Objects A and B are at the origin. If object A moves to #(-4 ,7 )# and object B moves to #(2 ,-3 )# over #4 s#, what is the relative velocity of object B from the perspective of object A?

2 Answers
Mar 3, 2018

Net displacement of #A# from origin is #sqrt(4^2 + 7^2)=8.061 m# and the net displacement of #B# from origin is #sqrt(2^2+3^2)=3.605m#

considering that they moved with constant velocity,we get, velocity of #A# is #8.061/4=2.015 ms^-1# and that of #B# is #3.605/4=0.090 ms^-1#

So,we have got velocity vector of #A# (let, #vec A#) and velocity vector of #B# (let, #vec B#)

So,reative velocity of #B# w.r.t #A# is #vec R = vec B - vec A=vec B + (-vec A)#

Now,the angle between the two vectors is found as follows,enter image source here

So,angle between #vec B# and #-vec A# is #3.955^@#

So,#|vec R| = sqrt(0.90^2 + 2.015^2 + 2*0.90*2.015*cos 3.955)=2.914 ms^-1#

Mar 3, 2018

Answer:

The relative velocity is #=<3/2, -5/4> ms^-1#

Explanation:

www.slideshare.net

The absolute velocity of #A# is #v_A=1/4 <-4,7> = <-1, 7/4>#

The absolute velocity of #B# is #v_B=1/4 <2,-3> = <1/2,-3/4>#

The relative velocity of #B# with respect to #A# is

#v_(B//A)=v_B - v_A#

#=<1/2,-3/4> -<-1, 7/4>#

#= <1/2+1, -3/4-7/4>#

#= <3/2, -5/4>#