Question

dupa intailnirea a doua nave maritime, una dintre ele continua sa se deplaseze spre sud, iar a doua spre est. Dupa 2 ore de la intalnire, distanta dintre ele era egala cu 60 km. Determinati viteza fiecarei nave,daca se stie ca viteza primei nave este cu 6 km/h mai mare decat viteza navei a doua?

Answer 1

   
[tex]\displaystyle\\ \text{Nava 2 are viteza } (x)~Km/h\\ \text{Nava 1 are viteza }(x + 6)~ Km/h\\ t = 2~ore \\\\ d = V \cdot t\\\\ V_2 = x\\ V_1 = x + 6\\\\ AC = 2x\\ AB = 2(x + 6)\\\\ AB^2+AC^2=BC^2\\ (2x)^2 + [2(x+6)]^2 = 60^2\\ 4x^2 + (2x+12)^2 = 60^2\\ 4x^2 + 4x^2+2 \cdot 2x \cdot 12 + 144 = 60^2\\ 4x^2 + 4x^2+48x + 144 = 3600\\ 4x^2 + 4x^2+48x + 144 - 3600 = 0\\ 8x^2 + 48x - 3456 = 0 | : 8\\ x^2 + 6x - 432 = 0 [/tex]


[tex]\displaystyle\\ x_{12} = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-6\pm\sqrt{36+4\cdot 432}}{2}= \\\\ =\frac{-6\pm\sqrt{36+1728}}{2}=\frac{-6\pm\sqrt{1764}}{2}=\frac{-6\pm42}{2}= -3\pm 21\\\\ x_1 = -3+21 = 18\\ x_2 = -3 - 21 \ \textless \ 0 \text{ Solutie neacceptata.}\\ \Longrightarrow~~~x = 18\\\\ \text{Viteza navei 2 } = x = \boxed{\bf 18~Km/h}\\ \text{Viteza navei 1 } = 18 + 6 = \boxed{\bf 24~Km/h} \\\\ \text{Distantele parcurse de nave:}\\ D2 = 18 \cdot 2 = 36 Km\\ D1 = 24 \cdot 2 = 48 Km\\ [/tex]