Help Needed: Calculating Displacement with Physics Homework

[C42711]

Having trouble with my physics homework, if anyone could please help me with how to this, that would be awesome.

My teacher says we are supposed to find the X and Y componenets and then add them, but I don't know how to find them.

The Smith's go on vacation. They travel 16.0 km north/west; then turn 27 degrees to the left and go 135 km; then 294 km south/west. What is their displacement from home?

 

[Nytik]

Split the travel into three sections, one for each direction traveled.
In each section, the distance traveled forms the hypotenuse of a right-angled triangle, the other two sides being the x- and y-axis. You can use trigonometry to find the x- and y-sides of each triangle.
These separate components can then be summed and combined into a final triangle, the hypotenuse of which will give you the total displacement.

 

[Andrew Mason]

Having trouble with my physics homework, if anyone could please help me with how to this, that would be awesome.

My teacher says we are supposed to find the X and Y componenets and then add them, but I don't know how to find them.

The Smith's go on vacation. They travel 16.0 km north/west; then turn 27 degrees to the left and go 135 km; then 294 km south/west. What is their displacement from home?

To find the x and y components of a vector, $\vec{A}$, you have to first find the angle that it makes with the x axis.

Once you have the angle, the x component is $\vec{x} = |\vec{A}|\cos\theta \hat x$ and the y component is $\vec{y} = |\vec{A}|\sin\theta \hat y$

AM

 

[C42711]

[PLAIN]http://img208.imageshack.us/img208/6876/morephysics.png
By c42711 at 2011-10-30

So, I got this far, but I don't know the angle to use after the turn, is it just 27 degrees? That just doesn't seem right. I don't know the angle for the last vector either, if I guessed I would say 45 degrees, just from how it looks. Sorry, any help would be much appreciated. Thank you.

 

[Nytik]

Since they were originally traveling at an angle of 45 degrees, the new direction would be 45+27 degrees.
The angles in the final triangle are 45 degrees as you suspect.

 

[C42711]

I'm sorry, I am having a lot of trouble with this. I solved for all of my components. I got the Xs to equal -347 and Ys to equal -155.3, but they don't work with the Pythagorean theorem to find the displacement from their home. The reason I made them negatives was some were traveling left or down, but I could be wrong about making them negative. This is what my work looks like:

[PLAIN]http://img830.imageshack.us/img830/1572/djsh.png
By c42711 at 2011-10-30

 

[Nytik]

Those values are fine. They form the two shorter sides of a right-angled triangle, the hypotenuse of which is the displacement. (Make sure you also work out an angle with respect to some given direction.)

 

[C42711]

I'm sorry, I am really having trouble with this. I may have gotten it, would the final displacement be -380 km?

 

[Nytik]

Well, you don't need the minus sign, as the direction will be specified by the angle theta you determine in a moment.
To do this, it is probably easiest to use tan(theta) = opposite/adjacent.
Draw the final (resultant) triangle out to help determine which side of the triangle is which, and also where your theta is actually being defined from (e.g. 30 degrees south of west.)

 

[C42711]

I got this as my final triangle, but I don't know how to go about finding an angle. I also don't understand why it isn't negative when they were traveling to the left and down. I'm sorry, I'm really trying, just having a lot of trouble.

[PLAIN]http://img412.imageshack.us/img412/4790/12549976.png
By c42711 at 2011-10-30

 

[C42711]

I just redid the problem and got the X component to be 69, not -347.
Then I got my final displacement to be -139, but I don't know. I am so lost.

 

[Andrew Mason]

I just redid the problem and got the X component to be 69, not -347.
Then I got my final displacement to be -139, but I don't know. I am so lost.

Just do what your teacher suggests.

$|\vec{A}| = 16 km$; θ=135°

x component = 16cos(135) = -11.3
y component = 16sin(135) = 11.3

$|\vec{B}| = 135 km$; θ=135°+27°=162°

x component = 135cos(162) = -128.4
y component = 135sin(162) = 41.7

Do the same for the third vector. Then add all the x components and the y components to get the x and y components of the resultant displacement. Then work out the angle from those x and y components. How would you determine the magnitude of the resultant?

AM

 

[Andrew Mason]

Your post #11 is correct. All you have to do is find the angle of the resultant from the x axis. Draw the x axis. What angle in the triangle you have drawn is the same as the angle between the resultant and the x axis?

AM