Vectors - Find resultant displacement

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SUMMARY

The discussion focuses on calculating the resultant displacement of a spelunker who moves in multiple directions: 75 m north, 250 m east, 135 m at a 30° angle north of east, and 125 m south. The correct approach involves breaking down the vector components into East-West and North-South directions, using trigonometric functions to find the components of the 135 m vector. The final displacement is calculated using the Pythagorean theorem, yielding a magnitude of approximately 361.63 m, with the angle determined by arctan(y/x), adjusted to reflect the correct direction as North of East.

PREREQUISITES
  • Understanding of basic vector components and vector addition
  • Familiarity with trigonometric functions: sine and cosine
  • Knowledge of the Pythagorean theorem for calculating distances
  • Ability to use arctangent for angle determination in right triangles
NEXT STEPS
  • Practice vector decomposition using different angles and magnitudes
  • Learn how to apply the Pythagorean theorem in various contexts
  • Explore advanced vector operations, including cross and dot products
  • Study real-world applications of vectors in physics and engineering
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Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators looking to enhance their teaching methods in vector mathematics.

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Homework Statement



While exploring a cave, a spelunker starts at the entrance and moves the following distances. She goes 75 m north, 250 m east, 135 m at an angle 30° north of east, and 125 m south. Find the resultant displacement from the cave entrance.
Magnitude


Homework Equations



I do not quite understand vectors all that much. I am new to this and I do not know where to begin with that problem.

The Attempt at a Solution



I've tried drawing a picture but it doesn't seem right so I'm stuck there.
 
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Do you know about vector components and vector addition?
 
The only "hard" one is that "135 m at an angle 30° north of east". If you draw a picture of that, you should see a right triangle with hypotenuse of length 135 and angle 30° The "opposite side" (North) is 135 sin(30)= 135/2= 67.5 m. The "near side" (East) is 135 cos(30)= 135 sqrt(3)/2= 116.9 m

Now make a list of East-West and North-South distances treating "East" and "North" as positive, "West" and "South" as negative:
East-West North-South
0 75
250 0
67.5 117
0 -125
Finally add up each of those. If you like, once you have a total for "East-West" and "North-South" you can use the Pythagorean theorem to get the straight line distance and the arctan to get the angle so you can write the answer in the same form as the vectors given in the problem.
 
HallsofIvy said:
The only "hard" one is that "135 m at an angle 30° north of east". If you draw a picture of that, you should see a right triangle with hypotenuse of length 135 and angle 30° The "opposite side" (North) is 135 sin(30)= 135/2= 67.5 m. The "near side" (East) is 135 cos(30)= 135 sqrt(3)/2= 116.9 m

Now make a list of East-West and North-South distances treating "East" and "North" as positive, "West" and "South" as negative:
East-West North-South
0 75
250 0
67.5 117
0 -125
Finally add up each of those. If you like, once you have a total for "East-West" and "North-South" you can use the Pythagorean theorem to get the straight line distance and the arctan to get the angle so you can write the answer in the same form as the vectors given in the problem.
Thank you
 
Last edited:
I did what you said. I got 324.492 for the straight line distance and 78.08 degrees for the angle but when i input those values it said I was wrong. Any help on what I'm doing wrong?
 
How are you working out the angle? The resultant displacement can also be treated as a right angled triangle with the east-west and north-south components being the two non-hypotenuse sides. how would you then work the angle out?
 
I did it wrong but I ended up getting the displacement. But when I try to use trig to get the angle it comes out wrong.
 
If you show us what you're doing to get the angle then we can help.
 
Kurdt said:
If you show us what you're doing to get the angle then we can help.

Okay for I got 361.6275447 for the displacement. I know that this is a right triangle so I use arctan (y/x) y being 360.585526 and x being 27.43432879. Which I calculated when adding all the x-components and y-components. I did that but the angle did not work when I input it on the site where I do my work. The displacement however worked. The angle is supposed to be North of East which I don't understand.
 
  • #10
Your numbers seem slightly out for the components of the vectors. Your method for obtaining the angle is correct.
 
  • #11
Kurdt said:
Your numbers seem slightly out for the components of the vectors. Your method for obtaining the angle is correct.

I don't know what to do. I've tried doing it over and over again but it comes out incorrect.
 
  • #12
FalconKICK said:
I don't know what to do. I've tried doing it over and over again but it comes out incorrect.

Ok so you added the x (west-east) components: 250 + 135*cos(30)

and the y (south-north) components: 75 + 135*sin(30) - 125

So what do you get for each?
 
  • #13
I figured it out. For the angle we need North of East and what I was getting was South of East so what I did is subtract 90, since it's a 90 degree triangle, from the angle I was getting and I got the angle for North of East.
 

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