Question

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?

Answer 1

`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"`