# Objects A and B are at the origin. If object A moves to (-3 ,5 ) and object B moves to (-6 ,8 ) 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 22, 2017

v_(B"/"A) = 1.41"m"/"s"

#### Explanation:

We're asked to find the relative velocity of object $B$ from the perspective of object $A$, with given displacements and times.

I'm going to assume the speed is either constant during the displacement, or we're asked to find the relative velocity at $t = 3$ $\text{s}$ with the speeds being average.

What we can do first is find the components of the velocity of each object:

v_(Ax) = overbrace((-3"m"))^("x-coordinate of A")/underbrace((3"s"))_"time" = -1"m"/"s"

v_(Ay) = overbrace((5"m"))^"y-coordinate of A"/underbrace((3"s"))_"time" = 1.67"m"/"s"

v_(Bx) = overbrace((-6"m"))^"x-coordinate of B"/underbrace((3"s"))_"time" = -2"m"/"s"

v_(By) = overbrace((8"m"))^"y-coordinate of B"/underbrace((3"s"))_"time" = 2.67"m"/"s"

The equation here for the relative velocity of $B$ relative to $A$, which we'll denote as ${\vec{v}}_{B \text{/} A}$, is

${\vec{v}}_{B \text{/"A) =vec v_(B"/"O) + vecv_(O"/} A}$

If you're wondering why this is the equation, picture the two velocities you're adding like fractions being multiplied, where ${\vec{v}}_{B \text{/} O}$ is $\text{B"/"O}$, and ${\vec{v}}_{O \text{/} A}$ is $\text{O"/"A}$:

The velocity we want to find (${\vec{v}}_{B \text{/} A}$), which as a fraction becomes $\text{B"/"A}$, is the product of the two other fractions, because the $\text{O}$s cross-cancel:

$\text{B"/"A" = "B"/(cancel("O")) xx (cancel("O"))/"A}$

And so written similarly the equation is

${\vec{v}}_{B \text{/"A) =vec v_(B"/"O) + vecv_(O"/} A}$

Notice that we calculated earlier the velocity of $A$ with respect to the origin (which would be ${\vec{v}}_{A \text{/} O}$). The expression in the equation is ${\vec{v}}_{O \text{/} A}$; the velocity of the origin with respect to $A$.

These two expressions are opposites:

${\vec{v}}_{O \text{/"A) = -vecv_(A"/} O}$

And so we can rewrite the relative velocity equation as

${\vec{v}}_{B \text{/"A) =vec v_(B"/"O) - vecv_(A"/} O}$

Expressed in component form, the equations are

${v}_{B x \text{/"Ax) = v_(Bx"/"O) - v_(Ax"/} O}$

${v}_{B y \text{/"Ay) = v_(By"/"O) - v_(Ay"/} O}$

Plugging in our known values from earlier, we have

v_(Bx"/"Ax) = (-2"m"/"s") - (-1"m"/"s") = color(red)(-1"m"/"s"

v_(By"/"Ay) = (2.67"m"/"s") - (1.67"m"/"s") = color(green)(1"m"/"s"

Thus, the relative speed of $B$ with respect to $A$ is

v_(B"/"A) = sqrt((color(red)(-1"m"/"s"))^2 + (color(green)(1"m"/"s"))^2) = color(blue)(1.41"m"/"s"

For additional information, the angle of the relative velocity vector at $t = 3$ $\text{s}$ is

theta = arctan((v_(By"/"Ay))/(v_(Bx"/"Ax))) = (color(green)(1"m"/"s"))/(color(red)(-1"m"/"s")) = -45^"o" + 180^"o" = color(purple)(135^"o"

The ${180}^{\text{o}}$ was added to fix the calculator error; It is "up" and "to the left" of $A$, so the angle can't be $- {45}^{\text{o}}$, where it is "down" and "to the right".