Question

Objects A and B are at the origin. If object A moves to `(9 ,7 )` and object B moves to `(-8 ,-4 )` 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

`v_(B"/"A) = 6.75` `"m/s"`

`theta = 212.9^"o"` (at `t = 3` `"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 = 3` `"s"` with the speeds being average.

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

`v_(Ax) = overbrace((9"m"))^("x-coordinate of A")/underbrace((3"s"))_"time" = 3"m"/"s"`

`v_(Ay) = overbrace((7"m"))^"y-coordinate of A"/underbrace((3"s"))_"time" = 2.33"m"/"s"`

`v_(Bx) = overbrace((-8"m"))^"x-coordinate of B"/underbrace((3"s"))_"time" = -2.67"m"/"s"`

`v_(By) = overbrace((-4"m"))^"y-coordinate of B"/underbrace((3"s"))_"time" = -1.33"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) = (-2.67"m"/"s") - (3"m"/"s") = color(red)(-5.67"m"/"s"`

`v_(By"/"Ay) = (-1.33"m"/"s") - (2.33"m"/"s") = color(green)(-3.67"m"/"s"`

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

`v_(B"/"A) = sqrt((color(red)(-5.67"m"/"s"))^2 + (color(green)(-3.67"m"/"s"))^2) = color(blue)(6.75` `color(blue)("m/s"`

For additional information, the angle of the relative velocity vector at `t = 3` `"s"` is

`theta = arctan((v_(By"/"Ay))/(v_(Bx"/"Ax))) = (color(green)(-3.67"m"/"s"))/(color(red)(-5.67"m"/"s")) = 32.9^"o" + 180^"o" = color(purple)(212.9^"o"`

The `180^"o"` was added to fix the calculator error; It is "down" and "to the left" of `A` (negative values for `x`- and `y`-velocities), so the angle can't be `-45^"o"`, where it is "up" and "to the right" (calculator simplifies two negative values as positive).