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On the map, let the +x-axis point east and the +y-axis north. An airplane flies at 880km/h...

Question:

On the map, let the +x-axis point east and the +y-axis north. An airplane flies at 880km/h northwestward direction (i.e., midway between north and west). Visualize the components of the first velocity, and when the plane flies due south at the same velocity.

Components of a Vector:

Let us consider a vector A is acting on the origin at an angle {eq}\theta {/eq} from the X-axis.

Then the component of the vector along the:

  • X-axis {eq}\Rightarrow \vec A cos\theta {/eq}
  • Y-axis {eq}\Rightarrow \vec A sin\theta {/eq}

Answer and Explanation:

Considering the +X-axis point east and the +Y-axis points north, an airplane flies at v = 880 km/h northwestward direction.


The angle of the North-West from the +X axis is {eq}\theta = 135^o. {/eq}


Thus we have the components of the velocity along the X and Y directions as

{eq}\begin{align*} v_x & = v \cos 135^o = 880 \cos 135^o \ km/h=-622.25 \ km/hr\\ v_y &= v \sin 135^o = 880 \sin 135^o \ km/h= 622.25 \ km/hr \end{align*} {/eq}


When the airplane travels due South (-Y axis) we have the angle measured from the +X axis is {eq}\theta = 270^o. {/eq}


Thus we have the components of the velocity along the X and Y directions as

{eq}\begin{align*} v_x & = v \cos 270^o = 880 \cos 270^o \ km/h=0\\ v_y &= v \sin 270^o = 880 \sin 270^o \ km/h= -880 \ km/hr \end{align*} {/eq}


Learn more about this topic:

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Vector Components: The Magnitude of a Vector

from UExcel Physics: Study Guide & Test Prep

Chapter 2 / Lesson 8
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