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
MetaPhysik
Jan 27, 2018

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

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