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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?

Text Solution

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It is a two dimensional motion. Therefore `veca BA` =acceleration of car B with respect to car A

`=vecaB-becaA`
Here, `vecaB`=acceleration of car `B=4m//s^2` (due north)
and `vecaA` = acceleration of car `A=2m//s^2` (due east)
`|veca BA|=sqrt((4)^2+(2)^2)=2sqrt5m//s^2`
and `prop=tan^(-1)(4/2)=tan^(-1)(2)`

Thus `vec BA` is `2sqrt5m//s^2` at an angle of `prop=tan^(-1)(2)` from west towards north.
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