Question

6. An airplane is taking off headed due north with an air speed of 173 miles per hour at an angle of 18° relative to the horizontal. The wind is blowing with a velocity of 42 miles per hour at an angle of S47°E. Find a vector that represents the resultant velocity of the plane relative to the point of takeoff. Let i point east, j point north, and k point up. (Show work)

Answer 1

The resultant velocity vector of the airplane relative to the point of takeoff is found by summing the component vectors of the airplane's velocity and the wind's velocity. Using trigonometry, we calculate the components of both vectors and add them together to solve for the total resultant velocity vector.

Step-by-step explanation:

To find the resultant velocity of the airplane relative to the point of takeoff, we need to consider both the airplane's velocity vector and the wind's velocity vector. The airplane's airspeed is vector A with a magnitude of 173 mph directed due north (0°), which translates to A = 173j. For the wind vector B, its speed is 42 mph at an angle of S47°E. We have to break this into components using the angles provided. The angle of the wind vector relative to the due east direction (positive i-axis) is 47°, hence, this vector translates to B = 42cos(47)i - 42sin(47)j.

Now, we sum the components of the two vectors to find the resultant velocity, C.
C = A + B

C = (0i + 173j) + (42cos(47)i - 42sin(47)j)

C = (42cos(47)i + (173 - 42sin(47))j)

With this result, we obtain a vector that indicates the plane's speed and direction relative to the point of takeoff, accounting for wind velocity.