Answer:
The rate of the jet in still air is 852 miles per hour. The rate of the wind is 85 miles per hour.
Step-by-step explanation:
Let suppose that jet travels uniformly, that is, at constant speed, the expressions for its travels against the wind and with the wind are, respectively:
Against the wind
[tex]v -u = \frac{\Delta x_{1}}{\Delta t_{1}}[/tex]
With the wind
[tex]v +u = \frac{\Delta x_{2}}{\Delta t_{2}}[/tex]
Where:
[tex]v[/tex] - Speed of the jet in still air, measured in miles per hour.
[tex]u[/tex] - Speed of wind, measured in miles per hour.
[tex]\Delta x_{1}[/tex], [tex]\Delta x_{2}[/tex] - Distances travelled by jet against the wind and with the wind, measured in miles.
[tex]\Delta t_{1}[/tex], [tex]\Delta t_{2}[/tex] - Times against the wind and with the wind, measured in hours.
By adding both expressions:
[tex]2\cdot v = \frac{\Delta x_{1}}{\Delta t_{1}}+\frac{\Delta x_{2}}{\Delta t_{2}}[/tex]
[tex]v = \frac{1}{2}\cdot \left(\frac{\Delta x_{1}}{\Delta t_{1}} + \frac{\Delta x_{2}}{\Delta t_{2}} \right)[/tex]
Given that [tex]\Delta x_{1} = 2301\,mi[/tex], [tex]\Delta t_{1} = 3\,h[/tex], [tex]\Delta x_{2} = 2811\,mi[/tex] and [tex]\Delta t_{2} = 3\,h[/tex], the speed of the jet is:
[tex]v = \frac{1}{2}\cdot \left(\frac{2301\,mi}{3\,h}+\frac{2811\,mi}{3\,h} \right)[/tex]
[tex]v = 852\,\frac{mi}{h}[/tex]
The rate of the jet in still air is 852 miles per hour.
Lastly, the rate of the wind is:
[tex]u = \frac{\Delta x_{2}}{\Delta t_{2}}-v[/tex]
[tex]u = \frac{2811\,mi}{3\,h}-852\,\frac{mi}{h}[/tex]
[tex]u = 85\,\frac{mi}{h}[/tex]
The rate of the wind is 85 miles per hour.
