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
Answers
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
The ground speed of the plane is 402 mph , Option B is the correct answer.
What is a Right Angled Triangle ?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
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