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When a parachute is deployed, it plays a crucial role in slowing down an object by increasing air resistance, or drag force. The parachute functions by creating a large surface area, which interacts with air particles to produce friction and subsequently reduce speed. This system is especially effective in high-velocity scenarios, such as slowing down a vehicle like a Chevrolet Camaro from high speeds.
In this example, the parachute attached to the Camaro has a diameter of 2.2 meters, providing a significant frontal area of approximately 3.801 square meters. Once deployed, it contributes a substantial portion to the overall drag force acting on the car, helping it decelerate efficiently.

• The drag coefficient of the parachute is another crucial factor, in this case, it's 1.5, which indicates its capacity to resist the motion through air.
• By increasing the drag force, the parachute aids in reducing the velocity of the car over time, enhancing safety by slowing down the vehicle sooner.

This dynamics of parachute is vital for understanding how drag forces can be maximized for controlled deceleration.

Newton's Second Law of Motion states that the force exerted on an object equals the mass of the object multiplied by its acceleration, or mathematically, \( F = ma \). This fundamental equation demonstrates how force, mass, and acceleration are interrelated. In practical terms, when you apply a net force to a mass, it will accelerate in the direction of that force.
In the scenario with the Camaro and the parachute, the forces at play cause acceleration in the opposite direction of motion, effectively being a deceleration. The car's mass is given as 2500 kg, and the net force acting on it is the total drag force created by the car and parachute drags combined:

• Car Drag Force: Depends on its coefficient (0.25) and the frontal area (1.2 m\(^2\)).
• Parachute Drag Force: Also incorporates its unique coefficient (1.5) and the parachute's area (3.801 m\(^2\)).

When we apply Newton's second law here, by setting the drag force \(-3.67575 \cdot v^2\) equal to the mass times the change in velocity over time (acceleration), we can solve for how the velocity changes over time. This expresses the deceleration force on the car due to the combined drag forces.

A differential equation is an equation that involves the derivatives of a function, describing how that function changes. In our context, we have a first-order differential equation relating velocity \( v \) and time \( t \) due to drag force effects.
In the exercise for the Camaro and parachute, we derived a key equation: \[-3.67575 \cdot v^2 = 2500 \cdot \frac{dv}{dt}.\]
This equation indicates how the velocity \( v \) decreases as a function of time due to the drag force acting on the car.

• The left side shows the negative deceleration resulting from combined drag, while the right side reflects the change in velocity over time.
• By rearranging this equation, we end up with an equation comfortable for integration, allowing the relationship between velocity, time, and deceleration to be solved.

Solving this differential equation involves integrating both sides over relevant limits, which offers us a solution for velocity as an expression dependent on elapsed time, ultimately demonstrating how the vehicle's speed alters due to external forces.

To find the velocity of the Camaro two minutes after the parachute deployment, we are primarily concerned with solving the final expression derived from the differential equation integration.
The essential steps involve integrating from the initial conditions — starting velocity of 110 m/s and time from 0 to 120 seconds. After performing integration, we arrive at the equation: \[-\frac{1}{v} + \frac{1}{110} = \frac{3.67575}{2500} \cdot 120.\]
Solving this equation, we determine that:

• Initial integration: creates a relationship between the inverse of current velocity and initial velocity.
• Final Calculation: involves rearranging and solving for \( v \) to determine the velocity at exactly 120 seconds.

After computation, it reveals a final velocity of approximately 3.70 m/s. This process of calculating velocity illustrates how differential equations translate into real-world applications, showcasing strong interaction between physical concepts like drag force and Newton's laws.