# Objects A and B are at the origin. If object A moves to (-2 ,2 ) and object B moves to (7 ,-5 ) 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.

May 25, 2017

#### Answer:

$3.80 \text{m"/"s}$, ${37.9}^{o}$ south of the horizontal

#### Explanation:

The components of the velocity of object A are

v_(Ax) = (-2"m")/(3 "s") = -0.667 "m"/"s"

v_(Ay) = (2"m")/(3 "s") = 0.667 "m"/"s"

and object B

v_(Bx) = (7"m")/(3 "s") = 2.33 "m"/"s"

v_(By) = (-5 "m")/(3"s") = -1.67 "m"/"s"

We're trying to find the velocity of object B with respect to object A. We'll call this velocity ${v}_{B \text{/} A}$, the velocity of object B with respect to the origin ${v}_{B \text{/} O}$, and the velocity of object A relative to the origin ${v}_{A \text{/} O}$.

The equation for relative velocity, using these frames of reference, is

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

And remembering that ${\vec{v}}_{A \text{/"B) = -vec v_(B"/} A}$, this equation becomes

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

Since we're solving for ${\vec{v}}_{B \text{/} A}$,

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

Or, in terms of components,

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

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

Now, let's plug in our known velocity components:

v_(Bx"/"Ax) = 2.33"m"/"s" - -0.667"m"/"s" = 3.00 "m"/"s"

v_(By"/"Ay) = -1.67"m"/"s" - 0.667 "m"/"s" = -2.33 "m"/"s"

Thus, the magnitude of the velocity of object B with respect to object A is

v_(B"/"A) = sqrt((3.00"m"/"s")^2 + (-2.33"m"/"s")^2) = color(red)(3.80 "m"/"s"

And the direction of the velocity of object B with respect to object A is

phi = arctan ((-2.33"m"/"s")/(3.00 "m"/"s")) = color(blue)(-37.9^o

There are technically two directions that satisfy this, each opposite to each other in a coordinate plane (the other angle is thus ${142}^{o}$). We have to choose the one that points from object A to object B, which if you were to plot their coordinates, would indeed be $- {37.9}^{o}$.

The direction is more easily found in this case, as we also could have just found the inverse tangent of the differences in the $y$-positions over the difference in the $x$-position of the two objects:

$\phi = \arctan \left(\left(2 \text{m"--5"m")/(-2"m"-7"m}\right)\right) = - {37.9}^{o}$