Car A has an acceleration of `2m//s^2` due east and car B,`4 m//s^2.` due north. What is the acceleration of car B with respect to car A?

Explanation

It is a two dimensional motion. Therefore,
`a_(BA)`= acceleration of car B with respect to car A
`=a_(B)-a_(A)`

Here, `a_(B)` = acceleration of car, `B = 4 m//s^(2)` (due north)
and `a_(A)` = acceleration of car `A = 2 m//s^(2)` (due east)
`|a_(BA)|=sqrt((4)^(2)+(2)^(2))`
`=2sqrt(5)m//s^(2)`
and `alpha = tan^(-1)((4)/(2))=tan^(-1)(2)`
Thus, `a_(BA)` is `2sqrt(5)m//s^(2)` at an angle of `alpha = tan^(-1)(2)` from west towards north.