• vaironl
In summary, the student is having trouble with vector diagrams and is having more difficulty with vectors than with calculations (trigonometry). The student is struggling to identify which angles to take the sine and cosines for when working with vectors, which is causing confusion. The student is given information sufficient to get the distance from point B to D, but is having difficulty doing so. The student is advised to use the component method to solve the problem, and to use normal vector addition to get the distance from point B to D.

#### vaironl

Hello All,

I'm taking a university physics course and while dealing with the introduction to vectors I believe I'm having more trouble at drawing diagrams than with the calculations (trigonometry) and therefore not having a clear diagram is causing some confusion.

The problem is as follows: Oasis B is located 25 km east from oasis A. A camel starts from oasis A and travels 19.6 km in the direction 14.7 degrees south of east. Then it walks 8 km directly to the north. How far is the camel from oasis B?

My plan is the following:

Add the total distance the camel traveled and label it as a vector (Vector R).

Then subtract the distance the camel traveled (Vector R) from oasis a to b (Vector D).

This results in the final vector F. F = D-R

I’m attaching images of my attempt, but to summarize my result I found that the camel is located Vector F: (6.05km, -3.03 km) at an angle of 333 degrees from the horizon/east.

Solution:

This actually includes the first diagram attempt, which my instructor pointed out as incorrect since I scribbled too much detail into it (huge image, not sure if it will load direct link just in case http://i.imgur.com/0jG2Yja.png?1): http://1. Homework Statement Hello All, I'm taking a university physics course and in dealing with the introduction to vectors I believe I'm having more trouble at drawing diagrams than with the calculations (trigonometry) and therefore not having a clear diagram is causing some confusion. The problem is as follows: Oasis B is located 25 km east from oasis A. A camel starts from oasis A and travels 19.6 km in the direction 14.7 degrees south of east. Then it walks 8 km directly to the north. How far is the camel from oasis B? My plan is the following: Add the total distance the camel traveled and label it as a vector (Vector R). Then subtract the distance the camel traveled (Vector R) from oasis a to b (Vector D). This results in the final vector F. I’m attaching images of my attempted solutions, but to summarize my result I found that the camel is located Vector F: (6.05km, -3.03 km) at an angle of 333 degrees from the horizon/east. Solution: This actually includes the first diagram attempt which my instructor pointed out as incorrect since I scribbled too much detail into it. [ATTACH=full]199655[/ATTACH]

#### Attachments

• nOZ2cSs.png
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You should be resolving the vectors into components in the x and y directions. Is that what you did? What is your question?

Chet

You don't need a precise diagram if you just draw a sketch and use the cosine and sine rules.
Breaking down into components can also help - though this method can be prone to rounding errors... I suspect it will be easiest in this case though.

Looking at your diagrams and working - it looks like you had a go using the component method.
Your main trouble seems to have been identifying which angles to take the sine and cosines for.
I'll have a closer look and see if I can figure out how to hep best - however: it will be best for you to redo the working, typing it out here rather than posting a scan.

• Chestermiller
OK - let's make some points.
point A is the Oasis A, point B is Oasis B.

It helps to be careful with notation when you are just learning - I'm going to try to give you a simple, formal, way to do this below:

The vector pointing from point P to point Q I'll write like this:
$$\vec r_{PQ} = (\text{km East}, \text{km North})$$
...so the vector pointing from A to B is ##\vec r_{AB} = (25,0)##, and the vector pointing from B to A is ##r_{BA}=(-25,0)=-\vec r_{AB}##

(Notice how I have not assumed that anyone of the points is at the origin of any coordinate system - though that can help.)

The camel travels from point A to point C, and then to point D. The two-leg nature of the journey is the second complication, so when you sketch this out, you should draw in the labels clearly. (It was quite difficult to read your working.)

You are given ##\vec r_{CD} = (0,8)## and information sufficient to get ##\vec r_{AC}##
You have been asked to find ##|\vec r_{BD}|## (the distance from B to D is the length of the vector from B to D).

Now the problem has been translated into maths, it should be easier to talk about.
The component method looks good for this:

Normal vector addition should show you... $$\vec r_{AB}+\vec r_{BD}=\vec r_{AD}$$ ... so you just need the expression for ##\vec r_{AD}##.
You should already know how to do that.

That should give you someplace to start from.

Last edited:
Chestermiller said:
You should be resolving the vectors into components in the x and y directions. Is that what you did? What is your question?

Chet

Hello,

What I attempted to do here is to break each of the vectors into individual components. The sheet I scanned, which isn't visible in the post has all of the work done, but as suggested I'll try my best to type it in.

Simon Bridge said:
OK - let's make some points.
point A is the Oasis A, point B is Oasis B.

It helps to be careful with notation when you are just learning - I'm going to try to give you a simple, formal, way to do this below:

The vector pointing from point P to point Q I'll write like this:
$$\vec r_{PQ} = (X\text{km East}, Y\text{km North})$$
...so the vector pointing from A to B is ##\vec r_{AB} = (25,0)##, and the vector pointing from B to A is ##r_{BA}=(-25,0)=-\vec r_{AB}##

(Notice how I have not assumed that anyone of the points is at the origin of any coordinate system - though that can help.)

The camel travels from point A to point C, and then to point D. The two-leg nature of the journey is the second complication, so when you sketch this out, you should draw in the labels clearly. (It was quite difficult to read your working.)

You are given ##\vec r_{CD} = (0,8)## and information sufficient to get ##\vec r_{AC}##
You have been asked to find ##|\vec r_{BD}|## (the distance from B to D is the length of the vector from B to D).

Now the problem has been translated into maths, it should be easier to talk about.
The component method looks good for this:

Normal vector addition should show you... $$\vec r_{AB}+\vec r_{BD}=\vec r_{AD}$$ ... so you just need the expression for ##\vec r_{AD}##.
You should already know how to do that.

That should give you someplace to start from.

Thanks for the detailed level of explanation, Mr. Simon. I first want to apologize for a bad quality scan and messy work.

I'll try to rewrite my work here.

Reinstating my thoughts I added the vectors that represented the total distance traveled by the camel: ##\vec r_{AD} = \vec r_{AC} + \vec r_{CD}##
I then subtracted the total distance traveled by the camel from the distance of Oasis A to Oasis b: ##\vec r_{AB} - \vec r_{AD} ##
This results in the distance the camel is from Oasis B.

Work:
Distance traveled by camel
##\vec r_{AD} = \vec r_{AC} + \vec r_{CD}##

To obtain the x component of ##\vec r_{AC}## I multiplied the magnitude of the distance traveled by the angle. The angle is south from east therefore it's negative.

19.6km*cos(-14.7) = 18.95km

The y component
19.6km*sin(-14.7) = -4.97km

Therefore vector ##\vec r_{AC} = (18.95km,-4.97km)##

Now using the same method for the calculations for ##\vec r_{CD}##

The x component
8km*cos(90) = 0km

The y component
8km*sin(90) = 8km

##\vec r_{CD} = (0,8) ##

Therefore, ##\vec r_{AD} = \vec r_{AC} + \vec r_{CD} = (18,95km, 3.03km)##

Now to find the distance from the camel to oasis B

Oasis A to B
##\vec r_{AB} = (25, 0) ##

##\vec r_{AB} - \vec r_{AD} = (6.05km, -3.03km)##

I found the magnitude to be 6.77km at an angle of 333 degrees from the horizontal.

To obtain the x component of rAC I multiplied the magnitude of the distance traveled by the angle. The angle is south from east therefore it's negative.
 Oh the angle is negative ... apologies.
You should end up with a positive x component and a negative y component for the position of point C.

I'd have done this in one go:
The sketch would look like a number 4 on it's side.
I would use the sketch to check my calculations - checking signs etc.

I found the magnitude [camel to oasis B] to be 6.77km at an angle of 333 degrees from the horizontal.
... so the oasis ends up about 6.8km south and to the east of the camel?

You were only asked to find the distance - not the direction - but fair enough.
Did you have any questions left?

Notes: there is no x nor y in the problem statement ... only compass directions.
So do not use x and y unless you have defined them in your working.

All compass directions are horizontal ... 333deg from the horizontal literally means the camel ends up below the ground. You mean "333deg anti-clock from due east" or, better, "27deg south of east". In general, it is best practice to put the answer in similar terms to the question.

Simon Bridge said:
I'd have done this in one go:
The sketch would look like a number 4 on it's side.
I would use the sketch to check my calculations - checking signs etc.

... so the oasis ends up about 6.8km south and to the east of the camel?

You were only asked to find the distance - not the direction - but fair enough.
Did you have any questions left?

Notes: there is no x nor y in the problem statement ... only compass directions.
So do not use x and y unless you have defined them in your working.

All compass directions are horizontal ... 333deg from the horizontal literally means the camel ends up below the ground. You mean "333deg anti-clock from due east" or, better, "27deg south of east". In general, it is best practice to put the answer in similar terms to the question.

I apologize again for using incorrect notation. It seems then that the first diagram was incorrect, but the second was not. This happened because I sinply attempted to draw the camel travel distance without finding its individual components first and I tried to guess we're it would be. I guess my responses were partially correct as I obtained the same number, but simply said the camel is 6.77 km at an angle 333deg. from the horizon. I will use your suggested notation of 6.77 km, 27 deg. South of east.

By the way, I'm a student very interested in learning physics, but as this post shows I seem to have trouble learning even basic new concepts. Do you and other readers have suggestions for websites that explain these, similar, and even more advance topics that I might see in a university physics course? I have the time to read, and would love new and reliable resources. I'm already reading the course book, and using khan academy.

Thanks again

IMHO, you did great on this problem, right from the get go. I was even able to relate to your 333 degrees.

Chet