Vector components question

  • Thread starter Coco12
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Homework Statement



A cyclist head due west on a straight road at 5.6m/s. A northeast wind is blowing at 10m/s. What is the effective speed of the tailwind?(resultant)

Homework Equations


Cos 45 10
Sin 45 10



The Attempt at a Solution


Basically I broke it down into its x and y components. Then added them together to the Rx and Ry and used Pythagorean to find the resultant. I just have one question though: when adding the x components for the wind and the cyclist speed. Would the Rx be -5.6m/s + 7.1( 7.1 is derived from cos 45 degrees *10) = 1.5? Or would the 5.6 be positive?
 

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  • #2
CAF123
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Would the Rx be -5.6m/s + 7.1( 7.1 is derived from cos 45 degrees *10) = 1.5? Or would the 5.6 be positive?
To help you answer your question, it might help to write out explicit expressions for the vectors representing the cyclist and the wind velocities.
 
  • #3
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To help you answer your question, it might help to write out explicit expressions for the vectors representing the cyclist and the wind velocities.
I did. Rx=ax+bx
Ry=ay +by

I'm just wondering since it's 5.6 m/s due west which on a Cartesian plane would be a negative x whether I would incorporate the negative when trying to find rx
 
  • #4
CAF123
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I did. Rx=ax+bx
Ry=ay +by

I'm just wondering since it's 5.6 m/s due west which on a Cartesian plane would be a negative x whether I would incorporate the negative when trying to find rx
Yes, you can represent the velocity vector of the cyclist as ##\vec{C} = -5.6 \hat{x}## and that of the wind as ##\vec{W} = (10 \cos 45)\hat{x} + (10 \sin 45) \hat{y}##. Now, as you said, just add components to get the resultant.

From a more conceptual point of view, imagine a game of tug of war. Person A pulls to left with force 5.6N and person B to right with force 10cos45 ≈ 7.1 N. The person pulling to right wins, but only marginally. (winning by 7.1 - 5.6, not 7.1 + 5.6)
 
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