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The-flow-velocity-of-a-river-increases-linearly-with-the-distance-r-from-its-bank-and-has-its-maximum-value-v-0-in-the-middle-of-the-river-The-velocity-near-the-bank-is-zero-A-boat-which-can-move

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Question Number 18269 by Tinkutara last updated on 17/Jul/17
The flow velocity of a river increases linearly with the distance (r) from its bank and has its maximum value v_0 in the middle of the river. The velocity near the bank is zero. A boat which can move with speed u in still water moves in the river in such a way that it is always perpendicular to the flow of current. Find (i) The distance along the bank through which boat is carried away by the flow current, when the boat crosses the river. (ii) The equation of trajectory for the coordinate system shown. Assume that the swimmer starts from origin.
$$\mathrm{The}\:\mathrm{flow}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{river}\:\mathrm{increases} \\ $$$$\mathrm{linearly}\:\mathrm{with}\:\mathrm{the}\:\mathrm{distance}\:\left({r}\right)\:\mathrm{from}\:\mathrm{its} \\ $$$$\mathrm{bank}\:\mathrm{and}\:\mathrm{has}\:\mathrm{its}\:\mathrm{maximum}\:\mathrm{value}\:{v}_{\mathrm{0}} \:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{middle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{river}.\:\mathrm{The}\:\mathrm{velocity} \\ $$$$\mathrm{near}\:\mathrm{the}\:\mathrm{bank}\:\mathrm{is}\:\mathrm{zero}.\:\mathrm{A}\:\mathrm{boat}\:\mathrm{which}\:\mathrm{can} \\ $$$$\mathrm{move}\:\mathrm{with}\:\mathrm{speed}\:{u}\:\mathrm{in}\:\mathrm{still}\:\mathrm{water}\:\mathrm{moves} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{river}\:\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{always}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{the}\:\mathrm{flow}\:\mathrm{of} \\ $$$$\mathrm{current}.\:\mathrm{Find} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{The}\:\mathrm{distance}\:\mathrm{along}\:\mathrm{the}\:\mathrm{bank}\:\mathrm{through} \\ $$$$\mathrm{which}\:\mathrm{boat}\:\mathrm{is}\:\mathrm{carried}\:\mathrm{away}\:\mathrm{by}\:\mathrm{the}\:\mathrm{flow} \\ $$$$\mathrm{current},\:\mathrm{when}\:\mathrm{the}\:\mathrm{boat}\:\mathrm{crosses}\:\mathrm{the} \\ $$$$\mathrm{river}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{trajectory}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{coordinate}\:\mathrm{system}\:\mathrm{shown}.\:\mathrm{Assume} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{swimmer}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{origin}. \\ $$
Commented by Tinkutara last updated on 17/Jul/17
Commented by mrW1 last updated on 19/Jul/17
u_y =u y=ut u_x =((2y)/d)v_0 =((2v_0 )/d)y with 0≤y≤(d/2) u_x =((2(d−y))/d)v_0 =2v_0 (1−(y/d)) with (d/2)≤y≤d u_x =(dx/dt)=(dx/dy)×(dy/dt)=u(dx/dy) ⇒u(dx/dy)=((2v_0 )/d)y with 0≤y≤(d/2) ⇒u(dx/dy)=2v_0 (1−(y/d)) with (d/2)≤y≤d x=(v_0 /(ud))y^2 +C_1 C_1 =0 since x=0 at y=0 ⇒x=(v_0 /du)y^2 with 0≤y≤(d/2) at y=(d/2): x=((v_0 d)/(4u)) x=((2v_0 )/u)(y−(y^2 /(2d)))+C_2 at y=(d/2): ((v_0 d)/(4u))=((2v_0 )/u)[(d/2)−(d/8)]+C_2 C_2 =−((v_0 d)/(2u)) ⇒x=((2v_0 )/u)(y−(y^2 /(2d))−(d/4)) with (d/2)≤y≤d at y=d: x=((v_0 d)/(2u)) y=ut x=((v_0 u)/d)t^2 with 0≤t≤(d/(2u)) x=2v_0 (t−((ut^2 )/(2d))−(d/(4u))) with (d/(2u))≤t≤(d/u)
$$\mathrm{u}_{\mathrm{y}} =\mathrm{u} \\ $$$$\mathrm{y}=\mathrm{ut} \\ $$$$\mathrm{u}_{\mathrm{x}} =\frac{\mathrm{2y}}{\mathrm{d}}\mathrm{v}_{\mathrm{0}} =\frac{\mathrm{2v}_{\mathrm{0}} }{\mathrm{d}}\mathrm{y}\:\mathrm{with}\:\mathrm{0}\leqslant\mathrm{y}\leqslant\frac{\mathrm{d}}{\mathrm{2}} \\ $$$$\mathrm{u}_{\mathrm{x}} =\frac{\mathrm{2}\left(\mathrm{d}−\mathrm{y}\right)}{\mathrm{d}}\mathrm{v}_{\mathrm{0}} =\mathrm{2v}_{\mathrm{0}} \left(\mathrm{1}−\frac{\mathrm{y}}{\mathrm{d}}\right)\:\mathrm{with}\:\frac{\mathrm{d}}{\mathrm{2}}\leqslant\mathrm{y}\leqslant\mathrm{d} \\ $$$$ \\ $$$$\mathrm{u}_{\mathrm{x}} =\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{dx}}{\mathrm{dy}}×\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{u}\frac{\mathrm{dx}}{\mathrm{dy}} \\ $$$$\Rightarrow\mathrm{u}\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{2v}_{\mathrm{0}} }{\mathrm{d}}\mathrm{y}\:\mathrm{with}\:\mathrm{0}\leqslant\mathrm{y}\leqslant\frac{\mathrm{d}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{u}\frac{\mathrm{dx}}{\mathrm{dy}}=\mathrm{2v}_{\mathrm{0}} \left(\mathrm{1}−\frac{\mathrm{y}}{\mathrm{d}}\right)\:\mathrm{with}\:\frac{\mathrm{d}}{\mathrm{2}}\leqslant\mathrm{y}\leqslant\mathrm{d} \\ $$$$\mathrm{x}=\frac{\mathrm{v}_{\mathrm{0}} }{\mathrm{ud}}\mathrm{y}^{\mathrm{2}} +\mathrm{C}_{\mathrm{1}} \\ $$$$\mathrm{C}_{\mathrm{1}} =\mathrm{0}\:\mathrm{since}\:\mathrm{x}=\mathrm{0}\:\mathrm{at}\:\mathrm{y}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{v}_{\mathrm{0}} }{\mathrm{du}}\mathrm{y}^{\mathrm{2}} \:\mathrm{with}\:\mathrm{0}\leqslant\mathrm{y}\leqslant\frac{\mathrm{d}}{\mathrm{2}} \\ $$$$\mathrm{at}\:\mathrm{y}=\frac{\mathrm{d}}{\mathrm{2}}:\:\mathrm{x}=\frac{\mathrm{v}_{\mathrm{0}} \mathrm{d}}{\mathrm{4u}} \\ $$$$ \\ $$$$\mathrm{x}=\frac{\mathrm{2v}_{\mathrm{0}} }{\mathrm{u}}\left(\mathrm{y}−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2d}}\right)+\mathrm{C}_{\mathrm{2}} \\ $$$$\mathrm{at}\:\mathrm{y}=\frac{\mathrm{d}}{\mathrm{2}}: \\ $$$$\frac{\mathrm{v}_{\mathrm{0}} \mathrm{d}}{\mathrm{4u}}=\frac{\mathrm{2v}_{\mathrm{0}} }{\mathrm{u}}\left[\frac{\mathrm{d}}{\mathrm{2}}−\frac{\mathrm{d}}{\mathrm{8}}\right]+\mathrm{C}_{\mathrm{2}} \\ $$$$\mathrm{C}_{\mathrm{2}} =−\frac{\mathrm{v}_{\mathrm{0}} \mathrm{d}}{\mathrm{2u}} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{2v}_{\mathrm{0}} }{\mathrm{u}}\left(\mathrm{y}−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2d}}−\frac{\mathrm{d}}{\mathrm{4}}\right)\:\mathrm{with}\:\frac{\mathrm{d}}{\mathrm{2}}\leqslant\mathrm{y}\leqslant\mathrm{d} \\ $$$$\mathrm{at}\:\mathrm{y}=\mathrm{d}:\:\mathrm{x}=\frac{\mathrm{v}_{\mathrm{0}} \mathrm{d}}{\mathrm{2u}} \\ $$$$ \\ $$$$\mathrm{y}=\mathrm{ut} \\ $$$$\mathrm{x}=\frac{\mathrm{v}_{\mathrm{0}} \mathrm{u}}{\mathrm{d}}\mathrm{t}^{\mathrm{2}} \:\mathrm{with}\:\mathrm{0}\leqslant\mathrm{t}\leqslant\frac{\mathrm{d}}{\mathrm{2u}} \\ $$$$\mathrm{x}=\mathrm{2v}_{\mathrm{0}} \left(\mathrm{t}−\frac{\mathrm{ut}^{\mathrm{2}} }{\mathrm{2d}}−\frac{\mathrm{d}}{\mathrm{4u}}\right)\:\mathrm{with}\:\frac{\mathrm{d}}{\mathrm{2u}}\leqslant\mathrm{t}\leqslant\frac{\mathrm{d}}{\mathrm{u}} \\ $$
Commented by mrW1 last updated on 19/Jul/17

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