Question

A political party has to start its procession in an area where wind is blowing at a speed of 41.4 km/h and party flags on the vehicles are fluttering along north-east direction. If the procession starts with a speed of 40 km/h towards north, find the direction of flags on the vehicles.

• A. south of east
• B. north of east
• C. south of east
• D. north of east

Answer 1

The correct option is C

south of east

When the procession is stationary, the flags flutters along the north - east direction.It means wind is flowing along the north-east direction.The flags will start fluttering along the direction of relative velocity of wind w.r.t.procession.As soon as procession gets into motion

As the car will move along north with velocity $v_p$,the flags will flutter in the resultant direction of wind. We can say the flag will flutter along ${\overrightarrow{v}}_{WP}$,relative velocity of wind w.r.t the procession.

${\overrightarrow{v}}_{WP}={\overrightarrow{v}}_W+(-{\overrightarrow{v}}_P)$

${\overrightarrow{v}}_P=40\hat{j}$

${\overrightarrow{v}}_W=41.4cos45\hat{i}+41.4sin45\hat{j}$

${\overrightarrow{v}}_W=\frac{41.4}{\sqrt{2}}\hat{i}+\frac{41.4}{\sqrt{2}}\hat{j}$

${\overrightarrow{v}}_{WP}={\overrightarrow{v}}_W-{\overrightarrow{v}}_P$

= $\frac{41.4}{\sqrt{2}}\hat{i}+\frac{41.4\hat{j}}{\sqrt{2}}-40\hat{j}$

${\overrightarrow{v}}_{WP}=\frac{41.4}{\sqrt{2}}\hat{i}+(\frac{41.4}{\sqrt{2}}-40)\hat{j}$

$\tan \theta =\left(\frac{v_y}{v_x}\right)=\frac{(\frac{41.4}{\sqrt{2}}-40)}{\left(\frac{41.4}{\sqrt{2}}\right)}$

$tan\theta \approx \frac{-1}{3}$

$\theta =ta{n}^{-1}\left(\frac{-1}{3}\right)$

Where $\theta$ is the angle made south of east direction.