Răspuns:
c)20 m/s
Explicație:
[tex] \displaystyle\\ F = ma \implies a = \frac{F}{m}\\ \\ v_0 = 0\frac{m}{s}\\ \\ \text{In intervalul 0 - 1 secunde:}\\ \\ a = \frac{20}{2} = 10\frac{m}{s^2}\\ \\ v = v_0 + at \implies v = 0 + 10\cdot 1 = 10\frac{m}{s}\text{ (la secunda 1)}\\ \\ \text{In intervalul 1 - 3 secunde:} \\ \\ F(t) = -10t + 30\\ \\ a(t) = \frac{F(t)}{m} = \frac{-10t + 30}{2} = -5t + 15\\ \\ \text{Varianta de clasa a XII-a:}\\ \\ \Delta v = \int_{1}^{3}a(t)dt\\ \\ = \int_{1}^{3}(-5t+15)dt = \int_{1}^{3}-5t\:dt + \int_{1}^{3}15dt\\ \\ = -5\int_{1}^{3}t\:dt + 15\int_{1}^{3}dt\\ \\ = -5\cdot\frac{t^2}{2} + 15\cdot \frac{t}{1}\\ \\ = \frac{-5t^2}{2} + 15t\Biggr|_{1}^{3}\\ \\ = \frac{-5\cdot 3^2}{2} + 15\cdot 3 - \Big(\frac{-5\cdot 1^2}{2} + 15\cdot 1\Big)\\ \\ = \frac{-5\cdot 9}{2} + 45 + \frac{5}{2} - 15\\ \\ = \frac{-45+5}{2} + 30\\ \\ = \frac{-40}{2}+30\\ \\ = -20+30\\ \\ = 10\frac{m}{s}[/tex]
[tex]\text{Varianta de clasa a IX-a:}\\ \\ \text{Avem un triunghi dreptunghic cu catetele:}\\ c_1 = 3-1 = 2\\ \\ c_2 = a(1) = 10\\ \\ \Delta v = \text{Aria acestui triunghi} = \frac{2\cdot 10}{2} = 10\frac{m}{s} \\ \\ v_f = v + \Delta v = 10\frac{m}{s} + 10\frac{m}{s} = \boxed{20\frac{m}{s}}[/tex]
