Objects A and B are at the origin. If object A moves to #(3 ,7 )# and object B moves to #(1 ,-4 )# over #3 s#, what is the relative velocity of object B from the perspective of object A? Assume that all units are denominated in meters.

1 Answer
Jan 27, 2018

Answer:

To B, A is moving away at 3.73 m/s @#80^@# NE
To A, B is moving away at 3.73 m/s @#80^@# SW

Explanation:

#vec v_A = (Delta vecr)/(Deltat)= ⅓ (3,7) = (1, 7/3) #
#vec v_B = (Delta vecr)/(Deltat)= ⅓ (1,-4) = (1/3, -4/3) #

From B's point of view, A is moving away from B at this velocity:
#vecv_B rarr_A = vecv_A - vecv_B = (1,7/3)-(1/3, -4/3)= (2/3 , 11/3) #

The magnitude of this relative velocity is:
#|vecv_(BA)| = |vecv_A - vecv_B|= sqrt ( (2/3)^2+(11/3)^2)=sqrt(125)/3=3.73 m/s#
@ this angle:
#theta_(BA) =tan^-1( 11/3/2/3")=tan^-1(11/2)= 80^@ NE#

From A's point of view, B is moving away from A at this velocity:
#vecv_A rarr_B=vecv_B-vecv_A =(1/3, -4/3)-(1,7/3)=(-2/3 , -11/3)#
#|vecv_(AB)|=sqrt ( (-2/3)^2+(-11/3)^2)= 3.73m/s #
@angle:
#theta_(BA) =tan^-1( -11/3/-2/3")=tan^-1(11/2)= 80^@ SW#