Objects A and B are at the origin. If object A moves to #(6 ,2 )# and object B moves to #(-1 ,1 )# over #4 s#, what is the relative velocity of object B from the perspective of object A?

1 Answer
Jun 26, 2017

Answer:

#v_(B"/"A) = 1.77# #"m/s"#

#theta = 188^"o"# (at #t = 4# #"s"#)

Explanation:

We're asked to find the relative velocity of object #B# from the perspective of object #A#, with given displacements and times.

I'm going to assume the speed is either constant during the displacement, or we're asked to find the relative velocity at #t = 4# #"s"# with the speeds being average.

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

#v_(Ax) = overbrace((6"m"))^("x-coordinate of A")/underbrace((4"s"))_"time" = 1.50"m"/"s"#

#v_(Ay) = overbrace((2"m"))^"y-coordinate of A"/underbrace((4"s"))_"time" = 0.50"m"/"s"#

#v_(Bx) = overbrace((-1"m"))^"x-coordinate of B"/underbrace((4"s"))_"time" = -0.25"m"/"s"#

#v_(By) = overbrace((1"m"))^"y-coordinate of B"/underbrace((4"s"))_"time" = 0.25"m"/"s"#

The equation here for the relative velocity of #B# relative to #A#, which we'll denote as #vecv_(B"/"A)#, is

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

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 #A# with respect to the origin (which would be #vecv_(A"/"O)#). The expression in the equation is #vecv_(O"/"A)#; the velocity of the origin with respect to #A#.

These two expressions are opposites:

#vecv_(O"/"A) = -vecv_(A"/"O)#

And so we can rewrite the relative velocity equation as

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

Expressed in component form, the equations are

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

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

Plugging in our known values from earlier, we have

#v_(Bx"/"Ax) = (-0.25"m"/"s") - (1.50"m"/"s") = color(red)(-1.75"m"/"s"#

#v_(By"/"Ay) = (0.25"m"/"s") - (0.50"m"/"s") = color(green)(-0.25"m"/"s"#

Thus, the relative speed of #B# with respect to #A# is

#v_(B"/"A) = sqrt((color(red)(-1.75"m"/"s"))^2 + (color(green)(-0.25"m"/"s"))^2) = color(blue)(1.77# #color(blue)("m/s"#

For additional information, the angle of the relative velocity vector at #t = 4# #"s"# is

#theta = arctan((v_(By"/"Ay))/(v_(Bx"/"Ax))) = (color(green)(-0.25"m"/"s"))/(color(red)(-1.75"m"/"s")) = 8.13^"o"#

Remember that there are two angles that satisfy the inverse tangent calculation, each #180^"o"# apart. In this problem, we can see that the position of #B# relative to #A# is "down" and "to the left" coordinate-wise.

Thus, our angle should be in the third quadrant, which is "down" and "to the left" of the origin. The true angle therefore is

#8.13^"o" + 180^"o" = color(purple)(188^"o"#