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, you need to divide the displacement vector by the time interval. The displacement vector gives the change in position of the airplane, and dividing it by the time interval gives you the average velocity.
(a) To find the magnitude of the average velocity, you can use the formula:
Magnitude of average velocity = magnitude of displacement vector / time interval
In this case, the magnitude of the displacement vector is:
magnitude of displacement vector = sqrt((3850)^2 + (-2430)^2)
Simplifying this gives:
magnitude of displacement vector = sqrt(14822500 + 5904900)
= sqrt(20727400)
= 4549.68 m
So, the magnitude of the average velocity is 4549.68 m/15 s = 303.31 m/s.
(b) To find the angle, you can use the inverse tangent function. The angle can be calculated as:
angle = arctan(y-component / x-component)
In this case, the y-component is -2430 m and the x-component is 3850 m. Substituting these values into the formula gives:
angle = arctan(-2430 / 3850)
= -0.573 radians
Now, you need to convert this angle from radians to degrees. Since the angle is between -180° and +180°, you can convert it using the formula:
angle in degrees = angle in radians * (180° / pi)
Substituting the value of the angle in radians gives:
angle in degrees = -0.573 * (180° / pi)
= -32.82°
Therefore, the airplane flew at an angle of approximately -32.82° measured from the positive x-axis.