If a car A moves with acceleration $2m/s^2 $due east and car B moves $1m/s^2 $due north. What would be the acceleration of car B with respect to car A. Now , for this. The solution in my textbook is that they just took the resultant. I don’t understand why would we take the resultant of the two vectors as the value for relative velocity $ a_{BA}$.
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1$\begingroup$ Acceleration is a vector quantity and thus follows vector laws. To car A, B will seem to accelerate west at 2 m/s^2 and accelerate north at 1m/s^2 which u add by pythagoras theorem. $\endgroup$– PhysikerCommented Mar 29, 2021 at 1:42
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$\begingroup$ Is there any method like “Minimum distance of approach “ used here ? @SarthakGirdhar $\endgroup$– S.M.TCommented Mar 29, 2021 at 1:45
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$\begingroup$ Maybe there is. I havent heard that term before. $\endgroup$– PhysikerCommented Mar 29, 2021 at 2:02
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1 Answer
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Since east and north are perpendicular to each other I would like to break it into $i$ and $j$
$A_a$=2i
$A_b$=1j
$A_{b/a}=A_b-A_a$
=j-2i
To find the magnitude of this relative acceleration it seems as if we are taking the direct resultant
