(I) Horizontal Velocity Models
In Section 1.2, we considered motion where acceleration was constant or a function of \(t\), defined by \(\frac{dv}{dt} = a(t)\) and \(\frac{dx}{dt} = v\). Now we consider models that include resistance, such as resistance proportional to the velocity.
Suppose that a body moves through a resisting medium with resistance proportional to velocity \(v\), such that: $$\frac{dv}{dt} = -kv, \quad x(0) = x_0, \quad v(0) = v_0$$
(a) Find velocity \(v(t)\) and position \(x(t)\):
The differential equation \(\frac{dv}{dt} = -kv\) is separable:
\(\int \frac{dv}{v} = \int -k \, dt \implies \ln|v| = -kt + C_1\)
\(v(t) = C_2 e^{-kt}\). Since \(v(0) = C_2 = v_0\), the velocity function is \(v(t) = v_0 e^{-kt}\).
To find position, we use \(\frac{dx}{dt} = v_0 e^{-kt}\):
\(x(t) = \int v_0 e^{-kt} \, dt = -\frac{v_0}{k} e^{-kt} + C_3\)
Applying the initial condition \(x(0) = -\frac{v_0}{k} + C_3 = x_0 \implies C_3 = \frac{v_0}{k} + x_0\).
The resulting position function is: \(x(t) = -\frac{v_0}{k} e^{-kt} + \frac{v_0}{k} + x_0\).
(b) Conclude that the body travels only a finite distance and find that distance:
As time \(t\) approaches infinity, \(e^{-kt}\) approaches zero. Therefore:
$$\lim_{t \to \infty} x(t) = \frac{v_0}{k} + x_0$$
The total distance traveled as \(t\) approaches infinity is: \((\frac{v_0}{k} + x_0) - x_0 = \frac{v_0}{k}\).
(II) Vertical Motion
We now consider vertical motion subject to air resistance \(F_R\). By Newton's Second Law: \(m \frac{dv}{dt} = F_G + F_R\).
From empirical investigations, air resistance is modeled as \(F_R = -kv^p\) (where \(k > 0\)). The power \(p\) typically falls in the range \(1 \le p \le 2\):
- \(p = 1\) for relatively low speeds.
- \(p = 2\) for relatively high speeds.
- \(1 < p < 2\) for medium speeds.
Case 1: Velocity Proportional to Resistance (\(p = 1\))
The resistance is \(F_R = -kv\). The equation of motion is: \(m \frac{dv}{dt} = -kv - mg \implies \frac{dv}{dt} = -\frac{k}{m}v - g\).
Let \(\rho = \frac{k}{m}\) be the drag coefficient. The differential equation becomes:
This is a separable equation. Integrating gives: \(\int \frac{dv}{v + \frac{g}{\rho}} = \int -\rho \, dt \implies \ln|v + \frac{g}{\rho}| = -\rho t + C_1\).
This simplifies to \(v + \frac{g}{\rho} = Ce^{-\rho t}\). Using \(v(0) = v_0 \implies C = \frac{g}{\rho} + v_0\).
The velocity function is: \(v(t) = (v_0 + \frac{g}{\rho})e^{-\rho t} - \frac{g}{\rho}\).
The limiting speed or terminal speed \(|v_{\tau}|\) is found via:
The terminal velocity magnitude is \(|v_{\tau}| = \frac{g}{\rho} = \frac{mg}{k}\). We can rewrite velocity as \(v(t) = (v_0 - v_{\tau})e^{-\rho t} + v_{\tau}\).
Integrating velocity gives the position function \(y(t)\) with \(y(0) = y_0\):
\(y(t) = y_0 + v_{\tau}t + \frac{1}{\rho}(v_0 - v_{\tau})(1 - e^{-\rho t})\).
Consider a bolt fired straight upward with an initial velocity \(v_0 = 49\) meters per second from a crossbow at ground level. We take air resistance into account with \(\rho = 0.04\):
\(\frac{dv}{dt} = -0.04v - 9.8 = -0.04(v + 245)\) with \(y(0) = 0\) and \(v(0) = 49\).
Integrating: \(\int \frac{dv}{v + 245} = \int -0.04 \, dt \implies \ln|v + 245| = -0.04t + C_1\).
\(v + 245 = C_2 e^{-0.04t}\). Using \(v(0) = 49 \implies C_2 = 294\).
The velocity function is \(v(t) = 294e^{-0.04t} - 245\).
The position function is \(y(t) = \int (294e^{-0.04t} - 245) \, dt = -7350e^{-0.04t} - 245t + C_3\).
Using \(y(0) = -7350 + C_3 = 0 \implies C_3 = 7350\).
Position: \(y(t) = 7350 - 7350e^{-0.04t} - 245t\).
Maximum Height: Set \(v(t) = 0 \implies 294e^{-0.04t} = 245 \implies e^{-0.04t} = \frac{245}{294}\).
The time to reach maximum height is \(t_m = 25 \ln\left(\frac{294}{245}\right) \approx 4.558\) seconds. The maximum height is \(y(t_m) \approx 108.3\) meters.
Time Aloft: Set \(y(t) = 0\) and solve for \(t\), which gives approximately \(9.411\) seconds.
Case 2: Velocity Squared Proportional to Resistance (\(p = 2\))
The air resistance is \(F_R = \pm kv^2\). The sign of the resistance force is always opposite the direction of velocity, so \(F_R = -kv|v|\). The model is \(\frac{dv}{dt} = -g - \rho v|v|\) where \(\rho = \frac{k}{m} > 0\).
Upward Motion
During upward motion, velocity \(v > 0\), so the equation is: \(\frac{dv}{dt} = -g - \rho v^2 = -g \left(1 + \frac{\rho}{g}v^2\right)\).
Integrating: \(\int \frac{dv}{1 + \frac{\rho}{g}v^2} = \int -g \, dt \implies \sqrt{\frac{g}{\rho}} \arctan\left(\sqrt{\frac{\rho}{g}}v\right) = -gt + C_1\).
The velocity is \(v(t) = \sqrt{\frac{g}{\rho}} \tan\left(C - t\sqrt{\rho g}\right)\), where \(C = \arctan\left(v_0 \sqrt{\frac{\rho}{g}}\right)\).
Downward Motion
During downward motion, velocity \(v < 0\), so the equation is: \(\frac{dv}{dt} = -g + \rho v^2 = -g \left(1 - \frac{\rho}{g}v^2\right)\).
Integrating using the identity \(\int \frac{1}{1 - u^2} \, du = \tanh^{-1}(u) + C\):
\(\sqrt{\frac{g}{\rho}} \tanh^{-1}\left(\sqrt{\frac{\rho}{g}}v\right) = -gt + C\).
The velocity is \(v(t) = \sqrt{\frac{g}{\rho}} \tanh\left(C - t\sqrt{\rho g}\right)\), where \(C = \tanh^{-1}\left(v_0 \sqrt{\frac{\rho}{g}}\right)\).