Please help,
The displacement vector for a 15.0 second interval of a jet airplane's flight is (3850, −2430) m. (a) What is the magnitude of the average velocity? (b) At what angle, measured from the positive x axis, did the airplane fly during this time interval? Express the angle as a number between −180° and +180°.
Thanks a lot.
To find the magnitude of the average velocity, we can use the formula:
Average velocity = displacement / time
In this case, the displacement vector is (3850, -2430) m, and the time interval is 15.0 seconds.
(a) Magnitude of the average velocity:
To find the magnitude, we need to calculate the length of the displacement vector. We can use the Pythagorean theorem:
Magnitude = sqrt((x^2) + (y^2))
Here, x = 3850 and y = -2430. Plugging these values into the formula:
Magnitude = sqrt((3850^2) + (-2430^2))
Calculating this expression will give you the magnitude of the average velocity.
(b) Angle measured from the positive x-axis:
To find the angle, we can use trigonometry. The angle can be determined by taking the inverse tangent of the y-coordinate divided by the x-coordinate:
Angle = arctan(y / x)
In this case, x = 3850 and y = -2430. Plugging these values into the formula:
Angle = arctan(-2430 / 3850)
Calculating this expression will give you the angle in radians. To express it as a number between -180° and +180°, you can convert radians to degrees and ensure the result lies within the desired range.
I hope this explanation helps!