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  • A butterfly is flying with a velocity  in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is : &nbsp

A butterfly is flying with a velocity 4\sqrt{2} in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is :
   
Option: 1 3\; m
Option: 2 20\; m
Option: 3 12\sqrt{2}\; m
Option: 4 15\; m

Answers (1)

best_answer

V_{B}=4 \sqrt{2} \mathrm{~m} / \mathrm{s} \text { (along } N-E \text { direction) }

\begin{aligned} &\text { Velocity of Butterly}= \bar{\text{V}}_{\text{B}}=V_{B} \cos 45\; \; \hat{i}+V_{B} \sin 45(\hat{\jmath}) \\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =4\left(\frac{m}{s}\right) \hat{i}+4\left(\frac{m}{s}\right) \hat{j} \end{aligned}

\begin{aligned} \vec{V}_{\text {wind }} &=1(-\hat{\jmath}) \\ \vec V_{\text {Resultant }} &=\vec V_{\text {Butterfly }}+\vec{V}_{\text {wind }} \\ &=4(\hat{i})+3 \hat{\jmath} \end{aligned}

\begin{aligned} \left|\bar{V}_{\text {Resullant }}\right| &=\sqrt{3^{2}+4^{2}} \\ &=5 \end{aligned}

\begin{aligned} \tan \theta &=\frac{3}{4} \quad \theta=37^{\circ} \\ S_{\text {resultant }} &=\left(V_{\text {Resultant }}\right) \times t \\ &=5 \times 3= 15 \mathrm{~m} \end{aligned}

The correct option is (4).

Posted by

vishal kumar

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