Answer:
[tex]v_1=800\ Km/h\\v_2=600 \ Km/h[/tex]
Step-by-step explanation:
Constant Speed Motion
Let's assume the speeds of both planes are v1 and v2 respectively, the plane 2 slower than the plane 1, thus
[tex]v_1-v_2=200[/tex]
Solving for v1
[tex]v_1=200+v_2\ \ \text{ .................[1]}[/tex]
The time the faster plane took to arrive is
[tex]\displaystyle t_1=\frac{x}{v_1}[/tex]
The time the slower plane took to arrive is
[tex]\displaystyle t_2=\frac{x}{v_2}[/tex]
We know that
[tex]t_2-t_1=1.5[/tex]
Or, equivalently
[tex]\displaystyle \displaystyle \frac{x}{v_2}-\frac{x}{v_1}=1.5[/tex]
Using x=3600 and replacing [1]
[tex]\displaystyle \displaystyle \frac{3600}{v_2}-\frac{3600}{v_2+200}=1.5[/tex]
Operating
[tex]3600(v_2+200)-3600v_2=1.5v_2(v_2+200)[/tex]
Operating and simplifying
[tex]v_2^2+200v_2-480000=0[/tex]
Solving for v2:
[tex]v_2=600,\ v_2=-800[/tex]
We take the positive solution
[tex]\boxed{v_2=600 \ Km/h}[/tex]
Thus
[tex]v_1=600+200=800[/tex]
[tex]\boxed{v_1=800\ Km/h}[/tex]
