A plane leaves an airport traveling at 400 mph in the direction N 45° E. A wind is blowing at 40 mph in the direction N 45° W. What is the ground speed of the plane?
400 mph
402 mph
429 mph
440 mph

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ethanbennink3 ethanbennink3 · Answer 1 · 2021-02-23T23:32:16+01:00

Answer:

402 mph

Step-by-step explanation:

Draw a right triangle with 400 and 40 as the two legs.

You know this is a right angle because 45 degrees NE is perpendicular to 45 degrees NW. (imagine a compass)

So, the two side lengths are 400 and 40

Using the Pythagorean theorem, (a^2 + b^2 = c^2) you can see that the answer is 401.995, or 402

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ayoushivt ayoushivt · Answer 2 · 2022-07-14T12:41:13+02:00

The ground speed of the plane is 402 mph , Option B is the correct answer.

A right angle triangle is a triangle with one of its sides 90 degree.

It is given that

airplane is travelling at speed of 400mph in the direction N 45° E

wind is blowing at 40 mph in the direction N 45° W

The resultant of these two vectors will be the ground speed of the plane.

As they both are travelling in perpendicular directions ( NE is perpendicular to NW)

A right angled Triangle as attached in the image will be formed

Hypotenuse by Pythagoras theorem will give the ground speed of the plane

Hypotenuse = [tex]\sqrt { 40 ^2 + 400^2}[/tex]

Hypotenuse = 401.99 mph

Therefore the ground speed of the plane is 402 mph.

To know more about Right Angled Triangle

https://brainly.com/question/3770177

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