Question

Objects A and B are at the origin. If object A moves to `(0 ,4 )` and object B moves to `(9 ,8 )` 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

`3.14` `"m/s"`

Explanation:

We're asked to find the relative velocity of object `B` with respect to object `A`, given their positions after a time interval.

Let's first find the (assumed to be constant) velocity components of `A` and `B` with respect to the origin:

• `v_(Ax"/"O) = (4"m")/(3"s") = 1.33` `"m/s"`

• `v_(Ay"/"O) = (0"m")/(3"s") = 0` `"m/s"`

• `v_(Bx"/"O) = (9"m")/(3"s") = 3` `"m/s"`

• `v_(By"/"O) = (8"m")/(3"s") = 2.67` `"m/s"`

The equation here for the velocity of `B` with respect to `A` is given by

`v_(B"/"A) = v_(B"/"O) + v_(O"/"A`

Here, the velocity `v_(O"/"A` of the origin with respect to `A` is simply the negative of the velocity of `v_(A"/"O)`

Splitting this up into components, we have

`v_(Bx"/"Ax) = v_(Bx"/"O) + v_(O"/"Ax)`

`v_(By"/"Ay) = v_(By"/"O) + v_(O"/"Ay)`

So,

`v_(Bx"/"Ax) = 3"m/s" + (-1.33"m/s") = 1.67` `"m/s"`

`v_(By"/"Ay) = 2.67"m/s" + (-0"m/s") = 2.67` `"m/s"`

The magnitude of `vecv_(B"/"A)` is given by

`v_(B"/"A) = sqrt((v_(Bx"/"Ax))^2 + (v_(By"/"Ay))^2)`

`= sqrt((1.67"m/s")^2 + (2.67"m/s")^2) = color(red)(3.14` `color(red)("m/s"`

Thus, the relative speed of object `B` relative to `A` is `3.14` meters per second.