Determining Displacement, Adding Vectors

[killaI9BI]

Homework Statement

I've been stuck on this problem for hours. Any help you can provide is greatly appreciated.

From a lookout point, a hiker sees a small lake ahead of her. In order to get around it, she walks 4.5km in a straight line toward the end of the lake. She turns right making a 60° angle with her original path, and walks to a campsite 6.4km in the new direction. Determine her displacement from the lookout point when she has reached the campsite.

Homework Equations

c² = a² + b² -2ab cosθ
sin a/a = sin b/b

The Attempt at a Solution

c² = 4.52 + 6.42 – 2(4.5)(6.4)cos60
c² = 20.25 + 40.96 – 28.8
c² = 32.41
c = 5.69km

sin θ = (sin 60/5.69) X 6.4
sin θ = 0.974
θ = 76.9°

θ = 90 – 77 = 13°

The hiker is 5.7km 13° above horizontal, right of her original heading from the lookout when she reached the campsite.

The book's answer is 5.8km, 18° away from the horizontal from the lookout.

 

[jedishrfu]

Have you tried graphing the solution and then comparing it to your calculations to see where you may have gone wrong?

Sometimes it helps with understanding things in a geometrical sense.

 

[killaI9BI]

I have. I've literally spent hours on this question. I'm trying to keep this correspondence course rolling with no instructor while working a full-time job and I'm at the point where I think the book is wrong.

I'm guessing that's not the case but I just can't afford to spend any more time spinning my wheels. Can you see any flaws with my equation?

It seems to go wrong right off the bat. This equation doesn't appear to be difficult but my answer for R=5.69km isn't the same as the book's answer of 5.8km. The book must be wrong. It's a very expensive book so if it's wrong, I'd like some cash back!

Thank you for your time.

 

[HallsofIvy]

If you goes, on the second leg, 60 degrees from the from the first leg, the angle inside your triangle is 180- 60= 120 degrees, not 60.

 

[killaI9BI]

Thank you but the diagram in the book shows the inside angle is 60 degrees. Unless I've misunderstood what you're suggesting.

I'd post a pic but I don’t know how.

 

[killaI9BI]

figured out how to post a pic and also realized that I messed up the formatting of the "²"s in my attempt at a solution so here it goes again:

c² = 4.5² + 6.4² – 2(4.5)(6.4)cos60
c² = 20.25 + 40.96 – 28.8
c² = 32.41
c = 5.69km

sin θ = (sin 60/5.69) X 6.4
sin θ = 0.974
θ = 76.9°

θ = 90 – 77 = 13°

The hiker is 5.7km 13° above horizontal, right of her original heading from the lookout when she reached the campsite.

The book's answer is 5.8km, 18° away from the horizontal from the lookout.

Can anyone help?

thank you!

 

[NascentOxygen]

Your triangle work is right. We don't have any information that relates the triangle to North or East, so you can't give any bearing relative to the cardinal points.

There is perhaps a reference if we can take the grid pattern on the mud map to be aligned N upwards, but no such correspondence is indicated, so it would be unwise to blindly assume it. Usually.

Perhaps as an exercise you could assume the vertical lines are, in fact, aligned N-S and see what the displacement angle is relative to East? You'll observe that the 4.5 km leg is not precisely northerly, but the grid can help you.

 

[NascentOxygen]

The point others have made that the second leg is actually at 120 degrees to the direction of the first should be heeded, and brought to the attention of the person who set this homework task.

 

[killaI9BI]

This is a correspondence course of which there is no instructor. I don't believe they are looking for an answer relative to the cardinal points as the book's answer is "5.8km, 18° away from the horizontal from the lookout."

My answer is "The hiker is 5.7km 13° above horizontal, right of her original heading from the lookout when she reached the campsite."

Why are the answers different if my triangle work is right?

 

[NascentOxygen]

My answer was perhaps unclear. Let's approach this another way...

At what angle to the vertical is that 4.5 km leg?

 

[killaI9BI]

According to my protractor; 7 degrees from vertical.

I think if I could figure out why my answer is 0.1 km different than the book's answer, the rest would work itself out.

 

[NascentOxygen]

According to my protractor; 7 degrees from vertical.

Okay, we'll go with that for the moment.

Now knowing that the 4.5 km leg is not exactly aligned with the vertical grid lines, does that change your answer for the angle to the horizontal of the displacement line?

I think if I could figure out why my answer is 0.1 km different than the book's answer, the rest would work itself out.

Your answer for the magnitude is correct.

 

[killaI9BI]

Okay, we'll go with that for the moment.

Now knowing that the 4.5 km leg is not exactly aligned with the vertical grid lines, does that change your answer for the angle to the horizontal of the displacement line?Your answer for the magnitude is correct.

ahh I see your point.

I'll change my response to "When at the campsite, the hiker is 5.7km, 77° to the right of the lookout with respect to the 4.5 km path".

Thank you very much for the perspective.

 

[NascentOxygen]

I'll change my response to "When at the campsite, the hiker is 5.7km, 77° to the right of the lookout with respect to the 4.5 km path".

BUT the textbook wants your answer to be expressed as degrees above the horizontal. So what would your answer be now?

 

[killaI9BI]

Well that's how they answered the question so I suppose it wouldn't hurt to follow suit.

I know the hyp is 5.69 but I'm having a difficult time trying to figure out one of the other angles.

Can you give me a hint?

 

[NascentOxygen]

Can you give me a hint?

I thought I had?

How many degrees is that acute angle between the "not quite vertical" path and the horizontal grid line at its base?

 

[killaI9BI]

lol

I thought there must be a more scientific and accurate way than using my protractor.

 

[NascentOxygen]

lol

I thought there must be a more scientific and accurate way than using my protractor.

There usually is. You generally can't rely on diagrams being sketched accurately to scale, instead you should use the lengths marked on. For example, if you can see what the x-step is for a particular y-step, you can use trig and say the angle is that whose tan = Δy / Δx. In your case, the map is marked on a grid to allow you to estimate distances off that grid. A protractor should give a similar answer because the figure here is sketched to scale.

 

[killaI9BI]

Please excuse me but I'm just not certain I understand correctly.

Are you saying that I should use the protractor answer or is there another way? We seem to have already established that my answer isn't going to match the answer provided in the back of the book (5.8km vs 5.7km).

If I use my protractor to determine the angle from horizontal, it says 27 degrees above horizontal.

I can't believe how long it's taking me to answer this one... I'm going to need to bring a sleeping bag for my exam.

 

[NascentOxygen]

Are you saying that I should use the protractor answer or is there another way? We seem to have already established that my answer isn't going to match the answer provided in the back of the book (5.8km vs 5.7km).

If I use my protractor to determine the angle from horizontal, it says 27 degrees above horizontal.

Instead of using a protractor to measure the angles, you could use another means, viz., trigonometry, to calculate the angles. This involves using sine, cosine, or tangent to find angles when you know lengths of the sides. To determine those lengths of sides, you can "count squares" (and partial squares) on the grid.

However, this seems new to you, so we are going to have to accept that using your protractor will suffice here. It's an improvement on your original answer, anyway.

The reason you don't get the book's answer is because they deliberately drew it not to scale, so that you won't get close to the right answer if you just take measurements off their drawing.

Good luck with your exam. Coffee and a cut lunch might be more appropriate?

 

[killaI9BI]

Thanks again for your help

 

[NascentOxygen]

However, this seems new to you, so we are going to have to accept that using your protractor will suffice here. It's an improvement on your original answer, anyway.

The reason you don't get the book's answer is because they deliberately drew it not to scale, so that you won't get close to the right answer if you just take measurements off their drawing.

You could try using a pencil to extend that 6.4km line to the length it should be were it drawn to an identical scale (i.e., mm/ km) as the 4.5km line, and then use your protractor to measure the angle you seek.

Just check that the 60° vertex is accurately drawn, else measurements off the drawing will be meaningless.

 

[killaI9BI]

I'm thinking my answer of 77 degrees to the right of the 4.5km path would be the most accurate answer I could give. Would you disagree with that?

I realize the book's answer is in degrees above horizontal but can't there be more than one correct answer?

 

[NascentOxygen]

I'm thinking my answer of 77 degrees to the right of the 4.5km path would be the most accurate answer I could give. Would you disagree with that?

Yes, that can't be faulted.

I realize the book's answer is in degrees above horizontal but can't there be more than one correct answer?

There are multiple correct answers here.

That is, a couple of valid answers each of which should be awarded full marks, because the problem is badly devised/defined.

 

[killaI9BI]

I really appreciate your feedback.

Thank you!