Question

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.

Answer 1

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`