Objects A and B are at the origin. If object A moves to $(- 7, - 5)$ $(-7 ,-5 )$ and object B moves to $(1, - 1)$ $(1 ,-1 )$ over $8 s$ $8 s$ , 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 $(- 7, - 5)$ $(-7 ,-5 )$ and object B moves to $(1, - 1)$ $(1 ,-1 )$ over $8 s$ $8 s$ , 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 · 2016-06-16T22:06:02

$v_{AB} = \frac{\sqrt{5}}{2} \frac{m}{s}$ $v_"AB"=sqrt5/2" "m/s$

$v_{BA} = - \frac{\sqrt{5}}{2} \frac{m}{s}$ $v_"BA"=-sqrt5/2" "m/s$

Explanation:

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$when the object moves from origin to the point of a the position$ $"when the object moves from origin to the point of a the position"$ $of its will have changed$ $"of its will have changed"$
$figure above shows displacement of the object A$ $"figure above shows displacement of the object A"$

enter image source here

$the figure above shows displacement of the object B$ $"the figure above shows displacement of the object B"$

enter image source here

$the relative displacement of B from perspective of A is shown in$ $"the relative displacement of B from perspective of A is shown in "$
$figure$ $"figure"$

$the relative displacement may be calculated using Pythagoras or$ $"the relative displacement may be calculated using Pythagoras or"$
$cosine laws$ $"cosine laws"$

$Delta s=sqrt((1+7)^2+(-1+5)^2)=sqrt((8)^2+(4)^2)=sqrt(64+16)=sqrt(80)$
$Delta s=sqrt 80=4sqrt5 " "m$

$v_"AB"=(Delta s)/(delta t)$

$v_"AB"=(4sqrt5)/8$

$v_"AB"=sqrt5/2" "m/s$

$v_"BA"=-sqrt5/2" "m/s$

enter image source here

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Objects A and B are at the origin. If object A moves to $(- 7, - 5)$ $(-7 ,-5 )$ and object B moves to $(1, - 1)$ $(1 ,-1 )$ over $8 s$ $8 s$ , 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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Explanation:

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