Question

Two airplanes leave an airport at the same time. The velocity of the first airplane is 740 m/h at a heading of 26.1 ◦ . The velocity of the second is 640 m/h at a heading of 109◦ . How far apart are they after 1.8 h?

Answer 1

The airplanes are 1649.84m apart.

Why?

To calculate how far are the airplanes apart, we need to find the angle between their displacements, then, we can use the law of cosines to solve the problem.

$Angle=109\°-26.1\°=82.9\°$

Now that we know the angle, we can use the law of cosines to find the distance between the two airplanes:

$c^{2}=a^{2}+b^{2}-2*a*b*Cos(\alpha )$

Also, we need to remember the formula to calculate distance:

$distance=velocity*time$

Substituting we have:

$a=740\frac{m}{h}\\\\b=640\frac{m}{h}\\\\\alpha =82.9\°\\\\time=1.8h$

$c^{2}=(740\frac{m}{h}*1.8h)^{2}+(640\frac{m}{h}1.8h)^{2}-2*(740\frac{m}{h} *1.8h)*(640\frac{m}{h}*1.8h)*Cos(82.9\°)\\\\c=\sqrt{1774224m^{2} +1327104m^{2} -379324.03m^{2} }=1649.84m$

Hence, we have that the airplanes are 1649.84m apart.

Have a nice day!