# Help Physics is killin me

1. Sep 18, 2004

### kimikims

Help!! Physics is killin me!!

I'm stuck on these problems.. can anyone help me??

------

A car travels 1.14km in the x- direction, then turns left 74.1 degrees to the original direction and travels an additional distance of 3.18km. Calculate the x component of the car's net displacement. Answer in units of km.

--------

A truck travels 1270m uphill along a road that makes a constant angle of 4.45 degrees with the horizontal.

1) Find the magnitude of the trucks horizontal component of displacement. Answer in units of m.

2) Find the magnitude of the trucks veritical component of displacement. Answer in units of m.

2. Sep 18, 2004

### PRodQuanta

$${x_f}-{x_i}$$ where f=final and i=initial

Make sure you label direction.

3. Sep 18, 2004

### recon

PRodQuanta, what exactly are you trying to imply here?

kimikims, you should have attempted the problems before posting (read the HW guidelines). Particularly which part of the problem is killing you?

All the questions you posted have to do with Trigonometry. Are you having difficulty understanding Trigonometry?

4. Sep 18, 2004

### arunma

For the first problem, just add these two vectors: (1.14 km, 0 degrees) + (3.18 km, 74.1 degrees)

Why does the second vector have a direction of 74.1 degrees? Keep in mind that the car turned left, which is in the positive y-direction (since it was traveling in the positive x-direction).

Also note that those vectors are in polar form. You need to convert to rectangular form using the following transformation.

x = r cos θ
y = r sin θ

Then you can add the two x's and the two y's. Well, try it out, you should get the right answer.

The second problem is the same. Here you have a vector's polar form (magnitude and angle with the x-axis). You're asked to find the x and y components. So you have to use the same transformations to find these components.

Both of these problems have a common theme: trigonometry and polar coordinates. If you're having trouble with this stuff, chances are that your problem is with the math, not the physics. I assume you're in algebra-based physics. Have you taken precalculus? If you have, you might want to review polar coordinates. You may also want to review just a little bit of geometry (specifically, the properties of triangles), because soon you'll be doing problems with blocks on inclined planes. If you can do the math, the physics is actually very easy.

5. Sep 18, 2004

### kimikims

I'm sorry, how do you find the x and y's of this problem?

6. Sep 18, 2004

### Pyrrhus

In my opinion, go open your book and review vectors, if you do not understand come back here again.

7. Sep 18, 2004

### recon

Why don't you try drawing out the path of the car and truck? Try to see that a there is a triangle shape in the figure. Just to give you a clue, the r in x = rcos(theta) is the hypotenuse of the triangle you're looking for.

8. Sep 18, 2004

### arunma

In each problem, x = r cos θ and y = r sin θ. The number r is the length of the vector you're dealing with.

When you find the x and y components of a vector, you're transforming the vector into Cartesian (rectangular) coordinates. The advantage here is that you can add x's or y's arithmetically (just don't add x to y!); you can't do that with polar coordinates.

9. Sep 18, 2004

### kimikims

Anyone know how to find a?

------

What is total x displacement (Xtotal)?

a^2 + b^2 = 3.18^2

Xtotal= 1.14 + a

Q = 74.1 degrees

cos Q = a/3.18

a = cos Q/ 3.18

a = ????

Xtotal = 1.14 + a

10. Sep 18, 2004

### recon

Code (Text):
/
/
/
----------------------/
This is the path on which the car travels. The triangle you are looking for is:

Code (Text):
/|
/ |
/  |
----------------------/___|

The angle that the hypotenuse makes to the base is 74.1 degrees. The hypotenuse has length 3.18 km. Let us name the horizontal base of the triangle x. Then,

$$cos (74.1) = \frac {x}{3.18}$$

$$x = cos(74.1)*3.18$$

Remember to add x to 1.14 km.

Last edited: Sep 18, 2004
11. Sep 18, 2004

### cdhotfire

i dont see how this is physics.

12. Sep 19, 2004

### Pyrrhus

Oh, but math is the language physics use!

13. Sep 19, 2004

### cdhotfire

Ya ya, but this is like stuff one learns in pre-cal. I wonder how this is a college class. W/e.

14. Sep 20, 2004

### recon

Not everyone here is American. For example, my countrymen use the words 'college', 'secondary school' and 'high school' interchangeably and would not know in which forum to post if they came here. Also, the word 'university' is used instead of 'college'.