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Chapter 2: Problem 72

Find the point of no return for an airport runway of \(1.50 \mathrm{mi}\) in length if a jet plane can accelerate at \(10.0 \mathrm{ft} / \mathrm{s}^{2}\) and decelerate at \(7.00 \mathrm{ft} / \mathrm{s}^{2} .\) The point of no return occurs when the pilot can no longer abort the takeoff without running out of runway. What length of time is available from the start of the motion in which to decide on a course of action?

Short Answer

Expert verified
Answer: The point of no return occurs when the sum of the distances traveled for acceleration and deceleration is equal to the length of the runway, which is calculated using the equation \(s_1 + \left(\frac{20}{14}s_1\right) = 1.50 * 5280\). The total time available to decide on a course of action is the sum of the times during acceleration and deceleration, represented as \(t = t_1 + t_2\).

Step by step solution

01

Convert the runway length to feet

To ensure consistency in units throughout the problem, we'll convert the runway length to feet.1 mi = 5280 ft. So, 1.50 mi = 1.50 * 5280 ft.
02

Find the distance traveled during acceleration

We'll use the kinematic equation \(v^2 = u^2 + 2as\) to find the distance traveled during acceleration, where \(v\) is the final velocity, \(u\) is the initial velocity (0 ft/s), \(a\) is the acceleration (10.0 ft/s²), and \(s\) is the distance traveled. We have \(v^2 = 2(10.0s)\).
03

Find the distance traveled during deceleration

Again, we'll use the kinematic equation \(v^2 = u^2 + 2as\) for deceleration, where \(v\) is the final velocity (0 ft/s), \(u\) is the initial velocity, \(a\) is the deceleration (-7.00 ft/s²), and \(s\) is the distance traveled. We have \(u^2 = 2(7.00s)\).
04

Solve for the initial velocity u

Notice that the plane's final velocity under acceleration is the same as its initial velocity during deceleration. Therefore, we can set the two equations equal to each other: \(2(10.0s_1) = 2(7.00s_2)\). Then, solve for \(u\): \(u = \sqrt{20s_1} = \sqrt{14s_2}\).
05

Find the point of no return

The point of no return occurs when the sum of the distances traveled for acceleration and deceleration is equal to the length of the runway. So, we have \((s_1 + s_2) = 1.50 * 5280\). Substitute \(u\) from the previous step and solve for \(s_1\): \(\frac{\sqrt{20s_1}}{\sqrt{14}} = \sqrt{s_2}\). Now, the equation is: \(s_1 + \left(\frac{20}{14}s_1\right) = 1.50 * 5280\).
06

Calculate the length of time available to decide on a course of action

To find the total time available, we'll use the equation \(s = ut + \frac{1}{2} at^2\) for both acceleration and deceleration. During acceleration, \(s = s_1\), \(u=0\), \(a= 10.0\), and the time is \(t_1\). During deceleration, \(s = s_2\), \(u = \sqrt{20s_1}\), \(a= -7.00\), and the time is \(t_2\). After finding \(t_1\) and \(t_2\), the sum of both times will be the total time available to decide on a course of action: \(t = t_1 + t_2\).

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