A boat crosses a river from one bank A to ano | vedclass

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Question

(C) To cross the river directly along the shortest path $AB$, the boat must head upstream at an angle $	heta$ with the line $AB$ such that the compon

Vedclass · Accepted Answer

$\frac{2D}{V_B\sqrt{3}}$

Vedclass · Answer

(C) To cross the river directly along the shortest path $AB$, the boat must head upstream at an angle $	heta$ with the line $AB$ such that the component of $V_B$ along the river cancels the water speed $V_W$.Let $	heta$ be the angle with the line $AB$. Then, $V_B \sin 	heta = V_W$.Given $V_B = 2V_W$, we have $2V_W \sin 	heta = V_W$, which implies $\sin 	heta = 1/2$, so $	heta = 30^\circ$.The effective velocity of the boat across the river is $V_{eff} = V_B \cos 	heta$.Substituting $	heta = 30^\circ$, $V_{eff} = V_B \cos 30^\circ = V_B (\sqrt{3}/2)$.The time taken $t$ to cross the distance $D$ is $t = D / V_{eff}$.Therefore, $t = D / (V_B \sqrt{3} / 2) = 2D / (V_B \sqrt{3})$.

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