Two airplanes A and B are flying horizontally at the same altitude, Plane B is southwest of plane A, 20km south and 20km west from A. If plane A is flying due west at 16 km/min and plane B is flying due north at 64/3 km/min. What will be their closest distance in km?

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The answer is, 426.67 km First we have to see the distance traveled by plane A, \[16 km \times20 \min=320 km\] Distance traveled by plane B, \[64/3km/\min \times 20 \min = 1280/3 km\] Now we have to find the shortest distance between them, \[320^{2}+1280/3^{2}\approx426.67 km\]

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Let plane A is at O
and plane B as shown.
distance covered by plane A in t minutes=16t
and distance covered by by plane B=64/3 t
PD=20-16t
CP=20-64/3t
distance CD=d
\[d=\sqrt{\left( 20-16t \right)^2+\left( 20-\frac{ 64 }{ 3}t \right)^2}\]
\~~
\[\frac{ ds }{ dt }=2(20-16t)(-16)+2(20-\frac{ 64 }{ 3 }t)(\frac{ -64 }{ 3})\]
\[=-640+512t-\frac{ 2560 }{ 3 }+\frac{ 8192 }{ 9 }t\]
\[=\frac{ 4608+8192 }{ 9 }t-\frac{ 1920+2560 }{ 3}\]
\[=\frac{ 12800 }{ 9 }t-\frac{ 4480 }{ 3}\]
\[\frac{ ds }{ dt}=0,gives\]
\[\frac{ 12800 }{ 9}t-\frac{ 4480 }{ 3}=0\]
\[t=\frac{ 4480 }{ 3 }\times \frac{ 9 }{ 12800}=\frac{ 28\times 3 }{ 80 }=\frac{ 21 }{ 20}~\min\]
\[\frac{ d^2s }{ dt^2}=\frac{ 12800 }{ 9}>0~at~t=\frac{ 21 }{ 20 }\]
Hence s is minimum at t=21/20
\~~~~
\~~~~
\[s=\left( \frac{ 64 }{ 20} \right)^2+\left( \frac{ -48 }{ 20 } \right)^2=\left( \frac{ 16 }{ 5 } \right)^2+\left( \frac{ -12 }{ 5} \right)^2\]
\[=\frac{ 256+144 }{ 25}\]
\[=\frac{ 400 }{ 25 }=16\]
\[d=\sqrt{s}=\sqrt{16}=4 ~km\]
shortest distance between planes = 4 km~~

Yeah it’s probably like a pretty big number or smth ion know why you even needa know that cause u ain’t gonna be a pilot

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