Calculating Cross-Country Ski Distance

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The discussion focuses on calculating the distance a cross-country skier is from the starting point after skiing in multiple directions defined by angles and distances. The skier's path is analyzed using vector components, where the angles are adjusted based on their directional references. The final position is determined by summing the x and y components of the vectors derived from the distances and angles. The Pythagorean theorem is then applied to compute the resultant distance from the starting point.

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  • Understanding of vector components in physics
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  • Familiarity with the Pythagorean theorem
  • Ability to manipulate angles in different directional contexts
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This discussion is beneficial for physics students, educators, and anyone interested in applying vector analysis to real-world scenarios, particularly in sports like cross-country skiing.

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A cross-country skier skis s1 in the direction theta1 west of south, then s2 in the direction theta2 north of east, and finally s3 in the direction theta3 south of west.

How far is the skier from the starting point? I drew it..and it looks really really weird, I haven't done physics in a couple years now, and I'm a bit rusty, any help on how to start this?
 
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I don't know how it could "look weird"- that would depend on what the numbers involved are! In general, it is a zig-zag path.

There are several different ways to do this. The way I would do it is to reduce to "components". Taking the positive x-component east and positive y-component n,
The components of a vector "of length s in direction θ (measured from due e) would be < r cos(θ), r sin(θ)>. The only problem is the angles.

If θ1 is "west of south" the angle measured from east is 270- θ1 and the vector is <s1cos(270-θ1), s1sin(270-θ1>.
If θ2 is measured "north of east", then it is already correct. The vector is <s2cos(θ2),s2sin(θ2)>.
If θ3 is measured "south of west", then we need to add 180 degrees to get the angle measured from east. The vector is <s3cos(θ3+ 180),s3sin(θ3+ 180)>

Adding those components, the vector from the intial position to the final position is
<s1cos(270-θ1)+s2cos(θ2)+s3cos(θ3+ 180), s1sin(270-θ1+s2sin(θ2)+s3sin(θ3+ 180)>

To find the distance from the initial position to the final position, use the Pythagorean formula: square the two components, add and then take the square root.

If you really are given only "s1, &theta;1", etc. then you will have a very complicated expression. If you are given specific numbers, substitute those first and then calculate.
 
i'm only given s1, theta1, s2, theta2, s3, theta3 :rolleyes:
 

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