# Homework Help: Physics-distance using vectors.

1. Sep 6, 2011

Physics--distance using vectors.

1. The problem statement, all variables and given/known data

An explorer in the dense jungles of equatorial Africa leaves his hut. He takes 40 steps at an angle 45 degrees north of east, then 83 steps at an angle 60 degrees north of west, then 50 steps due south. Assume his steps all have equal length. Save him from becoming hopelessly lost in the jungle by giving him the displacement, calculated using the method of components, that will return him to his hut.

What is the magnitude of the displacement that will return the explorer to his hut?

What is the direction of the displacement that will return the explorer to his hut?

2. Relevant equations

AxcosangleA =x-component
AysinangleA = y-component

3. The attempt at a solution

The problem is not the math, the problem is that I need help finding the angles I'm suppose to be measuring. I'm never sure which angle or how to find the measurement of half the angles.

Ax = (40.0)(cosign 45) = 28.28
Ay = (40.0)(sin 45)= 28.28

bx =(83.0)(cosign 60) = -41.5
by =(83.0)(sin 60) = 71.9

cx =(50.0)(cosign ?) =
cy=(50.0)(sin ?) =

Then I know how to find magnitude, but I don't know the angles. Thanks!

Last edited: Sep 6, 2011
2. Sep 6, 2011

### Staff: Mentor

Re: Physics--distance using vectors.

One way to keep track of the angles is to make a sketch of your coordinate system, labeling the axes with the directions. Then you can refer your directional instructions (like 45° North of East) to that diagram and "see" what reference direction (axis) is meant.

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3. Sep 6, 2011

Re: Physics--distance using vectors.

Would C's angle be 90 degrees then?

I got 100.2 steps north and -13.2 steps west.

To find the magnitude, I got an answer of 100.9 for magnitude of the displacement that will return the explorer to his hut, is that right?

Last edited: Sep 6, 2011
4. Sep 6, 2011

### Staff: Mentor

Re: Physics--distance using vectors.

You can assume a reference direction of East if you wish and take the angle for C to be -90°. But since you are being asked to work with components, a direction that is parallel to an axis of the coordinate system (like "due south") is already a "pure" component. So "50 steps due south" is a vector with components <0, -50>.

So your three outward-bound displacement vectors, in component form, are:

a = <28.28, 28.28>
b = <-41.5, 71.88>
c = <0, -50>

where all units are in "steps".

How did you calculate the total displacement?

5. Sep 6, 2011

Re: Physics--distance using vectors.

Ok, thanks for a clearer understanding on angles. I added the wrong vectors so I got a wrong displacement.

I added the components and obtained -13.22 and 50.16 by adding the x and y components.

The square root of (13.22)2+(50.16)2 is 51.87. Is that the correct overall displacement this young fellow needs to travel back to his hut?

6. Sep 6, 2011

### Staff: Mentor

Re: Physics--distance using vectors.

Yes, that distance looks good. Now, how will you go about finding the direction (angle) that he must travel?

7. Sep 6, 2011

Re: Physics--distance using vectors.

arctan-1 (50.16/13.22) = 75.24 or is it -75 degrees north of west?

-13.22 is on the x-axis to its to the left,(west) and 50.16 is positive so its north. Doesn't sound right though.

I'm pretty sure its east of south by looking at it, but I dont understand how to get that number without a coordinate plane.

Last edited: Sep 6, 2011
8. Sep 6, 2011

### Staff: Mentor

Re: Physics--distance using vectors.

You've got the coordinates of the vector that describes his position away from home, <-13.22, 50.16>. As you stated, that represents a location that's slightly west and north of his home position, along an angle of about 75° from his home. In order to head homeward then, he has to follow a direction 180° to that (i.e. reverse direction), which is slightly east and south. (It's easy to see if you draw it: just move the arrow on the end of the displacement vector to its other end!).

If you want to reverse the direction of a vector given in component form, simply change the signs of the components. The vector that he must follow homeward is then <13.22, -50.16>. So the angle will end up at about 75° south of east.

9. Sep 6, 2011

Re: Physics--distance using vectors.

Alright thanks! Makes sense, just need to work on directions and angles more than anything else.

Wouldn't it be -75 degrees because 50.15/-13.22 is -3.794 * arctan-1 is -75 degrees?

10. Sep 6, 2011

### Staff: Mentor

Re: Physics--distance using vectors.

Yup. (keeping in mind that "75°" is approximate).

You have to be a bit careful with the arctan function when you're looking for a direction angle because the function is designed to return a "principal value" assuming a restricted domain. For example, if both vector components happened to be negative you would expect an angle in the third quadrant (where both x and y values are negative). But the minus signs cancel when you do the division before taking the arctan! So it returns an angle in the first quadrant instead. If only one component has a negative sign, how is arctan supposed to know if it was associated with the y-component or the x-component? It's always a good idea to check which quadrant you expect the angle to be BEFORE you start punching buttons!

There is another function that's common in computation-oriented computer languages and math packages called atan2. It takes two arguments instead of one: the x and the y components. It will always return a result that is correct according to quadrant for the supplied values.

11. Sep 6, 2011

Re: Physics--distance using vectors.

Alright, thanks for clearing that up as well. I will check out that math package !

Also, I typed in both 75 degrees and -75 degrees(only 2 sig figs allowed) and they were both wrong.

The question was: What is the direction of the displacement that will return the explorer to his hut(in degrees)? I thought for sure it was 75 or -75 degrees, is there an issue?

Does anyone know why?

Last edited: Sep 6, 2011
12. Sep 6, 2011

### Staff: Mentor

Re: Physics--distance using vectors.

I see it as 75° south of east, or -75°.

13. Sep 6, 2011

Re: Physics--distance using vectors.

I don't know, says I have one try remaining. =(

Maybe I used the wrong numbers to calculate tan-1? Or maybe I need to use cotan-1?

I'm using masterphysics.com to type my answers in, so does typing "-" for negative not work?

14. Sep 6, 2011

### Staff: Mentor

Re: Physics--distance using vectors.

No, tan-1 is correct. tan-1(y/x), where x and y are the displacements along the X and Y directions. Was the question text that you posted all of the text? No additional information about how the answer is to be specified (such as with respect to a particular reference direction)?

15. Sep 6, 2011

Re: Physics--distance using vectors.

What is the direction of the displacement that will return the explorer to his hut?

\theta = _____ degrees East of South

Everything except that blank is provided, which is where the answer goes.

16. Sep 6, 2011

### Staff: Mentor

Re: Physics--distance using vectors.

Ah! Well, that explains it. They want an angle that references the "south" direction. Your calculated value was 75° south of east, taking east as the reference direction. What's the angle between that and the southerly direction?

What's x?

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17. Sep 6, 2011