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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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    $\begingroup$ 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. $\endgroup$ – LoneAcademic Mar 29 at 1:42
  • $\begingroup$ Is there any method like “Minimum distance of approach “ used here ? @SarthakGirdhar $\endgroup$ – Srijan M.T Mar 29 at 1:45
  • $\begingroup$ Maybe there is. I havent heard that term before. $\endgroup$ – LoneAcademic Mar 29 at 2:02
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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

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