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Flying against the wind, an airplane travels 7210 kilometers in 7 hours. Flying with the wind, the same plane travels 4050 kilometers in 3 hours. What is the rate of the plane in still air and what is the rate of the wind?

Respuesta :

[tex]\begin{gathered} \text{distance = d} \\ \text{speed (rate) = v} \\ \text{time = t} \\ \text{airplane = a} \\ \text{wind = w} \end{gathered}[/tex][tex]\begin{gathered} \text{When the airplane travels aigains the wind} \\ d=v_a\cdot t-v_w\cdot t \\ 7210=7v_a-7v_w \end{gathered}[/tex][tex]\begin{gathered} \text{When the airplane travels with the wind} \\ d=v_a\cdot t+v_w\cdot t \\ 4050=3v_a+3v_w \end{gathered}[/tex]

Now we have 2 equations and 2 variables (va and vw). We can isolate a variable in one Equation and substitue in the other, as follow:

[tex]\begin{gathered} 7210=7v_a-7v_{w_{}_{}} \\ 4050=3v_a+3v_w \\ \text{Isolating va in the first equation: } \\ 7v_a=7210+7v_{w_{}} \\ v_a=\frac{7210+7v_{w_{}}}{7}=1030+v_{w_{}} \\ \text{Substituting in the second equation: } \\ 4050=3v_a+3v_w \\ 4050=3(1030+v_{w_{}})+3v_w \\ 4050=3090+3v_{w_{}}+3v_w \\ 4050=3090+6v_w \\ 4050-3090=6v_w \\ 960=6v_w \\ v_w=\frac{960}{6}=160\text{ km/h} \\ v_a=1030+v_{w_{}}=1030+160 \\ v_a=1190\text{ km/h} \end{gathered}[/tex]

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