Question

Objects A and B are at the origin. If object A moves to `(-4 ,7 )` and object B moves to `(2 ,-3 )` over `4 s`, what is the relative velocity of object B from the perspective of object A?

Answer 1

Net displacement of `A` from origin is `sqrt(4^2 + 7^2)=8.061 m` and the net displacement of `B` from origin is `sqrt(2^2+3^2)=3.605m`

considering that they moved with constant velocity,we get, velocity of `A` is `8.061/4=2.015 ms^-1` and that of `B` is `3.605/4=0.090 ms^-1`

So,we have got velocity vector of `A` (let, `vec A`) and velocity vector of `B` (let, `vec B`)

So,reative velocity of `B` w.r.t `A` is `vec R = vec B - vec A=vec B + (-vec A)`

Now,the angle between the two vectors is found as follows,

So,angle between `vec B` and `-vec A` is `3.955^@`

So,`|vec R| = sqrt(0.90^2 + 2.015^2 + 2*0.90*2.015*cos 3.955)=2.914 ms^-1`

Answer 2

The relative velocity is `=<3/2, -5/4> ms^-1`

Explanation:

![www.slideshare.net]()

The absolute velocity of `A` is `v_A=1/4 <-4,7> = <-1, 7/4>`

The absolute velocity of `B` is `v_B=1/4 <2,-3> = <1/2,-3/4>`

The relative velocity of `B` with respect to `A` is

`v_(B//A)=v_B - v_A`

`=<1/2,-3/4> -<-1, 7/4>`

`= <1/2+1, -3/4-7/4>`

`= <3/2, -5/4>`