Objects A and B are at the origin. If object A moves to $(- 4, - 9)$ $(-4 ,-9 )$ and object B moves to $(1, 8)$ $(1 ,8 )$ 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 $(- 4, - 9)$ $(-4 ,-9 )$ and object B moves to $(1, 8)$ $(1 ,8 )$ 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 · 2017-12-15T21:11:08

The relative velocity is $= < \frac{5}{8}, \frac{17}{8} > m s^{- 1}$ $= < 5/8, 17/8> ms^-1$

Explanation:

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The absolute velocity of $A$ $A$ is $v_{A} = \frac{1}{8} < - 4, - 9 > = < - \frac{1}{2}, - \frac{9}{8} > m s^{- 1}$ $v_A=1/8 <-4,-9> = < -1/2,-9/8> ms^-1$

The absolute velocity of $B$ $B$ is $v_{B} = \frac{1}{8} < 1, 8 > = < \frac{1}{8}, 1 > m s^{- 1}$ $v_B=1/8 <1,8> = < 1/8,1> ms^-1$

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

$v_{B / A} = v_{B} - v_{A} = < \frac{1}{8}, 1 > - < - \frac{1}{2}, - \frac{9}{8} >$ $v_(B//A)=v_B-v_A= < 1/8,1> - < -1/2,-9/8>$

$= < \frac{1}{8} + \frac{1}{2}, 1 + \frac{9}{8} >$ $= <1/8+1/2, 1+9/8 >$

$= < \frac{5}{8}, \frac{17}{8} > m s^{- 1}$ $= < 5/8, 17/8> ms^-1$

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Objects A and B are at the origin. If object A moves to $(- 4, - 9)$ $(-4 ,-9 )$ and object B moves to $(1, 8)$ $(1 ,8 )$ 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 $(- 4, - 9)$ $(-4 ,-9 )$ and object B moves to $(1, 8)$ $(1 ,8 )$ 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.