The speed of the commercial jet is [tex]540mi/h[/tex] while the speed of the private airplane is [tex]330mi/h[/tex]
Let's name the commercial jet as cj and private airplane as pa, so we know the following:
It takes the commercial jet 1.1 hours for the flight, so:
[tex]t_{cj}=1.1h[/tex]
It takes the private airplane 1.8 hours for the flight, so:
[tex]t_{pa}=1.8h[/tex]
The speed of the commercial jet is 210 miles per hour faster than the speed of the private airplane:
Let's name the speed of the commercial jet as [tex]v_{cj}[/tex] and the speed of the private airplane as [tex]v_{pa}[/tex], then:
[tex]v_{cj}=v_{pa}+210[/tex]
From physics we know that:
[tex]v=\frac{d}{t} \\ \\ Where: \\ \\ v: \ speed \\ \\ d: \ distance \\ \\ t: \ time[/tex]
Since the distance from Denver to phoenix is unique, then:
[tex]d_{cj}=d_{pa}=d[/tex]
Thus, from the equation [tex]v_{cj}=v_{pa}+210[/tex] and given the relationship [tex]v=\frac{d}{t}[/tex] we have:
[tex]v_{cj}=v_{pa}+210 \\ \\ \frac{d}{t_{cj}}=\frac{d}{t_{pa}}+210 \\ \\ \\ Plug \ in \ t_{cj}=1.1 \ and \ t_{pa}=1.8 \ then: \\ \\ \frac{d}{1.1}=\frac{d}{1.8}+210 \\ \\ Isolating \ d: \\ \\ d(\frac{1}{1.1}-\frac{1}{1.8})=210 \\ \\ \frac{35}{99}d=210 \\ \\ d=\frac{99\times 210}{35} \\ \\ d=594miles[/tex]
Finally, the speeds are:
[tex]\bullet \ v_{cj}=\frac{d}{t_{cj}} \\ \\ v_{cj}=\frac{594}{1.1} \therefore \boxed{v_{cj}=540mi/h} \\ \\ \\ \bullet \ v_{pa}=\frac{d}{t_{pa}} \\ \\ v_{pa}=\frac{594}{1.8} \therefore \boxed{v_{pa}=330mi/h}[/tex]
