Question

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

Answer 1

It is a two dimensional motion. Therefore,

${\overrightarrow{a}}_{BA}$ = acceleration of car B with respect to car A

$={\overrightarrow{a}}_B-{\overrightarrow{a}}_A$

Here, ${\overrightarrow{a}}_B$ = acceleration of car

$B=4m/s^2$ (due north)

and ${\overrightarrow{a}}_A$ = acceleration of car $A=2m/s^2$ (due east)

$\left|{\overrightarrow{a}}_{BA}\right|=\sqrt{{\left(4\right)}^2+{\left(2\right)}^2}=2\sqrt{5}m/s^2$

and $\alpha ={\tan }^{-1}\left(\frac{4}{2}\right)={\tan }^{-1}\left(2\right)$

Thus, ${\overrightarrow{a}}_{BA}is2\sqrt{5}m/s^2$ at an angle of $\alpha ={\tan }^{-1}\left(2\right)$ from west towards north.