# 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?

Mar 3, 2018

Net displacement of $A$ from origin is $\sqrt{{4}^{2} + {7}^{2}} = 8.061 m$ and the net displacement of $B$ from origin is $\sqrt{{2}^{2} + {3}^{2}} = 3.605 m$

considering that they moved with constant velocity,we get, velocity of $A$ is $\frac{8.061}{4} = 2.015 m {s}^{-} 1$ and that of $B$ is $\frac{3.605}{4} = 0.090 m {s}^{-} 1$

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 $\vec{R} = \vec{B} - \vec{A} = \vec{B} + \left(- \vec{A}\right)$

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

So,angle between $\vec{B}$ and $- \vec{A}$ is ${3.955}^{\circ}$

So,$| \vec{R} | = \sqrt{{0.90}^{2} + {2.015}^{2} + 2 \cdot 0.90 \cdot 2.015 \cdot \cos 3.955} = 2.914 m {s}^{-} 1$

Mar 3, 2018

The relative velocity is $= < \frac{3}{2} , - \frac{5}{4} > m {s}^{-} 1$

#### Explanation:

The absolute velocity of $A$ is ${v}_{A} = \frac{1}{4} < - 4 , 7 > = < - 1 , \frac{7}{4} >$

The absolute velocity of $B$ is ${v}_{B} = \frac{1}{4} < 2 , - 3 > = < \frac{1}{2} , - \frac{3}{4} >$

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

${v}_{B / A} = {v}_{B} - {v}_{A}$

$= < \frac{1}{2} , - \frac{3}{4} > \prec - 1 , \frac{7}{4} >$

$= < \frac{1}{2} + 1 , - \frac{3}{4} - \frac{7}{4} >$

$= < \frac{3}{2} , - \frac{5}{4} >$