The displacement vector for a 15.0 second interval of a jet airplane's flight is (3150, −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°
a) magitude velocity= 1/15 * sqrt(3150^2+2430^2)
b) what is arc tan -2430/3150
To find the answers to these questions, we need to break down the given information and use some basic concepts of vector operations.
(a) The magnitude of the average velocity is a measure of how fast an object moves on average during a certain time interval. It is calculated using the formula:
Magnitude = (Total displacement) / (Total time taken)
In this case, the total displacement is given as (3150, -2430) m, which means the airplane moved 3150 meters in the positive x-direction and -2430 meters in the negative y-direction. The total time taken is given as 15.0 seconds.
To calculate magnitude, we can use the Pythagorean theorem, which states that the magnitude of a vector with components (x, y) is given by √(x^2 + y^2).
Using this formula, we can find the magnitude of the average velocity by plugging in the values:
Magnitude = √((3150)^2 + (-2430)^2) m
(b) To find the angle at which the airplane flew during this time interval, we can use trigonometry. The angle can be calculated using the formula:
Angle = arctan(y/x)
In this case, we have the x-component as 3150 m and the y-component as -2430 m. Plugging these values into the arctan function, we can find the angle.
So, to summarize:
(a) Magnitude of the average velocity = √((3150)^2 + (-2430)^2) m
(b) Angle = arctan((-2430) / 3150)