Adding vectors algebraically to find displacement

[KevinFuerst]

Hi everyone,
I am taking a physics course this year and I find myself already looking for help unfortunately. I think I understand the basics of adding vectors using A_x = A cosθ , A_y = A sinθ, etc. I am getting confused with the 3rd leg of the trip though. Since it is due west wouldn't that mean there is no vertical displacement? How can work that into my equation?

Homework Statement

Homework Equations

A_x = A cosθ
A_y = A sinθ
...
R = √R_x² + R_y²

The Attempt at a Solution

I have tried using the previous formulas to solve it but i get an answer of ≈320km. the back of the book says it should be 245km. could someone walk me though this? Thank you!

 

[kushan]

Hey Kevin ,
In the picture I have uploaded you can see
that
(*note the symbols in bold are vectors)

r =a + b

, subsequently
R= c+r

 

[CAF123]

What you want is [itex]\vec{R} = \vec{a} + \vec{b} + \vec{c}. [/itex]
Write out the vector equations for each of [itex] \vec{a}, \vec{b}, \vec{c} [/itex] and then add them together. This will give an eqn for [itex] \vec{R} [/itex]

Using [itex] |\vec{R}| = \sqrt{R_x^2 + R_y^2} [/itex] will give you it's length.
Can you find the angle from this?

 

[KevinFuerst]

thanks for the quick response but I am not quite following you guys. Still new to this.
Here is my work, maybe you can show me what i am doing wrong that way. Once i get the right distance i'll have no problem finding the angle. I under stand how to do that part.

A_x = A cos(30) = 175km (.866) = 151.55km
A_y = A sin(30) = 175 (.5) = 87.5km
B_x = B sin(20) =150km (.342) = -51.3km
B_y = B cos(20) = 150 (.939) = 140.96km
C_x = C cos(110) = -190km (-.342) = 64.98km
C_y = C sin(110) = 190 (.939) = 178.54km

√165.23² + 407² = √192951.4 ≈ 439km

 

[kushan]

Hey kelvin ,
As the diagram and the question says , the vector C is parallel to X axis , which mean there is no change in the y component of total displacement .
On the side note , What does the figure mean by 110 degrees ? what is it .

 

[kushan]

Try solving
Cx as 190
and Cy as 0

 

[CAF123]

Cx = C cos(110) = -190km (-.342) = 64.98km
Cy = C sin(110) = 190 (.939) = 178.54km

You have the components correct for vectors [itex] \vec{a} [/itex] and [itex] \vec{b},[/itex]however the angles are wrong for the components of [itex] \vec{c} [/itex]

What angle does [itex] \vec{c} [/itex] make with the positive [itex] x [/itex] direction?
There is no vertical component to [itex] \vec{c} [/itex]

 

[KevinFuerst]

Try solving
Cx as 190
and Cy as 0

thank you so much. i understand that now. it will help me on a previos one i thought i had right too.

You have the components correct for vectors [itex] \vec{a} [/itex] and [itex] \vec{b},[/itex]however the angles are wrong for the components of [itex] \vec{c} [/itex]

What angle does [itex] \vec{c} [/itex] make with the positive [itex] x [/itex] direction?
There is no vertical component to [itex] \vec{c} [/itex]

vector c is at 0 or 180 degrees isn't it?

 

[kushan]

180 you can call it .

 

[CAF123]

Vector [itex]\vec{c} [/itex] is at angle of 180^o with respect to the positive [itex] x [/itex] axis. Therefore the vector eqn for [itex]\vec{c} [/itex] is,
[itex] \vec{c} = (190cos180)\hat x = -190\hat x [/itex]

 

[KevinFuerst]

oh okay! that's where that 110 came from. i was trying to figure out the angle so i decided to try the interior angle of ΔABC

i had a very similar problem there i found the interior angle like that and coincidently the resultant was extremely close either way. I have now corrected both and i can't thank you two enough.