# Objects A and B are at the origin. If object A moves to (-3 ,1 ) and object B moves to (-6 ,3 ) 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.

Jun 6, 2016

From A's perspective B would increase the distance from A with a speed $\frac{\sqrt{13}}{3}$ m/sec, always moving in the same direction.

#### Explanation:

We have to assume that A and B objects move with constant speeds, each in its own direction, from the perspective of an observer positioned at the origin of coordinates.
So, every second each object covers the same distance.
Assume, object A's positions at seconds $1 ,$2, etc. are ${A}_{1}$, ${A}_{2}$ etc. and positions of object B at these moments are ${B}_{1}$, ${B}_{2}$ etc.

Obviously, all points ${A}_{n}$ lie on the same line originated at $O \left(0 , 0\right)$ as well as all ${B}_{n}$.
Also, since the speed of each object is constant,
$O {A}_{1} = {A}_{1} {A}_{2} = {A}_{2} {A}_{3} = \ldots$
$O {B}_{1} = {B}_{1} {B}_{2} = {B}_{2} {B}_{3} = \ldots$
A simple theorem of geometry tells that triangles $\Delta {A}_{1} {B}_{1} O$ and $\Delta {A}_{n} {B}_{n} O$ are similar for any $n$ and, therefore,
${A}_{1} {B}_{1} | | {A}_{2} {B}_{2} | | {A}_{3} {B}_{3} \ldots$
That means that vector expressing the relative velocity of object B from the perspective of object A has always the same direction and magnitude.

Let $U$ be a vector of velocity (of constant magnitude and direction) of object A, its X- and Y-components are ${U}_{X}$ and ${U}_{Y}$.

Analogously, $V$ is a vector of velocity (also of constant magnitude and direction) of the object B with components ${V}_{X}$ and ${V}_{Y}$.

If i and j are unit vectors along, correspondingly, the X-axis and the Y-axis, object A's position as a function of time $t$ can be described as
${U}_{X} \cdot t \cdot$ i $+ {U}_{Y} \cdot t \cdot$ j .
Analogously, object B's position is described by as
${V}_{X} \cdot t \cdot$ i $+ {V}_{Y} \cdot t \cdot$ j .

Relative position of object B from the perspective of object A (that is if we imagine object A to be an origin of a system of coordinates moving with it) is expressed as a difference between the two vectors above, that is

(V_X*t* i $+ {V}_{Y} \cdot t \cdot$ j) - (U_X*t* i $+ {U}_{Y} \cdot t \cdot$ j) =

$= \left[\left({V}_{X} - {U}_{X}\right) \cdot t\right]$ i $+ \left[\left({V}_{Y} - {U}_{Y}\right) \cdot t\right]$ j

The above implies that object B moves relatively to object A with constant speed towards the same direction. Relative velocity is a vector $\left({V}_{X} - {U}_{X} , {V}_{Y} - {U}_{Y}\right)$

Obviously, in meters per second ($\frac{m}{\sec}$),
${U}_{X} = - \frac{3}{3} = - 1$
${U}_{Y} = \frac{1}{3}$
${V}_{X} = - \frac{6}{3} = - 2$
${V}_{Y} = \frac{3}{3} = 1$

The relative movement of object B is described by a vector of velocity $W = V - U$ with components
${W}_{X} = {V}_{X} - {U}_{X} = - 2 - \left(- 1\right) = - 1$
${W}_{Y} = {V}_{Y} - {U}_{Y} = 1 - \frac{1}{3} = \frac{2}{3}$

Knowing the components of this vector of relative velocity of object B from the perspective of object A, we can easily determine the magnitude:
$| W | = \sqrt{{W}_{X}^{2} + {W}_{Y}^{2}} = \sqrt{{\left(- 1\right)}^{2} + {\left(\frac{2}{3}\right)}^{2}} = \frac{\sqrt{13}}{3}$

So, from A's perspective B would increase the distance from A with a speed $\frac{\sqrt{13}}{3}$ m/sec.