For the given problems, both the planes are 1486 m apart after 2.4 h.
Explanation:
Let us consider two airplanes as A1 and A2. We know that the
[tex]\text { Distance covered in a time }=\text { velocity of airplanes } \times \text { given time }[/tex]
The distance covered by airplane A1 is
[tex]\text { Distance covered by } A 1=750 \times 2.4=1800 \mathrm{m}[/tex]
Similarly, the distance covered by airplane A2 is
[tex]\text { Distance covered by } A 2=620 \times 2.4=1488 \mathrm{m}[/tex]
As the airplanes take off at an angle from the base, so they will be exhibiting X and Y components of displacements.
X components of both the airplanes
= [tex](1800 \times \sin 51.3)-(1488 \times \sin 163)=1404.77-435.04=969.7[/tex]
Y components of both the airplanes
= [tex]=(1800 \times \cos 51.3)-(1488 \times \cos 163)=1125.43+0.956=1126[/tex]
The net displacement after 2.4 hours will be equal to the distance at which both planes are airplanes to each other at 2.4 hours.
[tex]\text { Distance apart }=\sqrt{X^{2}+Y^{2}}[/tex]
Therefore, Net displacement is
[tex]=\sqrt{(969.7)^{2}+(1126)^{2}}=\sqrt{940318.09+1267876}=\sqrt{2208194.09}=1485.99 \mathrm{m}[/tex]
So, both the planes are 1486 m apart from each other after 2.4 hours.
