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(B) The velocity of the boat in still water is $v_b = 27 \ km \ h^{-1}$.The angle with the river flow is $	heta = 150^{\circ}$.The component of the b

Vedclass · Accepted Answer

$112.5 \ m$

Vedclass · Answer

(B) The velocity of the boat in still water is $v_b = 27 \ km \ h^{-1}$.The angle with the river flow is $	heta = 150^{\circ}$.The component of the boat's velocity perpendicular to the river flow is $v_{\perp} = v_b \sin(150^{\circ}) = 27 	imes \sin(150^{\circ}) = 27 	imes \sin(180^{\circ} - 30^{\circ}) = 27 	imes \sin(30^{\circ}) = 27 	imes 0.5 = 13.5 \ km \ h^{-1}$.To convert this to $m \ s^{-1}$, we multiply by $\frac{5}{18}$:$v_{\perp} = 13.5 	imes \frac{5}{18} = 3.75 \ m \ s^{-1}$.The time taken to cross the river is $t = 0.5 \ 	ext{minute} = 30 \ s$.The width of the river $d$ is given by $d = v_{\perp} 	imes t$.$d = 3.75 \ m \ s^{-1} 	imes 30 \ s = 112.5 \ m$.

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