# Question concerning displacement and vectors

• UhhRandomUser
In summary, the car is driven a total of 215 km west and 85 km southwest, resulting in a displacement from the point of origin of approximately 231.5 km at a direction of 40.6 degrees south of west. To solve this problem, it is recommended to break down the southwest vector into its southern and western components and use trigonometric functions to find the missing angle and displacement. It is also suggested to search for resources on vector components to gain a better understanding of this fundamental physics concept.
UhhRandomUser
[1] A car is driven 215 km west and then 85 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram.
[2] I have no idea what equations are necessary for this problem. I reread the chapter, but this problem in particular is stumping me.

[3] Originally, I thought I could make a triangle by using the vector of the 215 km west, the 85 km southwest, and the distance between the origin and the head of the 85 km southwest vector as sides. I thought if I could split the big triangle into two right triangles, I could use trigonometric functions to find the displacement, to no avail...

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Welcome to physics forums. I would suggest focusing on the "85 km southwest" part first.

Can you break it up into it's southern and western components?

It's a problem in trigonometry, if you drew the correct figure. Remember to use the directions of the individual vectors to find at least one angle in the triangle. You already have the lengths of two of the sides:

http://en.wikipedia.org/wiki/Solution_of_triangles

This is great advice so far!

While you could do vector addition by creating a triangle using the 215km W, 85km SW, and a hypotenuse connecting the origin to the SW tip, this could be tricky without CAD or unneeded trig functions.

I also recommend breaking up the SW vector into its S and W components. This way, when doing vector addition, you'll have a nice right triangle.

Do you see how to break the SW vector up? Imagine that's all you have for a moment, having it start at the origin. You know it's purely SW, so you know what angle it makes from the x axis. Knowing this angle and the length if the vector, you can draw (and calculate the lengths of) sides to create a triangle with the vector as the hypotenuse. Keep in mind you want one side purely in an x direction, and one in a y!

This divides the vector into its components.

I just figured out how to reply, sorry about that! And @Nathanael I can make it 85 km south and 85 km west correct?
@SteamKing thanks for the link, I'm checking it out now!
and @amadinger I thiiiiink I understood that a bit, thanks! :D

UhhRandomUser said:
I just figured out how to reply, sorry about that! And @Nathanael I can make it 85 km south and 85 km west correct?
@SteamKing thanks for the link, I'm checking it out now!
and @amadinger I thiiiiink I understood that a bit, thanks! :D

If you have a huge square that has sides that are 85 km long, how far is it from one corner to the corner that is diagonally opposite?

## 1. What is displacement?

Displacement refers to the change in position or location of an object. It is a vector quantity, meaning it has both magnitude (size or length) and direction.

## 2. How is displacement different from distance?

Distance is the actual length of the path traveled by an object, while displacement is the straight-line distance between the initial and final positions of the object.

## 3. What is a vector?

A vector is a quantity that has both magnitude and direction. It is represented by an arrow, with the length of the arrow representing the magnitude and the direction of the arrow indicating the direction.

## 4. How do you calculate displacement?

Displacement can be calculated by subtracting the initial position from the final position. The resulting value will be a vector with both magnitude and direction.

## 5. Can displacement ever be negative?

Yes, displacement can be negative if the object moves in the opposite direction of its initial position. This indicates that the object has moved in the opposite direction of its initial position, or that it has moved backwards.

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