One plane leaves the airport and travels west at 490 mph. A second plane leaves 2 hours later and travels north at 540 mph. At what rate is the distance between the two planes changing at when the second plane has been traveling for 3 hours

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WaywardDelaney WaywardDelaney · Answer 1 · 2019-11-15T07:18:51+01:00

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