a plane flying at 90 degrees and 100 ms is blown toward 180 degrees at 50 ms. What is the resultant velocity and direction?
I see right off the angle will satisfy this equation based on velocities
TanTheta= 50/100 where theta is S of E.
To find the resultant velocity and direction of the plane, you need to use vector addition.
First, let's break down the initial velocity into its horizontal and vertical components. The plane's initial velocity at 90 degrees can be broken down into Vx = 0 m/s (horizontal component) and Vy = 100 m/s (vertical component).
Next, break down the velocity of the wind. The wind is blowing towards 180 degrees with a velocity of 50 m/s. Again, this can be broken down into horizontal and vertical components. The wind's velocity can be represented as Vwx = -50 m/s (horizontal component, negative because it's in the opposite direction) and Vwy = 0 m/s (vertical component, no vertical component).
Now, add the respective components together to find the resultant velocity of the plane.
Horizontal component: Vx + Vwx = 0 m/s + (-50 m/s) = -50 m/s
Vertical component: Vy + Vwy = 100 m/s + 0 m/s = 100 m/s
Finally, use the Pythagorean theorem to find the magnitude of the resultant velocity:
Resultant velocity = √((-50 m/s)^2 + (100 m/s)^2) = √(2500 m^2/s^2 + 10000 m^2/s^2) = √12500 m^2/s^2 ≈ 111.8 m/s
To find the direction of the resultant velocity, you can use the inverse tangent function:
Direction = arctan(Vy / Vx) = arctan(100 m/s / -50 m/s) = arctan(-2) ≈ -63.4 degrees (in the fourth quadrant, counterclockwise from the positive x-axis).
Therefore, the resultant velocity of the plane is approximately 111.8 m/s at an angle of -63.4 degrees (counterclockwise from the positive x-axis).