# What is the displacement of the car from the point of origin

I just started a physics class and im totally confused, here's my problems for tonight if anybody can help it would be greatly appreciated. Im not necessarily looking just for answers but help getting them too:

1. A car is driven 130 km west and then 55 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)?
Magnitude: (km)
Direction: (° south of west )

2. A delivery truck travels 21 blocks south, 17 blocks east and 18 blocks north. What is its final displacement from the origin? Assume the blocks are equal length.
Magnitude: 17.3 blocks (i figured that part out)
Direction: ° (counterclockwise from east is positive)

3. Vector V1 is 8.17 units long and points along the -x axis. Vector V2 is 4.58 units long and points at +35.0° to the +x axis.
(a) What are the x and y components of each vector?
V1x=
V1y=
V2x=
V2y=
(b) Determine the sum V1 + V2.
Magnitude:
Direction: ° (counterclockwise from the +x axis is positive)

4. An airplane is traveling v = 810 km/h in a direction 38.5° west of north.
(a) Find the components of the velocity vector in the northerly and westerly directions.
____ km/h (north)
____ km/h (west)
(b) How far north and how far west has the plane traveled after 7 h?
____ km/h (north)
____ km/h (west)

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Tide
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You might start by telling us what you did so far. :-)

i did, look at #2, i got the magnitude. its all i could figure out ive been trying for hours.

i just realized i shoulda posted this in the homework section... didnt notice it before :/ sorry, maybe it will get moved

TriumphDog1 said:
1. A car is driven 130 km west and then 55 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)?
Magnitude: (km)
Direction: (° south of west )
Well, this one is quite simple. You are adding vectors. My teacher said to write the vectors in polar form and then in rectangular. Once that is done, you can add them. That will give you displacement.

$$\vec{Vector}=(A,\theta_A)$$
$$\vec{A}=(130,180~degrees)$$
$$\vec{B}=(55,45~degrees)$$

I don't know how to make the degree symbol in latex.

Ok, now you have the polar. Now, you need to convert that into rectangular and add.

$$x=Acos\theta_A$$
$$y=Asin\theta_A$$
$$A_x=130cos(180)=130*-1=-130$$
$$A_y=130sin(180)=130*0=0$$
$$A(-130,0)$$
$$B_x=55cos(45)=55*\frac{\sqrt{2}}{2}=-39$$
$$B_y=55sin(45)=55*\frac{\sqrt{2}}{2}=-39$$
$$B(39,39)$$

$$A+B=(A_x+B_x,A_y+B_y)$$
$$A+B=(-91,39)$$

Now, you have to convert back to polar to get your displacement.

$$A=\sqrt{x^2+y^2}$$
$$\theta_A=arctan(\frac{y}{x})$$
$$A=\sqrt{-91^2+39^2}=\sqrt{8281+1521}=\sqrt{9802}=99$$
$$\theta_A=arctan(\frac{39}{-91})=arctan(.43)=23$$

So, the displacement is 99km 23 degrees south of west or 157 degrees. I think... :rofl:

*edit: Changed angle of second vector...forgot it's the angle from the previous vector and not from origin. :rofl: Amazing how one little thing can screw up the whole answer*

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this homework is online and it checks the work auto, said that was wrong :( im completely lost with what you did too.

Hmmm...well I was assigned a problem just like this except it was 125km west and 65km south and that is not much of a difference and I got the right answer using the method I just showed.

I converted from polar to rectangular and then added the rectangular units and converted that back to polar. I'm pretty sure that's how you do it. I mean...I did it that way and got it right on my homework.

pervect
Staff Emeritus
TriumphDog1 said:
I just started a physics class and im totally confused, here's my problems for tonight
Did you ever hear of the pythagorean theorem? Think about it for a bit, and see if you can figure out how to apply it to figuring out the distance between two points. (If you have to, look it up: use google, http://www.google.com).

eeek...I think i did it right now. I edited my last post...I'm such a screw up!!

pervect said:
Did you ever hear of the pythagorean theorem? Think about it for a bit, and see if you can figure out how to apply it to figuring out the distance between two points. (If you have to, look it up: use google, http://www.google.com).
yes i know what it is and how to use it, thats how i came up with part of the answer for #2, but i cant seem to figure out how to apply it to anything else.
h8er, you have 2 degrees listed which one is correct?
// edit:
i tried em both and the displacement and direction are both wrong :(

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Hahahah...well then the online thing is wrong...I know I did them right. Both of the angles are the same just one is direction according to a compass and the other is just angle on an x-y coordinate system.

pervect
Staff Emeritus
TriumphDog1 said:
yes i know what it is and how to use it, thats how i came up with part of the answer for #2, but i cant seem to figure out how to apply it to anything else.
h8er, you have 2 degrees listed which one is correct?
// edit:
i tried em both and the displacement and direction are both wrong :(
OK, now that I found this thread again, lets take problem one.....

If you go 55 miles southwest, how many miles did you go south, and how many miles did you go west?

Once you've figured that out, you should be able to solve problem #1 like you did #3.

The other thing you need to know is some trigonometry. This is that if you have a right triangle like the one below

(hmmm, can't make the ascii diagram come out right, well, this is close)

PHP:
adjacant side
-------
\ a      |
\      |
\    | opposite side
\  |
\|
the sine of the angle a is the opposite side over the hypotenuse, the cosine of the angle a is the adjacant side over the hyptenuse

ive figured it all out thx for the help guys

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