Question

Question number 6


6. An airplane is flying with velocity of 70 km/h in north east direction. The wind is blowing 30 km/h from north to south. What is the resultant displacement of the airplane in 4 hours.
(Ans 212 km making angle $21.5°$ with east )

Answer 1

Dear student

$$\begin{gathered} Fromthediagramasshowninthefigure \\ {\overrightarrow{V}}_{plane}={\overrightarrow{V}}_{plane,air}+{\overrightarrow{V}}_{air,ground} \\ {\overrightarrow{V}}_{plane}=\frac{70}{\sqrt{2}}\hat{i}+\frac{70}{\sqrt{2}}\hat{j}-30\hat{j} \\ displacementin4houris \\ \overrightarrow{s}={\overrightarrow{V}}_{plane}\times t \\ \overrightarrow{s}=(\frac{70}{\sqrt{2}}\hat{i}+\frac{70}{\sqrt{2}}\hat{j}-30\hat{j})\times 4 \\ \overrightarrow{s}=(\frac{70}{\sqrt{2}}\hat{i}+\frac{70}{\sqrt{2}}\hat{j}-30\hat{j})\times 4 \\ \overrightarrow{s}=\frac{70\hat{i}+(70-30\sqrt{2})\times 4}{\sqrt{2}} \\ magnitudeofthedisplacementis \\ s=\left(\sqrt{{70}^2+(70-30\sqrt{2}{)}^2}\right)\times 2\sqrt{2}=212.79km \\ anglemadebythedisplacementwiththeeastis \\ \tan \theta =\frac{(70-30\sqrt{2})}{70} \\ \theta ={\tan }^{-1}\frac{(70-30\sqrt{2})}{70}=21.49° \\ Regards \end{gathered}$$