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.

1 Answer
Jul 18, 2018

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#