Question

Displacement from origin?

An object is displaced by 60.0m at a bearing of 60.0∘. If the object then moved 30.0m due north, what is the final magnitude of displacement of the object from its origin?

Answer 1

`color(blue)(30sqrt(7)m)`

Explanation:

I am taking the bearing of `60.0^@` as being from north. This gives an angle of `30^@` with the horizontal ( x axis ).

We will call this vector `veca`

Using cosine and sine functions:

`veca=60cos(30)+60sin(30)=(60sqrt(3))/2+30`

Due north is `90^@` to horizontal ( x axis ):

Calling this `vecb`

`vecb=30cos(90)+30sin(90)=0+30`

Displacement from origin is:

`||veca+vecb||`

`veca+vecb=((60sqrt(3))/2+30+(0+30))`

`=(60sqrt(3))/2+60`

`||veca+vecb||=sqrt(((60sqrt(3))/2)^2+(60)^2)=30sqrt(7)m~~79.37m`

Answer 2

`"displacement -- origin to object" = 79.4 m`

Explanation:

I made up these sketches of the 1st and 2nd locations of the object due to these displacements.

The origin is bottom left, marked `O`. `D_1 and D_2` are the locations after the 1st and 2nd displacements. I added 2 lines to each to make triangles. The horizontal part of the triangle is straight East from `O`. The vertical part of the triangle is straight South of the location of the object in the 1st and 2nd cases.

Look at the triangle of Case 1. The direction of the displacement was `60^@` East of North, so the angle inside the triangle at `O` is `30^@`. The angle East of `O` (bottom-right) is `90^@` so the angle where the object is sitting (top) is a #60^@ angle. So the triangle is a 30-60-right triangle.

Basic details about the relative lengths of the sides of a 30-60-right triangle:

• If you consider the length of the side opposite the `30^@` angle to be 1 unit, call it `1l`, then the length of the hypotenuse will be `2l`.
• The length of the 3rd side is `sqrt(3)""l` or `1.732l`. (I marked those relationships inside the 30-60-right triangle.)

We know that the length of the hypotenuse is 60 m. Using the above details about 30-60-right triangles we can say that the side opposite the `30^@` angle is 1/2 the length of the hypotenuse, or `30 m`. And we can say that the side opposite the `60^@` angle is `1.732l = 1.732*30 m = 51.96 m`. (I marked the lengths, in meters, outside the 30-60-right triangle.)

Now look at the triangle of Case 2. It is not a 30-60-right triangle. But we know 2 of the sides: the horizontal side is still 51.96 m and the vertical side is `30 m+30 m=60 m`.

Those 2 dimensions, and Pythagoras, allow us to find the final magnitude of displacement of the object from its origin.

`"displacement -- origin to object" = sqrt(51.96^2 + 60^2) m`

`"displacement -- origin to object" = 79.4 m`

I hope this helps,
Steve