Lista 1 - Pedro Henrique Barbosa de Lima

12/10/21\n1ª lista de Exercício Mecânica Geral 2\nAluno: Pedro Henrique Pacheco de Lima\nx² = v² * t²\nv² = 0.11.dt\nv² = 0.11z² + c\n\nv² = 0.05 - 30² = 15 m/s (alpares a 30º)\nd\t \nv² = -[1, + 5] dt : \nv² = -t² + 5t + c\n30\n\nv² = -t² + 5t + c\n30\n\n∫[v] = 30.0, v² = 15 m/s\n\nc = -3\n\n- para v² = 0\n\n-t² + 5t - 3 = 0\nλ = 25 - (4.9)(7.5) = 15\n\nt = 0.5 ± √(5)\n[1' = 133.06947] > t = 30\n\nΔS₁ = ∫ 0 30 a + 1/2 * dt : ΔS₁ = 0.1 t³ + c |_{0}^{6}\n\nΔS₁( t=0) = 0 : c = 0 12/10/21\nΔS₂ = ∫ [ -t² + 5t - 7.5 ]dt : ΔS₂ = -t³ + 5t² - 7.5t + c\n\np/t = 30º ΔS₂ = 450m\n\nΔSᶜ = -30³ + 5 * 30² - 75t + c 2c = 750\n\nΔS₂ = -t³ + 5t² - 75t + 750\n\nΔS₂( t=133.094) = -133.094³ + 5 * 133.094² - 75 * 133.094 + 750\n\nΔS₂ = 3857.1063m. d istance_total\n\nΔS₁ - p/t = ΔS₂\n\nΔS₁ = -133.094³ + 5*133.094² - 75*133.094 + 750\n\nΔS = 4500\n 17/10/21\n3) V₀ = 20m/s y = 0.05 ²m θ = 60º\n\nNa horizontal: x = V₀ * t\ny₀ = V₀ * cosθ * t\ny₀ = 0.05 * t\n\nNa vertical: y² = V₀ * sinθ - g * t\nV₀ * sin²θ = (20 * sin(60º)²) - 3.981(t)\n\nV₀ * t² = -1.491(0.05²)\n\nV₀² = -2V₀g + g * t² = -19.62 * 0.05²\n\n-2√3 - 5.981t + 5g₀cosθ(0.05²) = 0\n\nx = 33.9722836.24 = 17.42663407m\n\ny = 0.05x² = 0.05(47.48863107) = 15.28591331m\n\nVy² = 20 * (9.81)² = 2.9781 ( 15 0.2 5814333) V₀ = 0.46612259521m p no ponto b:\n\\( p = \\left(\\frac{(16.2^2 + 0^2)}{(4.2^2 + 0^2)}\\right)^{3/2} = 8 Km\\)\nV = 200 k\\h\nano ponto b: a = \\frac{u^2}{200} = \\frac{25 k}{h^2}\n1°) u_b = 2 Km/h = 20 m/s\n\\( \\Delta s = 320 m \\ t = 20 s \\)\na_{max} = 0.75 mi/s^2\n\\( \\frac{\\Delta s}{t} \\)\nS_1 \\rightarrow\\ α_0.92\nS_2\n\n12 hipótese \\ t = 12 s, S_f = -160 m\n\\( \\Delta S = V_0 \\ t + a t^2 = \\frac{1}{2} a (t_b^2) \\rightarrow \\frac{b}{2} = 20 \\cdot 12 + a \\cdot \\frac{2^2}{2}\\)\na = 0.75 m/s^2\n\\( a_{f} = 0.75 = a_{max} \\leq S_f = 136m \\ t = 12s \\)\nU_2 = V_0 + a t; V_f = 20 - 0.75 \\cdot 12 = 11 m/s\n - No regundo, instantâneo.\n\\( t = 10s, \\Delta S = 134m \\ V_0 = 11 \\ V_f = 20 \\ a = ? \\)