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.

1 Answer
May 25, 2017

Answer:

#3.80 "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"/"A)#, the velocity of object B with respect to the origin #v_(B"/"O)#, and the velocity of object A relative to the origin #v_(A"/"O)#.

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

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

And remembering that #vec v_(A"/"B) = -vec v_(B"/"A)#, this equation becomes

#vec v_(A"/"O) = vec v_(B"/"O) - vec v_(B"/"A)#

Since we're solving for #vec v_(B"/"A)#,

#vec v_(B"/"A) = vec v_(B"/"O) - vec v_(A"/"O)#

Or, in terms of components,

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

#v_(By"/"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 ((2"m"--5"m")/(-2"m"-7"m")) = -37.9^o#