Physics Calculating Distance and Direction

[Anonymous1135]

Homework Statement

A diver explores a shallow reef off the coast of Belize.
She initially swims 90.0 m north, makes a turn to the east
and continues for 200.0 m, then follows a big grouper for
80.0 m in the direction 30° north of east. In the meantime,
a local current displaces her by 150.0 m south. Assuming
Chapter 2 | Vectors 101
the current is no longer present, in what direction and how
far should she now swim to come back to the point where
she started?

Homework Equations

Answer is 240.2m and 2.2° South of West

The Attempt at a Solution

Ok so in my attempt I got

90.0m N ----------------------------- 90cos90 = 0 90sin0=90
200.0m E -----------------------------200cos200=200 200sin0=0
80.0m 30° North of East------------ 80cos30 = 69.28 80sin30=40
150m S ---------------------------------150cos270=0 150sin270=150

so 269.28i - 20 j

sqrt 269.28^2+20^2 = 270.02 in the direction 4.24° South of North Ive done this problem several time same result I have an idea where the issue is. Any help would be appreciated. This is not a homework question I am just doing some problem to get a head when spring semester starts

 

[Orodruin]

Go through each step and make sure you are using the correct angle in your computations and add the correct components.

I suggest you write down the vector you get on the form xi+yj in each step. It will make it easier to read your post.

South of North

There is no such thing in terms of bearings. Do you understand how the ”x of y” designation works?

 

[Anonymous1135]

Sorry, typo I didn't mean South of North I mean South of East since the x component is positive and and the y component is negative which make its in the 4th quadrant

I though my component were correct

90.0m N ----------------------------- 90cos90 = 0 90sin0=90-----------(0i + 90j)

200.0m E -----------------------------200cos200=200 200sin0=-------(200i + 0j)

80.0m 30° North of East---------- 80cos30 = 69.28 80sin30=40---(69.28i + 40j)

150m S --------------------------------150cos270=0 150sin270=150--(0i - 150j)

 

[Orodruin]

80.0 m in the direction 30° north of east. In the meantime,

Are you sure this does not say east of north?

 

[Anonymous1135]

This is how its wording from the book

If Calculated as East of North I get the answer 240.17 but the direction now it 2.2 North of East

 

[Orodruin]

This is how its wording from the book

Then it is a typo.

If Calculated as East of North I get the answer 240.17 but the direction now it 2.2 North of East

Which vector is directed 2.2° North of East and what does the question ask you for?

 

[Anonymous1135]

Then it is a typo.Which vector is directed 2.2° North of East and what does the question ask you for?

Well the way I came up with 2.2° north of east was adding the xy which gave me 240i+9.28j which puts it in the first quadrant and arctan ( 9.28/240) = 2.2°

 

[Orodruin]

Well the way I came up with 2.2° north of east was adding the xy which gave me 240i+9.28j which puts it in the first quadrant and arctan ( 9.28/240) = 2.2°

This does not answer the question I asked you. Please answer the question in #6 completely.

 

[Anonymous1135]

Not sure what you mean by which vector is directed 2.2° North of East and the question ask me which direction and how far she has to swim back to her starting point.

 

[haruspex]

adding the xy which gave me 240i+9.28j which puts it in the first quadrant

With the regard to the diver's course, what point does that (x,y) represent and with respect to what point as origin? As a vector representing a net movement, what movement?

 

[Anonymous1135]

Sorry I don't follow

 

[haruspex]

Sorry I don't follow

You calculated the x and y coordinates of some point. What point? Answer this in terms of the diver's journey.
Those coordinates only mean anything relative to an origin. What point is your origin? Again, identify the point in the diver's journey, not just "(0,0)".

 

[Orodruin]

the question ask me which direction and how far she has to swim back to her starting point.

Indeed, but this is not what you have computed.

Let me ask you this without vectors, just one dimension. I walk 30 m to the right. How far and in what direction do I have to walk to get back to where I started?

 

[Anonymous1135]

Indeed, but this is not what you have computed.

Let me ask you this without vectors, just one dimension. I walk 30 m to the right. How far and in what direction do I have to walk to get back to where I started?

You would walk back 30m to the left to get to where you started

 

[haruspex]

You would walk back 30m to the left to get to where you started

Right. If you still don't get what you did wrong in post #7, think about what exactly the (x,y) you calculated means... it is the vector that takes the diver from where to where? Or try to answer my questions in post #12.