Question

a commercial jet and a private airplane fly from Denver to phoenix. it takes the commercial jet 1.1 hours for the flight, and it takes the private airplane 1.8 hours. the speed of the commercial jet is 210 miles per hour faster than the speed of the private airplane. Find the speed of both airplanes​

Answer 1

Answer:

The speed of the commercial jet is $540mi/h$ while the speed of the private airplane is $330mi/h$

Step-by-step explanation:

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:

$t_{cj}=1.1h$

It takes the private airplane 1.8 hours for the flight, so:

$t_{pa}=1.8h$

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 $v_{cj}$ and the speed of the private airplane as $v_{pa}$, then:

$v_{cj}=v_{pa}+210$

From physics we know that:

$v=\frac{d}{t} \\ \\ Where: \\ \\ v: \ speed \\ \\ d: \ distance \\ \\ t: \ time$

Since the distance from Denver to phoenix is unique, then:

$d_{cj}=d_{pa}=d$

Thus, from the equation $v_{cj}=v_{pa}+210$ and given the relationship $v=\frac{d}{t}$ we have:

$v_{cj}=v_{pa}+210 \\ \\ \frac{d}{t_{cj}}=\frac{d}{t_{pa}}+210 \\ \\ \\ Plug \ in \ t_{cj}=1.1 \ and \ t_{pa}=1.8 \ then: \\ \\ \frac{d}{1.1}=\frac{d}{1.8}+210 \\ \\ Isolating \ d: \\ \\ d(\frac{1}{1.1}-\frac{1}{1.8})=210 \\ \\ \frac{35}{99}d=210 \\ \\ d=\frac{99\times 210}{35} \\ \\ d=594miles$

Finally, the speeds are:

$\bullet \ v_{cj}=\frac{d}{t_{cj}} \\ \\ v_{cj}=\frac{594}{1.1} \therefore \boxed{v_{cj}=540mi/h} \\ \\ \\ \bullet \ v_{pa}=\frac{d}{t_{pa}} \\ \\ v_{pa}=\frac{594}{1.8} \therefore \boxed{v_{pa}=330mi/h}$