Question

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.

Answer 1

`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`