Chapter 3.2, Problem 84PS

Solution

The velocity of a jet with a wing area of $5.5\times {10}^3$ square feet and a lift of $1.4\times {10}^7$ Newtons.

  $444$ feet per second

Given:

The velocity v in feet per second of a jet can be modeled by $v=8.8\sqrt{\frac{L}{A}}$ where A is the area of the wings in square feet and L is the lift in Newtons.

Concept Used:

• Power of a power property: ${\left(x^a\right)}^b=x^{a\cdot b}$
• Product of power property: $a^m\cdot a^n=a^{m+n}$
• Quotient of power property: $\frac{a^m}{a^n}=a^{m-n}$
• Product property of radicals: $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}$
• Quotient property of radicals: $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$
• The exponent form of the nth root of a number a is given by $\sqrt[n]{a}=a^{\frac{1}{n}}$

Calculation:

In order to find the velocity of the jet, substitute $A=5.5\times {10}^3\text{ and }L=1.4\times {10}^7$ into the given formula and simplify further as shown below,

  $\begin{array}{c}v=8.8\sqrt{\frac{L}{A}}\\ =8.8\sqrt{\frac{1.4\times {10}^7}{5.5\times {10}^3}}\\ =8.8\sqrt{\frac{1.4}{5.5}\times \frac{{10}^7}{{10}^3}}\\ =8.8\sqrt{\frac{1.4}{5.5}\times {10}^{7-3}}\\ =8.8\sqrt{\frac{1.4}{5.5}\times {10}^4}\\ =8.8\cdot \sqrt{\frac{1.4}{5.5}}\cdot \sqrt{{10}^4}\\ =8.8\cdot \sqrt{\frac{1.4}{5.5}}\cdot \sqrt{{\left({10}^2\right)}^2}\\ =8.8\cdot \sqrt{\frac{1.4}{5.5}}\cdot {10}^2\\ \approx 444\end{array}$

Therefore, the velocity of the jet is about $444$ feet per second.