A car makes a displacement of 100 m towards east and then 200 m towards north. Find the magnitude and direction of the resultant.

$\begin{array}{l}{\overline{\mathrm{S}}}_1=100\widehat{\mathrm{i}};{\overline{\mathrm{S}}}_2=200\widehat{\mathrm{j}}\\ \therefore \overline{\mathrm{S}}={\overline{\mathrm{S}}}_1+{\overline{\mathrm{S}}}_2=100\widehat{\mathrm{i}}+200\widehat{\mathrm{j}}\\ \therefore \mathrm{S}=\left|\overline{\mathrm{S}}\right|=\sqrt{(100{)}^2+(200{)}^2}\\ =100\sqrt{5}=100\times 2.237=223.7\mathrm{m}\end{array}$
 

$\begin{array}{l}\therefore \mathrm{Tan}\mathrm{\theta }=\frac{\mathrm{y}\text{ component of displacement }}{\mathrm{x}\text{ component of displacement }}=\frac{200}{100}=2\\ \mathrm{here}, \mathrm{\theta } \mathrm{is} \mathrm{angle} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x}-\mathrm{axis}\\ \mathrm{\theta }={\mathrm{Tan}}^{-1}\left(2\right)\text{ along N of E }(\because \mathrm{i},\mathrm{j}\text{ are }+\mathrm{ve})\end{array}$