Respuesta :
Answer:
The rate at which both the plane changing distance is 522 mph
Step-by-step explanation:
Given as :
The west going plane travel distance (Dw) = 490 mile
The west going plane leave airport time at (Tw) = T hour
The north going plane travel distance (Dn) = 540 mile
The north going plane leave airport time at (Tn) = ( T + 2 ) hour
Let the Rate at which second plane changes distance = S mph
As per question the north going plane travel fro 3 hours
So, Speed of north going plane (Sn) = [tex]\frac{Diatance}{Time}[/tex]
Or, Sn = [tex]\frac{Dn}{Tn}[/tex]
Or, Sn = [tex]\frac{540}{3}[/tex]
∴ Sn = 180 mph
Now ,∵ North going plane travel for 3 hours so,
Tn = 3 h
Or, T + 2 = 3 h
So, T = 3 - 2 = 1 h
∴ The time cover by west gong plane = 1 h
So, Speed of west going plane (Sw) = [tex]\frac{Diatance}{Time}[/tex]
Or, Sw = [tex]\frac{Dw}{Tw}[/tex]
Or, Sw = [tex]\frac{490}{1}[/tex]
∴ Sw = 490 mph
So, The rate at which both plane changing = [tex]\sqrt{(180)^{2} + (490)^{2}}[/tex]
Or, The rate at which both plane changing = [tex]\sqrt{272,500}[/tex]
∴ The rate at which both plane changing = 522 mph
Hence The rate at which both the plane changing distance is 522 mph Answer
