# Objects A and B are at the origin. If object A moves to (6 ,2 ) and object B moves to (-1 ,1 ) over 4 s, what is the relative velocity of object B from the perspective of object A?

Jun 26, 2017

${v}_{B \text{/} A} = 1.77$ $\text{m/s}$

$\theta = {188}^{\text{o}}$ (at $t = 4$ $\text{s}$)

#### 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 $t = 4$ $\text{s}$ with the speeds being average.

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

v_(Ax) = overbrace((6"m"))^("x-coordinate of A")/underbrace((4"s"))_"time" = 1.50"m"/"s"

v_(Ay) = overbrace((2"m"))^"y-coordinate of A"/underbrace((4"s"))_"time" = 0.50"m"/"s"

v_(Bx) = overbrace((-1"m"))^"x-coordinate of B"/underbrace((4"s"))_"time" = -0.25"m"/"s"

v_(By) = overbrace((1"m"))^"y-coordinate of B"/underbrace((4"s"))_"time" = 0.25"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) = (-0.25"m"/"s") - (1.50"m"/"s") = color(red)(-1.75"m"/"s"

v_(By"/"Ay) = (0.25"m"/"s") - (0.50"m"/"s") = color(green)(-0.25"m"/"s"

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

v_(B"/"A) = sqrt((color(red)(-1.75"m"/"s"))^2 + (color(green)(-0.25"m"/"s"))^2) = color(blue)(1.77 color(blue)("m/s"

For additional information, the angle of the relative velocity vector at $t = 4$ $\text{s}$ is

theta = arctan((v_(By"/"Ay))/(v_(Bx"/"Ax))) = (color(green)(-0.25"m"/"s"))/(color(red)(-1.75"m"/"s")) = 8.13^"o"

Remember that there are two angles that satisfy the inverse tangent calculation, each ${180}^{\text{o}}$ apart. In this problem, we can see that the position of $B$ relative to $A$ is "down" and "to the left" coordinate-wise.

Thus, our angle should be in the third quadrant, which is "down" and "to the left" of the origin. The true angle therefore is

${8.13}^{\text{o" + 180^"o" = color(purple)(188^"o}}$