# Objects A and B are at the origin. If object A moves to #(9 ,7 )# and object B moves to #(-8 ,-4 )# 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.

##### 1 Answer

#### Answer:

#### Explanation:

We're asked to find the *relative velocity* of object

I'm going to assume the speed is either constant during the displacement, or we're asked to find the relative velocity at *average*.

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

The equation here for the relative velocity of

If you're wondering why this is the equation, picture the two velocities you're adding like fractions being multiplied, where

#vecv_(B"/"O)# is#"B"/"O"# , and#vecv_(O"/"A)# is#"O"/"A"# :The velocity we want to find (

#vecv_(B"/"A)# ), which as a fraction becomes#"B"/"A"# , is the product of the two other fractions, because the#"O"# s cross-cancel:

#"B"/"A" = "B"/(cancel("O")) xx (cancel("O"))/"A"# And so written similarly the equation is

#vecv_(B"/"A) =vec v_(B"/"O) + vecv_(O"/"A)#

Notice that we calculated earlier the velocity of *with respect to the origin* (which would be *the velocity of the origin with respect to #A#*.

These two expressions are **opposites**:

And so we can rewrite the relative velocity equation as

Expressed in component form, the equations are

Plugging in our known values from earlier, we have

Thus, the relative speed of

For additional information, the angle of the relative velocity vector at

#t = 3# #"s"# is

#theta = arctan((v_(By"/"Ay))/(v_(Bx"/"Ax))) = (color(green)(-3.67"m"/"s"))/(color(red)(-5.67"m"/"s")) = 32.9^"o" + 180^"o" = color(purple)(212.9^"o"# The

#180^"o"# was added to fix the calculator error; It is "down" and "to the left" of#A# (negative values for#x# - and#y# -velocities), so the angle can't be#-45^"o"# , where it is "up" and "to the right" (calculator simplifies two negative values as positive).