Respuesta :
Answer:
The answer to the question is;
The equation of motion of the object is
x(t) = 88.29·t + 794.6[tex]e^{-\frac{t}{9} }[/tex] -794.6
Explanation:
The force due to air resistance = 50 N-sec/m
dividing by the mass we get 50 N-sec/m /(450 kg) = 1/9 m-s/s²m =1/9 s⁻¹
Therefore F₁ = 450*9.81 = 4414.5 N, F₂ = -bv(t) = -50v
Formula is [tex]m\frac{dv}{dt} = mg-bv(t)[/tex]
[tex]450\frac{dv}{dt} = 450*9.81-50v[/tex]
[tex]\frac{dv}{dt} = 9.81-v/9[/tex]
[tex]\int\limits^.\frac{dv}{9.81-v/9} = \int\limits^ .dt[/tex]
=[tex]-9\int\limits^.\frac{du}{u} = \int\limits^ .dt[/tex]
-9㏑[9.81-v/9] = t +c
㏑[9.81-v/9] = -t/9 +c
9.81-v/9 =C[tex]e^{-\frac{t}{9} }[/tex]
-v/9 = C[tex]e^{-\frac{t}{9} }[/tex] - 9.81
put v(t) = C[tex]e^{-\frac{t}{9} }[/tex] + 88.29
V0 =0 = C[tex]e^{0}[/tex] + 88.29 ⇒ C = -88.29
v(t) = 88.29 - 88.29·[tex]e^{-\frac{t}{9} }[/tex]
Equation of motion
x(t) = [tex]\int\limits^ . v(t)dt[/tex] = 88.29·t + 794.6[tex]e^{-\frac{t}{9} }[/tex] +C₁
x = 0 c₁ = -794.6
x(t) = 88.29·t + 794.6[tex]e^{-\frac{t}{9} }[/tex] -794.6
