Question
A boat crosses a river from part $A$ to part $B$ which are just on opposite side. The speed of the water is $v_{w}$ and that of boat is $v_{h}$ relative to still water. Assume $v_{b}=2 v_{w}$. What is the time taken by the boat? If it has to cross the river directly on the $A B$ line.(a) $\frac{2 D}{v_{b} \sqrt{3}}$(b) $\frac{\sqrt{3} D}{2 v_{b}}$(c) $\frac{D}{v_{b} \sqrt{2}}$(d) $\frac{D \sqrt{2}}{v_{b}}$
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A boat is trying to cross a river from point A to point B, which are directly opposite each other. The speed of the water is $v_{w}$ and the speed of the boat is $v_{b}$ relative to still water. It is given that $v_{b}=2 v_{w}$. Show more…
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A boat travels at a speed of $v_{\mathrm{BW}}$ relative to the water in a river of width $D .$ The speed at which the water is flowing is $v_{\mathrm{W}}$ a) Prove that the time required to cross the river to a point exactly opposite the starting point and then to return is $T_{1}=2 D / \sqrt{v_{B W}^{2}-v_{W}^{2}}$ b) Also prove that the time for the boat to travel a distance $D$ downstream and then return is $T_{1}=2 D v_{\mathrm{B}} /\left(v_{\mathrm{BW}}^{2}-v_{\mathrm{w}}^{2}\right)$
A boat travels at a speed of $v_{\mathrm{BW}}$ relative to the water in a river of width D. The speed at which the water is flowing is $v_{\mathrm{W}}$ a) Prove that the time required to cross the river to a point exactly opposite the starting point and then to return is $T_{1}=2 D / \sqrt{v_{\mathrm{BW}}^{2}-v_{\mathrm{W}}^{2}}$ b) Prove that the time for the boat to travel a distance $D$ downstream and then return is $T_{1}=2 D v_{\mathrm{BW}} /\left(v_{\mathrm{BW}}^{2}-v_{\mathrm{W}}^{2}\right)$
Let $l$ be the distance covered by the boat $A$ along the river as well as by the boat $B$ acrc the river. Let $v_{0}$ be the stream velocity and $v^{\prime}$ the velocity of each boat with respect water. Therefore time taken by the boat $A$ in its journey $$ t_{A}=\frac{l}{v^{\prime}+v_{0}}+\frac{l}{v^{\prime}-v_{0}} $$ and for the boat $B$ $$ t_{B}=\frac{l}{\sqrt{v^{\prime 2}-v_{0}^{2}}}+\frac{l}{\sqrt{v^{\prime 2}-v_{0}^{2}}}=\frac{2 l}{\sqrt{v^{\prime 2}-v_{0}^{2}}} $$ Hence, $$ \frac{t_{A}}{t_{B}}=\frac{v^{\prime}}{\sqrt{v^{\prime 2}-v_{0}^{2}}}=\frac{\eta}{\sqrt{\eta^{2}-1}}\left(\text { where } \eta=\frac{v^{\prime}}{v}\right) $$ On substitution $\quad t_{A} / t_{B}=1 \cdot 8$
Physical Fundamentals Of Mdchanics
Kinematics
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