Problem 23 - Calculus Early Transcendentals

Question

The velocity (in \(\mathrm{mi} / \mathrm{hr}\) ) of an airplane flying into a headwind is given by \(v(t)=30\left(16-t^{2}\right),\) for \(0 \leq t \leq 3 .\) Assume that \(s(0)=0\) and \(t\) is measured in hours. a. Determine and graph the position function, for \(0 \leq t \leq 3\) b. How far does the airplane travel in the first \(2 \mathrm{hr} ?\) c. How far has the airplane traveled at the instant its velocity reaches 400 mi/hr?

Step-by-Step Solution

Answer

Answer: The airplane travels 240 miles in the first 2 hours. When its velocity reaches 400 mi/hr, it has traveled 30(16sqrt(80/3)-320/9) miles.

Step 1: Find the position function

To find the position function, we need to integrate the given velocity function \(v(t)=30(16-t^{2})\) with respect to time. $$s(t) = \int v(t) \, dt = \int 30(16-t^2) \, dt$$ Now, we have to integrate the function with respect to \(t\). $$s(t) = 30\int (16-t^2) \, dt = 30\left(16t-\frac{t^3}{3}\right) + C$$ Since \(s(0)=0\), we can determine the constant \(C\). $$0 = 30\left(16(0)-\frac{(0)^3}{3}\right) + C \Rightarrow C=0$$ So, the position function is: $$s(t) = 30\left(16t-\frac{t^3}{3}\right)$$

Step 2: Graph the position function

To graph the position function, we can use a graphing software or tool to plot the function for the given interval, \(0 \leq t \leq 3\). The graph of the function will show how the position changes with respect to time during the flight.

Step 3: Determine the distance in the first 2 hours

To determine how far the airplane travels in the first 2 hours, we need to evaluate the position function at \(t=2\). $$s(2) = 30\left(16(2)-\frac{(2)^3}{3}\right) = 30(32-\frac{8}{3})=960-\frac{240}{3}=\frac{720}{3}$$ The airplane travels 240 miles in the first 2 hours.

Step 4: Find the distance when velocity reaches 400 mi/hr

To find the time when the velocity reaches 400 mi/hr, we need to solve the equation \(v(t)=30(16-t^{2})=400\) for \(t\). $$30(16-t^{2})=400 \Rightarrow 16-t^2=\frac{400}{30} \Rightarrow t^2=16-\frac{400}{30} \Rightarrow t^2=\frac{80}{3}$$ Now, we need to take the square root of both sides to find \(t\). $$t = \sqrt{\frac{80}{3}}$$ To find the distance when the velocity reaches 400 mi/hr, we need to evaluate the position function at this time. $$s\left(\sqrt{\frac{80}{3}}\right) = 30\left(16\sqrt{\frac{80}{3}}-\frac{(\sqrt{\frac{80}{3}})^3}{3}\right)$$ $$s\left(\sqrt{\frac{80}{3}}\right) = 30\left(16\sqrt{\frac{80}{3}}-\frac{320}{9}\right)$$ The airplane has traveled 30(16sqrt(80/3)-320/9) miles at the instant its velocity reaches 400 mi/hr.

Key Concepts

Velocity Function

When we talk about motion along a straight line, the velocity function plays a pivotal role in understanding how quickly an object moves over time. In our specific problem, the velocity of the airplane, represented as \( v(t) = 30(16 - t^2) \), describes how the speed changes with time \( t \). The coefficients in the equation indicate factors relating to speed reduction due to a headwind.

This function is quadratic, signifying that velocity doesn't just decrease linearly—it decreases in a way that the effect intensifies over time. Understanding such characteristics is essential since it helps infer when the airplane might slow considerably or accelerate. Looking at this function can give insights into the real-life situation of an airplane heading into a wind resistance which slows it down quadratically with time.

Integration

Integration is the mathematical process that allows us to transition from the velocity function to the position function, \( s(t) \). By integrating \( v(t) \), we are essentially aggregating all the tiny bits of velocity over time to determine where the object is located at a given point.

With the velocity function \( v(t) = 30(16 - t^2) \), we integrate with respect to \( t \) to find:

• \( s(t) = \int 30(16-t^2) \; dt \).

After computation, this results in \( s(t) = 30(16t - \frac{t^3}{3}) + C \). The constant \( C \) (from the indefinite integral) gets determined by initial conditions; in this exercise, since \( s(0) = 0 \), we find that \( C = 0 \). Thus, understanding integration is crucial as it transforms velocity into position, giving us a clearer picture of the airplane's actual path over time.

Distance Calculation

Distance calculations are about deriving how far an object has traveled within certain boundaries. By using the position function, you can simply substitute the given time to find how far the airplane has traveled. For instance, to determine the distance in the first 2 hours, you evaluate \( s(2) \), resulting in:

• \( s(2) = 30(16 \times 2 - \frac{2^3}{3}) \).

Upon solving, you find that the distance is 240 miles.

Moreover, considering velocity conditions like attaining 400 mi/hr involves solving the velocity equation \( 30(16-t^2) = 400 \) to find \( t \), and then using this \( t \) to calculate the respective distance \( s(t) \). These calculations are essential for understanding both instantaneous and accumulated travel distances under given conditions in practical scenarios.

Graphing Functions

Graphing functions offers a visual interpretation that complements analytical solutions. For our task, graphing the position function \( s(t) = 30(16t - \frac{t^3}{3}) \) for \( 0 \leq t \leq 3 \) shows how the airplane’s position changes over time.

This graph will typically be a smooth curve due to the cubic term in the function, indicating how the airplane's position varies nonlinearly. Tools like graphing calculators or software can help make such plots easily, aiding in visualizing crucial points like starting and stopping distances or significant changes in acceleration.

Through graphing, one can better understand the relationship between time and distance, seeing the impact of the headwind and changes in velocity—offering a practical way to predict and analyze motion trends in realistic contexts.