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Car A has an acceleration of 2m//s^2 due...

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?

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The correct Answer is:
A
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)=2sqrt5 m//s^2`
and `alpha=tan^-1(4/2)=tan^-1(2)`
Thus, `a_(BA)` is `2sqrt5 m//s^2` at an angle of `alpha=tan^-1(2)` west towards north.

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