An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each oth
er. One airplane is 70 miles from the point and has a speed of 420 miles per hour. The other is 240 miles from the point and has a speed of 1440 miles per hour.
(a) At what rate is the distance between the planes changing? in mph?
(b) How much time does the controller have to get one of the airplanes on a different flight path?
(a) At what rate is the distance between the planes changing? in mph?
(b) How much time does the controller have to get one of the airplanes on a different flight path?
2 answers:
Answer:
(a) ðz/ðt = 1500miles/hr
(b) 10minutes
Step-by-step explanation:
Let x rep plane one
Let y rep plane two
ðx/ðt = 1440miles/hr = speed of plane one
ðy/ðt = 420miles/hr = speed of plane 2
ðz/ðt = ? = rate of change of distance btw two plane moving at right angle to meet at point o...
See attachment
Since x² + y² = z²... equation 1
So therefore (240² + 70²) ^½= z = 250 miles = distance between two planes
Differential rate of equation 1 gives
(2x* ðx/ðt ) + (2y* ðy/ðt) = 2z* ðz/ðt
Substituting to get ðz/ðt
(2*240*1440) + (2*70*420) = 2*250*ðz/ðt
750000 = 500 ðz/ðt
ðz/ðt = 1500miles/hr
B.
Since v = d/t where d = distance = 250miles
t = time = ? = time needed for controller to divert one of e planes , v = 1500miles/hr
t = d/v = 250/1500 = 10minutes
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Two similar pyramid have base area of 12.2 cm² and 16 cm².
surface area of the larger pyramid = 56 cm²
find out the surface area of the smaller pyramid
To proof =
Let us assume that the surface area of the smaller pyramid be x.
as surface area of the larger pyramid is 56 cm²
Two similar pyramid have base area of 12.2 cm² and 16 cm².
by using ratio and proportion
we have
ratio of the base area of the pyramids : ratio of the surface area of the pyramids
x = 12.2 ×56×
by solvingthe above terms
we get
x =42.7cm²
Hence the surface area of the smaller pyramid be 42.7cm²
Hence proved