Question

What is the particle's displacement from r1 to r2?

The position vector of a particle is initially r1=(-3.0m) `hati` + (4.0m) `hatj` and then later is r2=(9.0m) `hati` + (-3.5m) `hatj`.

Please have me an explanation.

Answer 1

Explanation:

The position vector of a particle is initially r1=(-3.0m) `hati` + (4.0m) `hatj` and then later is r2=(9.0m) `hati` + (-3.5m) `hatj`.

Let `O` be the orgin in X-Y plane A and B represent the initial and final positions of the particle.

`vec(OA)=vecr_1=(-3.0m) hati + (4.0m) hatj`

`vec(OB)=vecr_2=(9.0m) hati + (-3.5m) hatj`

By triangle law of vector addition we have
Displacement vector
`vec(AB)=vec(OB)-vec(OA)`

`=(9-(-3))mhati+((-3.5)-4)mhatj`

`=(12)mhati-(7.5)mhatj`