Objects A and B are at the origin. If object A moves to $(- 2, 7)$ $(-2 ,7 )$ and object B moves to $(6, - 5)$ $(6 ,-5 )$ over $3 s$ $3 s$ , what is the relative velocity of object B from the perspective of object A? Assume that all units are denominated in meters.

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Objects A and B are at the origin. If object A moves to $(- 2, 7)$ $(-2 ,7 )$ and object B moves to $(6, - 5)$ $(6 ,-5 )$ over $3 s$ $3 s$ , what is the relative velocity of object B from the perspective of object A? Assume that all units are denominated in meters.

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Answer 1 · 2016-03-05T09:35:32

Initially both A and B were at origin O (0,0)
after 3sec A reaches at (-2,7) and B at (6,-5)
So displacement vector of A = $- 2 ˆ i + 7 ˆ j$ $-2hati+7hatj$ m
velocity vector of A, ${\to V}_{A} = \frac{1}{3} (- 2 ˆ i + 7 ˆ j)$ $vecV_A=1/3(-2hati+7hatj)$ m/s
displacement vector of B = $6 ˆ i - 5 ˆ j$ $6hati-5hatj$ m
velocity vector of B, ${\to V}_{B} = \frac{1}{3} (6 ˆ i - 5 ˆ j)$ $vecV_B=1/3(6hati-5hatj)$ m/s
So the relative velocity of B from the perspective of A is given by
${\to V}_{B A} = {\to V}_{B} - {\to V}_{A} = \frac{1}{3} ((6 + 2) ˆ i - (5 + 7) ˆ j)$ $vecV_(BA) =vecV_B-vecV_A =1/3((6+2)hati-(5+7)hatj)$
$= \frac{1}{3} (8 ˆ i - 12 ˆ j)$ $=1/3(8hati-12hatj)$ m/s
$∣ ∣ ∣ {\to V}_{B A} ∣ ∣ ∣ = \frac{1}{3} \sqrt{8^{2} + 12^{2}} = \frac{\sqrt{208}}{3} = 4.81 m s^{- 1}$ $|vecV_(BA)| =1/3sqrt(8^2+12^2)=sqrt208/3=4.81ms^-1$

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Objects A and B are at the origin. If object A moves to $(- 2, 7)$ $(-2 ,7 )$ and object B moves to $(6, - 5)$ $(6 ,-5 )$ over $3 s$ $3 s$ , what is the relative velocity of object B from the perspective of object A? Assume that all units are denominated in meters.

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Objects A and B are at the origin. If object A moves to $(- 2, 7)$ $(-2 ,7 )$ and object B moves to $(6, - 5)$ $(6 ,-5 )$ over $3 s$ $3 s$ , what is the relative velocity of object B from the perspective of object A? Assume that all units are denominated in meters.