Problem 21 - Calculus Early Transcendentals

Question

The velocity (in miles/hour) of an air-plane flying into a headwind is given by \(v(t)=30\left(16-t^{2}\right),\) for \(0 \leq t \leq 3 .\) Assume that \(s(0)=0\). 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 \(400 \mathrm{mi} / \mathrm{hr} ?\)

Step-by-Step Solution

Answer

Answer: When the airplane's velocity is 400 mi/hr, it has traveled approximately 807.57 miles. During the first 2 hours, it travels 880 miles.

Step 1: Find the Position Function s(t)

To find the position function s(t), we need to integrate the velocity function v(t) with respect to time t and add a constant if necessary to satisfy the initial condition s(0) = 0. Given, \(v(t) = 30\left(16 - t^2\right)\), we now integrate to get s(t): \(s(t) = \int v(t) dt = \int 30\left(16 - t^2\right) dt\) Now integrate: \(s(t) = 30 \int \left( 16 - t^2 \right) dt = 30 \left(16t - \frac{t^3}{3}\right) + C\) Here, C is the constant of integration. To find its value, use the initial condition s(0) = 0: \(0 = 30 \left( 16 \cdot 0 - \frac{0^3}{3} \right) + C\) So, C = 0. The position function s(t) is: \(s(t) = 30 \left( 16t - \frac{t^3}{3} \right)\)

Step 2: Distance Traveled in the First 2 Hours

To find the distance the airplane travels in the first 2 hours, we will evaluate the position function s(t) at t = 2: \(s(2) = 30 \left( 16 \cdot 2 - \frac{2^3}{3} \right) = 30 \left( 32 - \frac{8}{3} \right)\) Now simplify and calculate the result: \(s(2) = 30 \left( \frac{96 - 8}{3} \right) = 30 \left(\frac{88}{3}\right) = 880\) So, the airplane travels 880 miles in the first 2 hours.

Step 3: Distance Traveled When Velocity Reaches 400 mi/hr

We need to find the time t at which the velocity reaches 400 mi/hr. To do that, we set v(t) = 400 and solve for t: \(400 = 30\left(16 - t^2\right)\) Now, divide by 30: \(\frac{400}{30} = 16 - t^2\) Then, simplify: \(\frac{40}{3} = 16 - t^2\) Now, solve for t^2: \(t^2 = 16 - \frac{40}{3}\) Finally, solve for t (noting that t should be positive since it is time): \(t = \sqrt{16 - \frac{40}{3}} \approx 1.825\) Now, we need to find the distance traveled at this instant by evaluating the position function s(t) at t ≈ 1.825: \(s(1.825) \approx 30 \left( 16 \cdot 1.825 - \frac{1.825^3}{3} \right)\) After calculating this expression, we get: \(s(1.825) \approx 807.57\) So, the airplane travels approximately 807.57 miles at the instant its velocity reaches 400 mi/hr.

Key Concepts

Velocity and Integration

In calculus, the velocity of an object as a function of time can be used to determine its position function. This process involves integration, which is essentially the reverse operation of differentiation. By integrating the velocity function, we can find the position function, often denoted by \(s(t)\), that describes the object’s position at any given time \(t\).

To illustrate, consider the velocity function given in the problem: \(v(t) = 30(16 - t^2)\). To find the position function \(s(t)\), we perform the integration \(\int v(t) \, dt\).

• First, substitute the velocity function into the integral: \(\int 30(16 - t^2) \, dt\).
• Next, apply the integration to each term: \(30 \left(16t - \frac{t^3}{3}\right) + C\), where \(C\) is the constant of integration.

Determining the constant of integration is essential to accurately describing the position function, especially when initial conditions are provided. We'll delve into these initial conditions later. For now, remember that integrating velocity gives you the position function, revealing much about the motion of objects over time.

Distance Traveled

Calculating the distance an object travels over a period involves evaluating the position function at different points in time. This tells us how far the object has moved within that interval.

For example, we determine how far the airplane traveled in the first 2 hours by evaluating the position function, \(s(t)\), at \(t = 2\):

• Substitute \(t = 2\) into the position function: \(s(2) = 30 \left(16 \cdot 2 - \frac{2^3}{3}\right)\).
• Simplify the expression: \(s(2) = 30 \left(32 - \frac{8}{3}\right)\).
• Calculate to find the distance: \(s(2) = 880\) miles.

This means the airplane has traveled 880 miles in the first 2 hours. The same method applies to finding distance at any other specified time, reflecting the integral nature of the calculation process. Just plug in the time and solve the expression, and you’ll have the distance traveled in that interval.

Initial Conditions in Calculus

Understanding the role of initial conditions in calculus is crucial when dealing with integration. Initial conditions provide specific values at the beginning of the problem, which allows us to find any constants of integration accurately.

In this exercise, the initial condition is \(s(0) = 0\), meaning that at time \(t = 0\), the initial position of the airplane is 0 miles.

• When integrating the velocity function to find the position function, a constant \(C\) is typically added: \(s(t) = 30\left(16t - \frac{t^3}{3}\right) + C\).
• To find \(C\), substitute the initial condition into the position function: \(0 = 30\left(16 \cdot 0 - \frac{0^3}{3}\right) + C\).
• From this, we find \(C = 0\).

Thus, the fully determined position function is \(s(t) = 30\left(16t - \frac{t^3}{3}\right)\). These initial conditions ensure the solution accurately reflects the scenario described in the problem, establishing a starting point for analyzing further movement. In practice, setting up initial conditions lets us solve real-world problems with greater precision and context.