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Mathematics 22 Online
Ellese:

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?

mariachiguy:

|dw:1718466769642:dw|

Alexis1415:

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, \[320^{2}+1280/3^{2}\approx426.67 km\]

surjithayer:

|dw:1718500354233:dw| 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

carry:

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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