# Objects A and B are at the origin. If object A moves to (0 ,-2 ) and object B moves to (5 ,4 ) over 8 s, what is the relative velocity of object B from the perspective of object A? Assume that all units are denominated in meters.

Aug 19, 2016

$= \left(\begin{matrix}\frac{5}{8} \\ \frac{3}{4}\end{matrix}\right)$ m/s

#### Explanation:

at $t = 0$, $\vec{O A} = \vec{O B} = \left(\begin{matrix}0 \\ 0\end{matrix}\right)$

So:
$\vec{{\left(A B\right)}_{0}} = \left(\begin{matrix}0 \\ 0\end{matrix}\right)$

At $t = 8$,
$\vec{{\left(A B\right)}_{8}} = \vec{{\left(A O\right)}_{8}} + \vec{{\left(O B\right)}_{8}}$

$= - \vec{{\left(O A\right)}_{8}} + \vec{{\left(O B\right)}_{8}}$

$= - \left(\begin{matrix}0 \\ - 2\end{matrix}\right) + \left(\begin{matrix}5 \\ 4\end{matrix}\right) = \left(\begin{matrix}5 \\ 6\end{matrix}\right)$

from A's perspective $\vec{{\left(A B\right)}_{8}}$ is the displacement of B from A at $t = 8$, ie holding point A fixed.

so if

${\vec{\Delta r}}_{A B} = \left(\begin{matrix}5 \\ 6\end{matrix}\right)$ m

then
${\vec{v}}_{A B} = \frac{{\vec{\Delta r}}_{A B}}{\Delta t} = \frac{1}{8} \left(\begin{matrix}5 \\ 6\end{matrix}\right)$

$= \left(\begin{matrix}\frac{5}{8} \\ \frac{3}{4}\end{matrix}\right)$ m/s