Find the ground speed of an airplane with air speed of 480km/hr and heading due east if a wind blowing south has a velocity of 40km/hr.
I did this:
plane: <480, 0>
wind: <0, -40>
And added them together to get <480, -40> and then used Pythagorean theorem to get a speed of about 481.664. Is this correct?
yep.
Well, well, well...looks like we have a flyboy in our midst, trying to calculate some speed! Bravo!
Now, let's take a closer look at your calculations. You added the vectors correctly, which is definitely a good start. The sum of the plane and wind vectors should indeed be <480, -40>. Kudos to you!
However, when it comes to finding the ground speed using the Pythagorean theorem, I'm sorry to burst your bubble, but it's not quite correct. Remember, we're dealing with vectors here, not just scalar quantities.
To find the ground speed, you need to find the magnitude of the resultant vector <480, -40>. So, grab your mathematical tools, and calculate the square root of...wait for it...(480^2 + (-40)^2). This should give you a ground speed of approximately 480.1 km/hr.
So, there you have it! The actual ground speed is about 480.1 km/hr. Keep soaring high with your math skills, my friend!
Yes, your approach is correct. To find the ground speed of the airplane, you need to consider the vector sum of the airspeed and the wind velocity.
You correctly represented the airspeed as a vector <480, 0> km/hr since the airplane is heading due east, which means it is going horizontally.
The wind velocity can be represented as a vector <0, -40> km/hr since it is blowing south, in the negative y direction.
To find the total velocity vector, you add the components of the airspeed and wind velocity vectors:
<480, 0> + <0, -40> = <480, -40> km/hr
Then, you can use the Pythagorean theorem to find the magnitude of the resulting vector:
Magnitude = sqrt((480^2) + (-40^2)) = sqrt(230,400 + 1,600) = sqrt(232,000) ≈ 481.664 km/hr
So, you correctly calculated the ground speed of the airplane as approximately 481.664 km/hr.
Your approach is correct, but there seems to be a small calculation error. Let's go through it step by step to find the correct answer.
First, let's represent the velocity vectors as you mentioned:
Plane's airspeed: <480, 0> km/hr (since it is heading due east)
Wind velocity: <0, -40> km/hr (since it is blowing south)
To find the ground speed, we need to add the vectors representing the plane's airspeed and the wind velocity. By adding the corresponding components, we get:
Ground speed = <480, 0> + <0, -40> = <480, -40> km/hr
Now, to find the magnitude of the ground speed, we can use the Pythagorean theorem. The magnitude of a vector is the square root of the sum of the squares of its components:
Magnitude of ground speed = √(480^2 + (-40)^2)
Calculating this, we get:
Magnitude of ground speed ≈ √(230,400 + 1,600)
≈ √232,000
≈ 481.67 km/hr (rounded to two decimal places)
So the correct ground speed of the airplane, with an airspeed of 480 km/hr and a wind blowing south at 40 km/hr, is approximately 481.67 km/hr.