Question

The 4200 km trip from new york to san francisco takes 7 h flying against the wind, but only 6 h returning. How do you find the speed of the plane in still air and the wind speed?

Answer 1

Plane speed `=514 2/7" Km/h" "~~514.286` to 3 decimal places
Wind speed `=85 5/7" Km/h "~~85.714` to 3 decimal places

Explanation:

To solve these equations we need to manipulate so that we have just 1 unknown and its relationship to some values.

Assumption: the speed (velocity) of the wind is constant

Let the speed of the plane be `p`
Let the speed of the wind be `w`
Let time going be in hours be `t_g->7h`
Let the time returning in hours be `t_r->6h`

Note that speed = `("distance")/("time")`

So `"distance "="time "xx" speed"`

Consider going: relative to the ground, the actual speed is `p-w`
Consider returning: relative to the ground the actual speed is `p+w`
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`color(blue)("Determine the ratio of wind speed to plane speed")`
We need this to change things so that, by substitution, we end up with just 1 unknown.

So for going we have:
`" "4200=7xx(p-w)" ".........Equation(1)`

And for returning we have:
`" "4200=6xx(p+w)" "............Equation(2)`

Equation both to each other through distance

`7xx(p-w)=4200=6xx(p+w)`

`7xx(p-w)=6xx(p+w)`

`7p-7w=6p+6w`

Subtract `6p` from both sides

`7p-6p=0+6w`

`p=6w" "...........Equation(3)`
`w=p/6" "...........Equation(4)`
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`color(blue)("Determining the actual speeds")`

Using Equation(3) substitute for `p` in Equation(1) giving

`4200=7(p+w)" "->" "4200=7(6w+w)`
`" "4200=49w`

`w=4200/49=85.71428.... = 85 5/7" Km/h"`

Using Equation(4) substitute for `w` in Equation(1) giving

`4200=7(p+w)" "->" "4200=7(p+p/6)`
`" "4200=49/6 p`

`p=(6xx4200)/49=514.2857... =514 2/7" Km/h"`