# Objects A and B are at the origin. If object A moves to (0 ,4 ) and object B moves to (9 ,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 14, 2017

$3.14$ $\text{m/s}$

#### Explanation:

We're asked to find the relative velocity of object $B$ with respect to object $A$, given their positions after a time interval.

Let's first find the (assumed to be constant) velocity components of $A$ and $B$ with respect to the origin:

• ${v}_{A x \text{/"O) = (4"m")/(3"s}} = 1.33$ $\text{m/s}$

• ${v}_{A y \text{/"O) = (0"m")/(3"s}} = 0$ $\text{m/s}$

• ${v}_{B x \text{/"O) = (9"m")/(3"s}} = 3$ $\text{m/s}$

• ${v}_{B y \text{/"O) = (8"m")/(3"s}} = 2.67$ $\text{m/s}$

The equation here for the velocity of $B$ with respect to $A$ is given by

v_(B"/"A) = v_(B"/"O) + v_(O"/"A

Here, the velocity v_(O"/"A of the origin with respect to $A$ is simply the negative of the velocity of ${v}_{A \text{/} O}$

Splitting this up into components, we have

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

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

So,

${v}_{B x \text{/"Ax) = 3"m/s" + (-1.33"m/s}} = 1.67$ $\text{m/s}$

${v}_{B y \text{/"Ay) = 2.67"m/s" + (-0"m/s}} = 2.67$ $\text{m/s}$

The magnitude of ${\vec{v}}_{B \text{/} A}$ is given by

v_(B"/"A) = sqrt((v_(Bx"/"Ax))^2 + (v_(By"/"Ay))^2)

= sqrt((1.67"m/s")^2 + (2.67"m/s")^2) = color(red)(3.14 color(red)("m/s"

Thus, the relative speed of object $B$ relative to $A$ is $3.14$ meters per second.