# Thread: [SOLVED] velocity - component form

1. ## [SOLVED] velocity - component form

Here is the problem:
An airplane is flying on bearing of 194° at 460 mph. Find component form of velocity of airplane.

Here is what I have done...is it correct?
194° => 166°
v = (460cos166°)i + (460sin166°)j
v = <460cos166°, 460sin166°>
v = <-446.3, 111.29>

Thanks!

2. Originally Posted by live_laugh_luv27
Here is the problem:
An airplane is flying on bearing of 194° at 460 mph. Find component form of velocity of airplane.

Here is what I have done...is it correct?
194° => 166°
v = (460cos166°)i + (460sin166°)j
v = <460cos166°, 460sin166°>
v = <-446.3, 111.29>

Thanks!
I assume that the x-axis of the coordinate system is pointing East and the y-axis is pointing North. If so

$\vec v = 460 \cdot (\cos(104^\circ\ ,\ \sin(104^\circ)$

3. Shouldn't a bearing of 194° correspond to an angle of 256° (or -104°) in standard position?

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4. Originally Posted by yeongil
Shouldn't a bearing of 194° correspond to an angle of 256° (or -104°) in standard position?

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How do you know what angle 194° corresponds to?

5. Originally Posted by live_laugh_luv27
How do you know what angle 194° corresponds to?
The way I learned it in navigation 0° = North, 90° = East, 180° = South, and 270° = West. So 194° is 14° to the west of south. Now in the Cartesian plane, the terminal side would be in Quadrant III (look at earboth's diagram). The negative y-axis is now 270°, so 14° to the west of this axis means you subtract 14° from 270°, getting 256°.

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6. so, would the answer be, <-111.28, -446.336>, or <111.28, 446.336>?

7. Originally Posted by live_laugh_luv27
so, would the answer be, <-111.28, -446.336>, or <111.28, 446.336>?
<-111.28, -446.336>. In standard representation the head of this vector would be in Quadrant III. The head of the other vector you list is in Quadrant I.

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8. great, thanks!