91Ó°ÊÓ

The body is subjected to two forces: the gravitational force and the resisting force due to the viscous medium. Let: - \(m\) be the mass of the body, - \(g\) be the acceleration due to gravity, - \(v(t)\) be the velocity of the body at time \(t\), - \(y(t)\) be the distance fallen by the body at time \(t\), - \(R\) be the resisting force due to the viscous medium. According to the problem, the resisting force \(R\) is proportional to the velocity \(v(t)\). Therefore, we can write: \[R = -kv(t)\] where \(k\) is a positive constant of proportionality and the negative sign indicates that the resisting force acts in the opposite direction of the velocity.

Apply Newton's second law of motion in the vertical direction: \[\begin{equation} m\frac{d^2y}{dt^2} = -kv(t) - mg \end{equation}\] Since the motion is in the direction of the positive y-axis and down, the gravitational force acts in the positive direction, giving a positive sign to the constant \(k\). Therefore, we can rewrite the equation as: \[\begin{equation} m\frac{d^2y}{dt^2} = mg - kv(t) \end{equation}\] Since \(v(t) = \frac{dy}{dt}\), we can write the equation in terms of \(v(t)\) as: \[\begin{equation} m\frac{dv}{dt} = mg - kv \end{equation}\]

We have a first-order linear ordinary differential equation (ODE): \[\frac{dv}{dt} + \frac{k}{m}v =g\] To solve the ODE, we can use an integrating factor: \(e^{kt/m}\). Multiply both sides of the ODE by the integrating factor: \[e^{kt/m}\frac{dv}{dt} + e^{kt/m}\frac{k}{m}v = ge^{kt/m}\] Now, the left side of the equation is the derivative of \(ve^{kt/m}\) with respect to \(t\). Hence, \[\frac{d}{dt}(ve^{kt/m}) = ge^{kt/m}\] Integrate both sides with respect to \(t\): \[\int \frac{d}{dt}(ve^{kt/m})dt = g \int e^{kt/m} dt\] \[ve^{kt/m} = \frac{gm}{k} e^{kt/m} + C_1\] To find \(v(t)\), we need to isolate it: \[v(t) = \frac{gm}{k} + C_1e^{-kt/m}\] As it's given that the body is falling from rest, i.e., the initial condition is \(v(0) = 0\). Using the initial condition, we can find the value of \(C_1\): \[0 = \frac{gm}{k} + C_1e^0\] \[C_1= -\frac{gm}{k}\] So, the expression for the velocity at any time \(t\) is: \[v(t) = \frac{gm}{k}(1 - e^{-kt/m})\]

To find the distance fallen at any time \(t\), we integrate the expression for \(v(t)\) with respect to \(t\): \[y(t) = \int v(t) dt = \int \frac{gm}{k}(1 - e^{-kt/m}) dt\] Solve the integration: \[y(t) = \frac{gm}{k^2}t + \frac{m^2}{k^2}(1-e^{-kt/m}) + C_2\] As the body falls from rest, the initial position is \(y(0) = 0\). Using the initial condition, we can find the value of \(C_2\): \[0 = \frac{gm}{k^2}(0) + \frac{m^2}{k^2}(1-e^{0}) + C_2\] \[C_2= 0\] So, the expression for the distance fallen at any time \(t\) is: \[y(t) = \frac{gm}{k^2}t + \frac{m^2}{k^2}(1-e^{-kt/m})\] Hence, we have found expressions for the body's velocity and distance fallen at any time \(t\): - Velocity: \(v(t) = \frac{gm}{k}(1 - e^{-kt/m})\) - Distance fallen: \(y(t) = \frac{gm}{k^2}t + \frac{m^2}{k^2}(1-e^{-kt/m})\)