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.

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_(Ax"/"O) = (4"m")/(3"s") = 1.33# #"m/s"#

• #v_(Ay"/"O) = (0"m")/(3"s") = 0# #"m/s"#

• #v_(Bx"/"O) = (9"m")/(3"s") = 3# #"m/s"#

• #v_(By"/"O) = (8"m")/(3"s") = 2.67# #"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"/"O)#

Splitting this up into components, we have

#v_(Bx"/"Ax) = v_(Bx"/"O) + v_(O"/"Ax)#

#v_(By"/"Ay) = v_(By"/"O) + v_(O"/"Ay)#

So,

#v_(Bx"/"Ax) = 3"m/s" + (-1.33"m/s") = 1.67# #"m/s"#

#v_(By"/"Ay) = 2.67"m/s" + (-0"m/s") = 2.67# #"m/s"#

The magnitude of #vecv_(B"/"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.