Math  /  Algebra

Question368. Mootorpaat sõitis jõel vastuvoolu 16 km ja pöördus siis tagasi, kulutades tagasiteel 40 minutit vähem aega kui liikumisel vastuvoolu. Leidke paadi kiirus seisvas vees, kui jõe voolu kiirus on 2 km/h2 \mathrm{~km} / \mathrm{h}.

Studdy Solution
Solve the equation: Multiply through by the common denominator (vb2)(vb+2)(v_b - 2)(v_b + 2) to clear the fractions: 16(vb+2)16(vb2)=23(vb2)(vb+2) 16(v_b + 2) - 16(v_b - 2) = \frac{2}{3} (v_b - 2)(v_b + 2)
Simplify the left side: 16vb+3216vb+32=23(vb24) 16v_b + 32 - 16v_b + 32 = \frac{2}{3} (v_b^2 - 4)
64=23(vb24) 64 = \frac{2}{3} (v_b^2 - 4)
Multiply both sides by 3 to eliminate the fraction: 192=2(vb24) 192 = 2(v_b^2 - 4)
Divide by 2: 96=vb24 96 = v_b^2 - 4
Add 4 to both sides: 100=vb2 100 = v_b^2
Take the square root of both sides: vb=100 v_b = \sqrt{100} vb=10 v_b = 10
The speed of the boat in still water is: 10km/h \boxed{10 \, \mathrm{km/h}}

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