# 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.

Jan 27, 2018

To B, A is moving away at 3.73 m/s @${80}^{\circ}$ NE
To A, B is moving away at 3.73 m/s @${80}^{\circ}$ 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:
${\vec{v}}_{B} {\rightarrow}_{A} = {\vec{v}}_{A} - {\vec{v}}_{B} = \left(1 , \frac{7}{3}\right) - \left(\frac{1}{3} , - \frac{4}{3}\right) = \left(\frac{2}{3} , \frac{11}{3}\right)$

The magnitude of this relative velocity is:
$| {\vec{v}}_{B A} | = | {\vec{v}}_{A} - {\vec{v}}_{B} | = \sqrt{{\left(\frac{2}{3}\right)}^{2} + {\left(\frac{11}{3}\right)}^{2}} = \frac{\sqrt{125}}{3} = 3.73 \frac{m}{s}$
@ 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:
${\vec{v}}_{A} {\rightarrow}_{B} = {\vec{v}}_{B} - {\vec{v}}_{A} = \left(\frac{1}{3} , - \frac{4}{3}\right) - \left(1 , \frac{7}{3}\right) = \left(- \frac{2}{3} , - \frac{11}{3}\right)$
$| {\vec{v}}_{A B} | = \sqrt{{\left(- \frac{2}{3}\right)}^{2} + {\left(- \frac{11}{3}\right)}^{2}} = 3.73 \frac{m}{s}$
@angle:
theta_(BA) =tan^-1( -11/3/-2/3")=tan^-1(11/2)= 80^@ SW