Question

A drag chute must be designed to reduce the speed of a 3000-lb dragster from 220 mph to \(50 \mathrm{mph}\) in \(4 \mathrm{sec}\). Assume that the drag force is proportional to the velocity. (a) What value of the drag coefficient \(k\) is needed to accomplish this? (b) How far will the dragster travel in the 4-sec interval?

Short answer

Question: Determine the drag coefficient of a dragster and the distance it will travel during a 4-second braking process. The initial velocity of the dragster is 220 mph, the final velocity is 50 mph, and the weight of the dragster is 3000 lbs. Answer: The drag coefficient is approximately 579.3, and the dragster will travel approximately 443.56 feet during the 4-second braking process.

Step-by-step solution

Determine initial and final velocities and weight of dragster

We are given the initial velocity \(v_i = 220\) mph, the final velocity \(v_f = 50\) mph, and the weight of the dragster \(W = 3000\) lb. First, we need to convert the velocities to ft/s (feet per second). 1 mph = \(5280 / 3600\) ft/s. So, \(v_i = 220 \times \frac{5280}{3600} \approx 322.67\) ft/s \(v_f = 50 \times \frac{5280}{3600} \approx 73.33\) ft/s

Determine acceleration of the dragster

Now, we will determine the acceleration required to achieve the decay in speed in the given timeframe. We will use the equation: \(v_f = v_i + at\) Where \(v_i\) is the initial velocity, \(v_f\) is the final velocity, \(a\) is the acceleration, and \(t\) is the time. We are given \(t = 4\) sec. Rearrange the equation to solve for the acceleration: \(a = \frac{v_f - v_i}{t}\) Now plug in the values: \(a = \frac{73.33 - 322.67}{4} \approx -62.335\) ft/s² The negative sign indicates that it's a deceleration.

Determine the required drag force

We can use Newton's second law (F = ma) to calculate the required drag force to achieve this acceleration. Since the drag force is acting opposite to the direction of motion, it has a negative sign as well. \(F_d = - ma\) Substitute the values: \(F_d = - (-62.335) \times 3000 \approx 186995\) lb

Determine the drag coefficient k

The drag force is proportional to the velocity, so we can write: \(F_d = -k v\) Now, we can solve for the drag coefficient \(k\) using the initial velocity and the required drag force at that velocity: \(k = \frac{-F_d}{v_i} \approx \frac{186995}{322.67} \approx 579.3\) So, the drag coefficient \(k \approx 579.3\).

Determine the dragster's position equation

Since the drag force is proportional to the velocity, we have a linear first-order differential equation of the form: \(-k v = ma\) We can solve this differential equation and find the equation for the position of the dragster as a function of time. \(v = v_i + at = v_i - kt\) Integrate with respect to time: \(x(t) = v_i t - \frac{1}{2}kt^2 + C\) At \(t=0\), \(x(0) = 0\), thus \(C = 0\). The equation becomes: \(x(t) = v_i t - \frac{1}{2}kt^2\)

Find the distance traveled in 4 seconds

Now, we will use the position equation to find the distance traveled by the dragster in the 4-sec interval: \(x(4) = 322.67 \times 4 - \frac{1}{2}(579.3)(4)^2 \approx 443.56\) ft. The dragster will travel approximately 443.56 feet during the 4-second interval. Summary: (a) The required drag coefficient is approximately \(k \approx 579.3\). (b) The dragster will travel approximately 443.56 feet during the 4-second interval.

Key concepts

Drag Force

Drag force is a resistance force that acts opposite to the direction of motion of an object moving through a fluid such as air or water. It is a crucial element for vehicles like dragsters, which encounter substantial air resistance at high speeds. Drag force can be expressed mathematically as
\[\begin{equation} F_d = \frac{1}{2} C_d \rho A v^2 \end{equation}\], where

• \[\begin{equation} C_d \end{equation}\] is the drag coefficient,
• \[\begin{equation} \rho \end{equation}\] is the density of the fluid,
• \[\begin{equation} A \end{equation}\] is the cross-sectional area, and
• \[\begin{equation} v \end{equation}\] is the velocity of the object relative to the fluid.

In the provided exercise, the drag force is proportional to velocity, which simplifies the equation as the other factors are considered constant or incorporated into a single proportionality constant,
\[\begin{equation} k \end{equation}\]. This relationship is key in designing systems like parachutes or drag chutes that counteract the motion of a vehicle to decrease its speed safely and effectively.

Deceleration

Deceleration refers to the decrease in the velocity of an object per unit of time, essentially negative acceleration. It is a rate at which an object slows down, commonly experienced when a vehicle, like a dragster, is coming to a stop. The formula for deceleration can be represented as
\[\begin{equation} a = \frac{\triangle v}{\triangle t} \end{equation}\], where

• \[\begin{equation} \triangle v \end{equation}\] is the change in velocity (final velocity minus initial velocity) and
• \[\begin{equation} \triangle t \end{equation}\] is the change in time.

The negative sign indicates the direction of the acceleration is opposite to the motion. In our exercise, calculating the deceleration was essential to determine the force needed to be exerted by the drag chute to achieve a specific decrease in the dragster's speed within a given timeframe.

Drag Coefficient

The drag coefficient,
\[\begin{equation} C_d \end{equation}\], is a dimensionless quantity that quantifies the drag or resistance of an object in a fluid environment. It represents how streamlined an object is and varies with the shape of the object and the nature of the fluid flow around it. In aerodynamics, the drag coefficient is used to model the impact of the aerodynamic drag on an object, such as a car or airplane. The formula used in the exercise assumes a direct proportionality between the velocity and the drag force, allowing for the calculation of a drag coefficient
\[\begin{equation} k \end{equation}\], which acts as a measure of the overall effect of drag on the dragster. This coefficient is invaluable for engineers when designing vehicles and their components to optimize performance and safety.

Newton's Second Law

Newton's second law of motion states that the force on an object is equal to the mass of the object multiplied by the acceleration of the object:
\[\begin{equation} F = ma \end{equation}\]. This law is a cornerstone of classical mechanics, explaining how the velocity of an object changes when it is subjected to external forces. In our example, the law is applied to relate the mass of the dragster to the deceleration produced by the drag force of the chute. The resulting force is what causes the change in the velocity of the dragster, and using this law allowed for the calculation of the drag coefficient needed to design the drag chute optimally.