Question

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

Answer 1

The rate at which both the plane changing distance is 522 mph

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) =
`(Diatance)/(Time)`

Or, Sn =
`(Dn)/(Tn)`

Or, Sn =
`(540)/(3)`

∴ 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) =
`(Diatance)/(Time)`

Or, Sw =
`(Dw)/(Tw)`

Or, Sw =
`(490)/(1)`

∴ Sw = 490 mph

So, The rate at which both plane changing =
`\sqrt{(180)^(2) + (490)^(2)}`

Or, The rate at which both plane changing =
`√(272,500)`

∴ The rate at which both plane changing = 522 mph

Hence The rate at which both the plane changing distance is 522 mph Answer