Question

Car A has an acceleration of $2m/s^2$ due east and car B, $4m/s^2$ due North. Acceleration of car B with respect to car A is

• A. $2\sqrt{5}$ at an angle ${\tan }^{-1}\left(2\right)$ North of West.
• B. $2\sqrt{5}$ at an angle ${\tan }^{-1}\left(2\right)$ West of North.
• C. $4\sqrt{2}$ at an angle ${\tan }^{-1}\left(\frac{1}{3}\right)$ North of West.
• D. $4\sqrt{2}$ at an angle ${\tan }^{-1}\left(\frac{1}{3}\right)$ West of North.

Answer 1

The correct option is A $2\sqrt{5}$ at an angle ${\tan }^{-1}\left(2\right)$ North of West.

This is a two - dimensional motion.

Let North and East directions by positive $y-$ axis and $x-$ axis respectively.
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 $=4\hat{j}m/s^2$
And $\overrightarrow{a_A}=$ acceleration of car A $=2\hat{i}m/s^2$
$\therefore \overrightarrow{a_{BA}}=(4\hat{j}-2\hat{i})m/s^2$

$\left|\overrightarrow{a_{BA}}\right|=\sqrt{(4{)}^2+(2{)}^2}=2\sqrt{5}m/s^2$
And $a={\tan }^{-1}\left(\frac{4}{2}\right)={\tan }^{-1}\left(2\right)$