval of \(25 \mathrm{s} :\) When its speed is 32 \(\mathrm{m} / \mathrm{s}\) , the car slows down at a rate of \(-0.42 \mathrm{m} / \mathrm{s}^{2},\) and when its speed is decreased to \(24 \mathrm{m} / \mathrm{s},\) it slows down at \(-0.30 \mathrm{m} / \mathrm{s}^{2} .\) (a) Find the coefficient of rolling friction and the air drag constant \(D .(b)\) At what constant speed will this car move down an incline that makes a \(2.2^{\circ}\) angle with the horizontal? (c) How is the constant speed for an incline of angle \(\beta\) related to the terminal speed of this sports car if the car drops off a high cliff? Assume that in both cases the air resistance force is proportional to the square of the speed, and the air drag constant is the same.

Step by step solution

01

Analyze the Forces

There are two primary forces acting against the car's motion: rolling friction and air resistance. The equation of motion can be written as \(-m a = f_r + f_d\), where \(f_r = \mu m g\) is the rolling friction and \(f_d = D v^2\) is the air resistance, where \(\mu\) is the coefficient of rolling friction, \(m\) is the mass of the car, and \(D\) is the air drag constant.

02

Use Given Conditions at 32 m/s

For \(v = 32 \ \text{m/s}\), the acceleration \(a = -0.42 \ \text{m/s}^2\). The equation becomes:\[-1350 \times (-0.42) = \mu \times 1350 \times 9.8 + D \times 32^2.\]This simplifies to:\[567 = 13230\mu + 1024D.\]

03

Use Given Conditions at 24 m/s

For \(v = 24 \ \text{m/s}\), the acceleration \(a = -0.30 \ \text{m/s}^2\). The equation becomes:\[-1350 \times (-0.30) = \mu \times 1350 \times 9.8 + D \times 24^2.\]This simplifies to:\[405 = 13230\mu + 576D.\]

04

Solve the System of Equations

We have two equations:1. \[567 = 13230\mu + 1024D.\]2. \[405 = 13230\mu + 576D.\]Subtract the second equation from the first to eliminate \(\mu\):\[162 = 448D\]Solve for \(D\):\[D = \frac{162}{448} = 0.3616.\]Substitute \(D\) back into one of the equations to find \(\mu\):\[405 = 13230\mu + 576 \times 0.3616\]\[405 = 13230\mu + 208.896\]\[196.104 = 13230\mu\]\[\mu = \frac{196.104}{13230} = 0.0148.\]

05

Calculate Constant Speed on Incline

For a constant speed on an incline, the force balance equation becomes:\[m g \sin(\theta) = \mu m g \cos(\theta) + D v^2.\]For \(\theta = 2.2^{\circ}\), and knowing \(\mu\) and \(D\) from earlier steps, solve for \(v\):\[1350 \times 9.8 \times \sin(2.2^{\circ}) = 1350 \times 9.8 \times \mu \times \cos(2.2^{\circ}) + D v^2.\]Solve numerically:\[1350 \times 9.8 \times \sin(2.2^{\circ}) = 1350 \times 9.8 \times 0.0148 \times \cos(2.2^{\circ}) + 0.3616 v^2\]Calculate and find \(v\).

06

Relate Incline Speed to Terminal Velocity

The terminal velocity \(v_t\) occurs when the net force is zero during free fall. For an incline angle \(\beta\), and similarly equating forces, we express \(v\) depending on \(v_t\):\[v^2 = \left(\frac{g \sin(\beta) + \mu g \cos(\beta)}{D}\right)\]The terminal velocity \(v_t = \sqrt{\frac{mg}{D}}\), thus relating \(v\) and \(v_t\) by the ratio of gravitational, frictional, and drag forces.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Air Resistance

Air resistance is a force that opposes the motion of an object through the air. It can usually be described as a drag force that increases with the speed of the object. The faster an object moves, the greater the air resistance it experiences. This force is proportional to the square of the object's velocity.
In mathematical terms, when dealing with air resistance, it's often written as:

• \( f_d = Dv^2 \)

Here, \( f_d \) is the force of air resistance, \( D \) is the air drag constant, and \( v \) is velocity. This relationship shows that if the speed doubles, the air resistance will quadruple.
Air resistance plays a crucial role when calculating the motion of objects, particularly at high speeds, like the sports car in our exercise. It becomes vital to determine how much it slows down the car alongside rolling friction. Notably, air resistance can significantly affect the car's motion on both flat surfaces and inclined planes.

Inclined Planes

Inclined planes are flat surfaces tilted at an angle, different from horizontal planes. When an object moves on an incline, the forces at play are different from those on flat ground due to the angle's effect on force components.
On an inclined plane, gravity causes a component of force to act along the plane's surface. This force tends to pull the object downward. The components of gravitational force can be broken down as follows:

• \( m g \cos(\theta) \) - This is the normal force perpendicular to the surface.
• \( m g \sin(\theta) \) - This is the parallel force pulling the object down the plane.

In the context of our exercise, understanding these components is essential to calculate the speed needed to maintain motion on a slope. The incline complicates the equations by adding new force components that we balance against friction and drag forces.

Terminal Velocity

Terminal velocity refers to the highest velocity an object reaches when the net force acting on it becomes zero during free fall. At this point, the forces of gravity and air resistance are equal, resulting in a constant speed.
When discussing terminal velocity, especially in cases involving a car, similar principles apply to free-falling objects:

• Air resistance equals gravitational force.
• No further acceleration occurs once terminal velocity is achieved.

To express terminal velocity in equations, we use:

• \( v_t = \sqrt{\frac{mg}{D}} \)

Where \( v_t \) denotes terminal velocity, \( m \) the mass, \( g \) the gravitational constant, and \( D \) the air drag constant. For the sports car, terminal velocity is significant when equating its motion either down a slope or when falling freely, balancing gravitational and drag forces.

Newton's Laws of Motion

Newton's laws of motion form the foundation for understanding the movement of objects, especially in dynamic situations involving multiple forces like rolling friction and air resistance.

• First Law (Inertia): An object remains at rest or in uniform motion unless acted upon by an unbalanced force.
• Second Law (F=ma): An object's acceleration is directly proportional to the net force and inversely proportional to its mass.
• Third Law: For every action, there is an equal and opposite reaction.

In our exercise, the second law is primarily at play. The balance of forces defines the car's acceleration and deceleration. Calculations involve forces like rolling friction \( \mu mg \) and air resistance \( Dv^2 \). Understanding these laws helps us solve for critical variables like acceleration rates efficiently, relying on known forces and principles governing motion.