What is the Total Displacement of a Car Driven in Multiple Directions?

[queenspublic]

Homework Statement

A car is driven east for a distance of 55 km, then north for 34 km, and then in a direction 30° east of north for 23 km. Draw the vector diagram and determine the total displacement of the car from its starting point.

Homework Equations

the answer = arc tan (ay/ax)

The Attempt at a Solution

The magnitude is 87 km.

But I don't understand how to find the direction (in degrees) counterclockwise from east. You have to use that tangent formula. But I don't know what numbers to plug in for the x and y component.

 

[Delphi51]

Have you seen the vector method where you find the x and y components and add them up separately to get the total? That method applies really nicely to this problem because the first two vectors are already just x or y vectors. So you only need to analyze the 3rd one. Sketch the 23 at 30 degrees E of N. Drop a vertical line from the end and make a horizontal line to the right from the beginning. This forms a triangle. Think of it as a horizontal vector (base of the triangle) plus a vertical or y vector (height of triangle). Figure out where the 30 degrees is - might be better to mark a 60 degree angle in the triangle. Use sine and cos to find the x and y sides.

Add up the x and y parts from all three vectors to get the total.

Finally, you need to convert this x and y total into polar form - distance and angle.
Again make a triangle under the vector. This time you need to find the hypotenuse and use the tangent to find the angle.

 

[queenspublic]

I still don't really understand. Do you want me to draw that mini-triangle (that I used to determine the magnitude with the 16.3 on each side) again?

Do I use 55cos30 and 34sin30 to find x component and y component?

I am so frickin clueless.

 

[Delphi51]

Do I use 55cos30 and 34sin30

No.
You have an arrow 23 km long at an angle of 30 degrees from vertical (north), or 60 degrees from horizontal (east-west). The horizontal part will therefore be 23*cos(60). Can you find this and the vertical part?

 

[queenspublic]

So the vertical is 23sin30?

 

[LowlyPion]

Resolve the vectors into their components and then simply add the components. This gives you the components of your resultant.

Resolving into components:
http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec5

Btw useful calculator for cross products:
http://hyperphysics.phy-astr.gsu.edu/hbase/vvec.html#vvc3

 

[queenspublic]

Please...i don't understand...

But thanks for the links. I just don't know what numbers I'm suppose to plug in.

 

[LowlyPion]

Draw the vectors on an x,y plane with east as positive x.

If you carefully draw each of the vectors then you should see how the components add together to yield your result.

V1 = 55 i + 0 j
V2 = 0 i + 34 j
V3 = 23*sin 30 i + 23*cos30 j

R = V1 + V2 + V3

 

[queenspublic]

why Delphi tell me to do 23cos60?

 

[queenspublic]

Thank you so much!

39 -- final answer

 

[queenspublic]

I understand it now. X components. Y components. Was totally confused until I read problem again. Add the i's and j's. Use arc tan of 54 over 66.5

 

[LowlyPion]

30° east of north is on the I quadrant side of the y-axis. The angle is with the y-axis as I read it.

On my graph it makes it sinθ for the i components and cosθ for the j.

 

[Delphi51]

23 cos(60) is the same as 23 sin(30).