Question

Two airplanes A and B are flying horizontally at the same altitude so that the position of B is southwest of A and 20 km to west an 20 km to the south of A. Suppose plane A is flying due west at 16 km/min and plane B is flying due north at 21.3 km/min. What is the closest distance

Answer 1

At minute 1 Plane B would have travelled 21.3 kilometers north
Plane A would have travelled 16 kilometers west

Answer 2

find the hypotenuse of the triangle to calculate the shortest distance between the planes
Using the formula
\[a^2+b^2=c^2\]

Answer 3

it would be \[1.3^2+4^2=c^2\]

Answer 4

\[17.96 =c^2\]
\[\sqrt{17.96}=c\]
\[c=4.23792402008\]

Answer 5

To solve this problem, we can use the Pythagorean theorem. Let's call the distance between planes A and B "d".

First, we can find the time it takes for each plane to travel 20 km. For plane A, we know it's flying due west at 16 km/min, so it takes 20/16 = 1.25 minutes to travel 20 km. For plane B, we know it's flying due north at 21.3 km/min, so it takes 20/21.3 = 0.94 minutes to travel 20 km.

Next, we can find the distance between plane A and the point directly south of plane B. This distance is simply 20 km.

Finally, we can use the Pythagorean theorem to find the distance between plane B and the point directly west of plane A. We have one leg of the right triangle as 20 km (the distance between plane A and the point directly south of plane B), and the other leg as the distance plane B travels in 1.25 minutes (since that's how long it takes plane A to travel 20 km west). We can find this distance by multiplying the speed of plane B by the time it takes to travel 20 km, which gives us 21.3 * 1.25 = 26.63 km. Using the Pythagorean theorem, we have:

d^2 = 20^2 + 26.63^2
d^2 = 400 + 710.32
d^2 = 1110.32
d = 33.33 km (rounded to two decimal places)

Therefore, the closest distance between the two planes is approximately 33.33 km.