Intro to physics vector problem

  • Thread starter lizette
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  • #1
lizette
Hi again,

This is the question:

An explorer is caught in a whiteout while returning to base camp. He was supposed to travel due north 5.6 km, but when the snow clears, he discovers that he is actually 7.8km at 50 degrees north of due east.

a)How far and b) in what direction must he now travel to reach base camp?

When I drew the sketch, it was a triangle. I think that I have the two values for two sides of the triangle: 5.6 km and 7.8 km. To determine how far, I think that I have to figure out the length of that other side.

90 degrees - 50 degrees = 40 degrees

I used 40deg as my angle to solve for the missing length of the triangle (a line connecting the location of where the explorer was lost to the location of the base camp).

So,

cos 40 = 5.6/d , where d is the distance
d = 7.31 km

Am I doing this correctly? Because I tried to do:
cos 50 = X/7.8 to see if I would get the distance from the origin to the base camp, but instead of getting 5.6 km I got 5.0 km, leading me to believe that my method was wrong.

Also, when they ask for direction, can I just say that the explorer has to move west? or should I include the degrees west he/she must travel?


Thanks.
 

Answers and Replies

  • #2
enigma
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Hi lizette!

Welcome to the forums.

Originally posted by lizette

An explorer is caught in a whiteout while returning to base camp. He was supposed to travel due north 5.6 km, but when the snow clears, he discovers that he is actually 7.8km at 50 degrees north of due east.

a)How far and b) in what direction must he now travel to reach base camp?

When I drew the sketch, it was a triangle. I think that I have the two values for two sides of the triangle: 5.6 km and 7.8 km. To determine how far, I think that I have to figure out the length of that other side.
You've got the right goal, but you're going about getting there wrong.

90 degrees - 50 degrees = 40 degrees

I used 40deg as my angle to solve for the missing length of the triangle
This only works if the angle at the camp is 90 degrees. If you draw the picture to scale, you'll see that it isn't.

This is the way I'd go about it:

Step 0) Define a coordinate system and draw a picture (which you did)

We'll call east the +X direction and north the +Y direction

Step 1) Break everything into components.

For the desired path, the explorer wanted to go 0km east, and 5.6km north, so the desired path vector Vd is ( 0 , 5.6 ) km.

His actual path was off. He travelled 7.8km, 50 degrees north of east. If you split that into X and Y components using Va sin 50 for Y and Va cos 50 for X, you'll get the right X and Y components (convince yourself that those are the correct sin's and cos's to get the right components).

You now have two (X,Y) vectors. If you subtract the actual position from the desired position, you'll have the correcting path vector. Using that path vector, you can find the magnitude (the distance) by using the Pythagorean theorem on the two components, and the angle by finding the arctan of the rise over the run (remember to check quadrants with the atan... you can be -and will be with this problem- in the wrong quadrant if you use the answer without correcting for quad.)

Hope that helps,
 
  • #3
HallsofIvy
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By the way, you quote the problem as saying "he discovers that he is actually 7.8km at 50 degrees north of due east."

That seems ambiguous to me. 7.8km and 50 degrees north of due east of what??? My first tendency was to interpret that as "from camp" but then the problem is trivial. I think the problem means that the explorer had TRAVELED 7.8 km, 50 degrees north of due east (instead of due north) so that is his position relative to his original position instead of relative to camp. As enigma said, draw a picture. You should notice that "50 degrees north of due east" is the same as "40 degrees east of due north" and since his camp is due north, one of the angles in your triangle will be 40 degrees exactly as you said. That was correct. When enigma noted "this only works for right triangles" he was referring to your "cos 40 = 5.6/d ".
You will need to use the cosine or sine laws for general triangles.
 

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