A boy runs on a horizontal road with a speed | vedclass

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(A) Given:Velocity of boy, $\overrightarrow{v_b} = 4 \ m \ s^{-1}$ and velocity of rain, $\overrightarrow{v_r} = 8 \ m \ s^{-1}$.The velocity of rain

Vedclass · Accepted Answer

$(1-\sqrt{2})^2$

Vedclass · Answer

(A) Given:Velocity of boy, $\overrightarrow{v_b} = 4 \ m \ s^{-1}$ and velocity of rain, $\overrightarrow{v_r} = 8 \ m \ s^{-1}$.The velocity of rain with respect to the boy is $\overrightarrow{v_{rb}} = \overrightarrow{v_r} - \overrightarrow{v_b}$.Let the vertical direction be the $y$-axis and the horizontal direction be the $x$-axis.Rain velocity components: $v_{rx} = v_r \sin(45^{\circ}) = 8 	imes \frac{1}{\sqrt{2}} = 4\sqrt{2} \ m \ s^{-1}$ and $v_{ry} = v_r \cos(45^{\circ}) = 8 	imes \frac{1}{\sqrt{2}} = 4\sqrt{2} \ m \ s^{-1}$.When the boy runs from west to east $(\overrightarrow{v_b} = 4 \hat{i})$:$\overrightarrow{v_{rb}} = (v_{rx} \hat{i} - v_{ry} \hat{j}) - (4 \hat{i}) = (4\sqrt{2} - 4) \hat{i} - 4\sqrt{2} \hat{j}$.$	an 	heta = \frac{|v_{rb,x}|}{|v_{rb,y}|} = \frac{4\sqrt{2} - 4}{4\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$.When the boy runs from east to west $(\overrightarrow{v_b} = -4 \hat{i})$:$\overrightarrow{v_{rb}} = (v_{rx} \hat{i} - v_{ry} \hat{j}) - (-4 \hat{i}) = (4\sqrt{2} + 4) \hat{i} - 4\sqrt{2} \hat{j}$.$	an \alpha = \frac{|v_{rb,x}|}{|v_{rb,y}|} = \frac{4\sqrt{2} + 4}{4\sqrt{2}} = \frac{\sqrt{2} + 1}{\sqrt{2}}$.Therefore, the ratio $\frac{	an 	heta}{	an \alpha} = \frac{(\sqrt{2} - 1)/\sqrt{2}}{(\sqrt{2} + 1)/\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$.Rationalizing the denominator: $\frac{(\sqrt{2} - 1)(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{2 + 1 - 2\sqrt{2}}{2 - 1} = 3 - 2\sqrt{2} = (\sqrt{2} - 1)^2$.

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