Objects A and B are at the origin. If object A moves to (-2 ,2 )(2,2) and object B moves to (7 ,-5 )(7,5) over 3 s3s, 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

3.80 "m"/"s"3.80ms, 37.9^o37.9o south of the horizontal

Explanation:

The components of the velocity of object A are

v_(Ax) = (-2"m")/(3 "s") = -0.667 "m"/"s"vAx=2m3s=0.667ms

v_(Ay) = (2"m")/(3 "s") = 0.667 "m"/"s"vAy=2m3s=0.667ms

and object B

v_(Bx) = (7"m")/(3 "s") = 2.33 "m"/"s"vBx=7m3s=2.33ms

v_(By) = (-5 "m")/(3"s") = -1.67 "m"/"s"vBy=5m3s=1.67ms

We're trying to find the velocity of object B with respect to object A. We'll call this velocity v_(B"/"A)vB/A, the velocity of object B with respect to the origin v_(B"/"O)vB/O, and the velocity of object A relative to the origin v_(A"/"O)vA/O.

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

vec v_(A"/"O) = vec v_(A"/"B) + vec v_(B"/"O)vA/O=vA/B+vB/O

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

vec v_(A"/"O) = vec v_(B"/"O) - vec v_(B"/"A)vA/O=vB/OvB/A

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

vec v_(B"/"A) = vec v_(B"/"O) - vec v_(A"/"O)vB/A=vB/OvA/O

Or, in terms of components,

v_(Bx"/"Ax) = v_(Bx"/"O) - v_(Ax"/"O)vBx/Ax=vBx/OvAx/O

v_(By"/"Ay) = v_(By"/"O) - v_(Ay"/"O)vBy/Ay=vBy/OvAy/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"vBx/Ax=2.33ms0.667ms=3.00ms

v_(By"/"Ay) = -1.67"m"/"s" - 0.667 "m"/"s" = -2.33 "m"/"s"vBy/Ay=1.67ms0.667ms=2.33ms

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"vB/A=(3.00ms)2+(2.33ms)2=3.80ms

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^oϕ=arctan(2.33ms3.00ms)=37.9o

There are technically two directions that satisfy this, each opposite to each other in a coordinate plane (the other angle is thus 142^o142o). 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^o37.9o.

The direction is more easily found in this case, as we also could have just found the inverse tangent of the differences in the yy-positions over the difference in the xx-position of the two objects:

phi = arctan ((2"m"--5"m")/(-2"m"-7"m")) = -37.9^oϕ=arctan(2m5m2m7m)=37.9o