Finding magnitude of a third displacement

[JennAshley19]

A particle undergoes three displacements. The first has a magnitude of 15m and makes and angle of 35 degrees with the positive x axis. The second has a magnitude of 7m and makes an angle of 163 degrees with the positive x-axis. After the third displacement the particle returns to its initial position.

Find the magnitude of the third displacment. Answer in units of m.

 

[PeterO]

A particle undergoes three displacements. The first has a magnitude of 15m and makes and angle of 35 degrees with the positive x axis. The second has a magnitude of 7m and makes an angle of 163 degrees with the positive x-axis. After the third displacement the particle returns to its initial position.

Find the magnitude of the third displacment. Answer in units of m.

Draw a diagram, and use components, both x- and y-components.

 

[JennAshley19]

I have the diagram I don't know what to plug the given into though.

 

[PeterO]

I have the diagram I don't know what to plug the given into though.

The two given displacement can be split into an x- component and a y-component

Like for example: a 10m displacement making 35 degrees with the positive x-axis will have an x-component of 10.cos 30 and a y-component of 10.sin30.

By adding the two x components and the two y components you will find the final displacement after those two.

Lets assume that is +5m x and +18 m y.

From there you need a displacement with component -5, x and -18, y to get back where you started. That displacement can be described by magnitude and angle with respect to the positive x-axis - even if the angle is something like 234 degrees.

 

[Xyius]

Remember magnitude means length of the vector, which also means hypotenuse of a triangle. If you are given the hypotenuse and the angle that it makes with the base of the triangle, you can calculate the x and y components using trigonometry. (The two legs of the triangle.) Remember every vector is essentially a triangle that can be treated in this way.

What you need to realize is that the TOTAL displacement is zero because the particle returns to its original position. You are asked to find the third displacement vector and are only given information about the first two. Since you know that the third vector returns the particle to its original position, the sum of the first two vectors is equal to the negative of the third vector since they all should add up to be zero.

Hope this helps.