Lesson 11 (09/18/2026) — Acceleration-Velocity Models (2.3)

Lesson 11 (09/18/2026) — Acceleration-Velocity Models (2.3)

Review:

\[ \frac{dP}{dt} = 2P - P^2 - h \]

Critical points: \(c = 1 \pm \sqrt{1-h}\)

1. \(h > 1\)

   No critical point

   • irrespective of initial condition, all fish are harvested/dead eventually

2. \(h = 1\)

   only 1 critical point

   • \(P(0) > 1 \rightsquigarrow\) population stabilizes to critical point
   • \(P(0) < 1 \rightsquigarrow\) all fish harvested or dead.

3. \(h < 1\)

   two critical points

   \(c_1 = 1 - \sqrt{1-h}, \quad c_2 = 1 + \sqrt{1-h}\)

   • \(c_1\) is unstable
   • \(c_2\) is stable

\(h = 1\) where Qualitative behavior of \(P(t)\) changes is called a Bifurcation point.

Bifurcation Diagram

Bifurcation Diagram is graph relating critical points and parameter \(h\)

\[ c = 1 \pm \sqrt{1-h} \implies 1-h = (c-1)^2 \] \[ h = 1 - (c-1)^2 \]

Review: Find velocity and position for a particle moving with constant acceleration \( a \).

\[ v'(t) = a(t) = a \]

integrate w.r.t \( t \)

\[ v(t) = at + v(0) \]

\( v(0) \) is initial velocity

\[ x'(t) = v(t) = at + v(0) \]

integrate w.r.t \( t \)

\[ x(t) = \frac{1}{2}at^2 + v(0)t + x(0) \]

\( x(0) \) is initial position

We shoot an arrow straight up from the ground with an initial velocity of \(49\text{ m/s}\)

1. What is the maximum height the arrow reaches?
2. At what time does the arrow hit the ground?

\(v(0) = 49\text{ m/s}, \quad x(0) = 0\)

\(a(t) = -g = -9.8\text{ m/s}^2 \quad [\text{acceleration due to gravity}]\)

\(v'(t) = -9.8\)

Integrate w.r.t \(t\)

\[ v(t) = -9.8t + v(0) \leadsto v(t) = -9.8t + 49\text{ m/s} \]

\[ x'(t) = v(t) = -9.8t + 49 \]

integrate w.r.t \(t\)

\[ x(t) = -4.9t^2 + 49t + x(0) \leadsto x(t) = -4.9t^2 + 49\text{ m} \]

max height attain when \(v(t) = 0 \leadsto v(t) = -9.8t + 49 = 0\)

\[ \implies t = 5\text{ secs.} \]

\(\text{max height} = x(5) = -4.9(5)^2 + 49(5) = 122.5\text{ m}\)

Hits the ground when \(x(t) = 0 \leadsto -4.9t^2 + 49t = 0 \leadsto \underline{t = 10\text{ secs}}\)

Question: What happens if there is air Resistance?

Guess: max height is less
Reaches ground faster.

Newton's 2^nd Law:

\[ F = m a \] \[ F_G + F_R = m \cdot \frac{dv}{dt} \]

\( F_G \rightarrow -mg \)

\( F_R \rightarrow \text{force due to resistan} \)
usually \( F_R = -k v^p, \quad 1 \le p \le 2 \)
\( k \) is resistance constant

\( \underline{\underline{p = 1}} \)

\( F_G = -mg, \quad F_R = -k v \)

\[ -mg - k v = m \frac{dv}{dt} \quad \leadsto \quad \frac{dv}{dt} = -g - \left(\frac{k}{m}\right) v \] \[ \frac{k}{m} = \rho = \text{drag constant}. \]

IVP when there is Resistance:

\[ \frac{dv}{dt} = -g - \rho v, \quad v(0) = v_0 \]

both linear & Seperable

\[ \int \frac{1}{\rho v + g} \, dv = \int -1 \, dt + C \] \[ \frac{1}{\rho} \ln(\rho v + g) = -t + C \] \[ \ln(\rho v + g) = -\rho t + \underbrace{\rho c}_{\text{new constant}} = -\rho t + d \] \[ \rho v + g = e^{-\rho t} \cdot \underbrace{e^d}_{\text{new const}} = A e^{-\rho t} \] \[ v(t) = \frac{A e^{-\rho t} - g}{\rho} \]

\[ v(t) = \frac{A e^{-\rho t} - g}{\rho} \]

Find \( A \), by plugging in \( v(0) = v_0 \)

\[ v_0 = v(0) = \frac{A e^0 - g}{\rho} \implies A = (\rho v_0 + g) \] \[ v(t) = \frac{(\rho v_0 + g) e^{-\rho t} - g}{\rho} = \left[\frac{\rho v_0 + g}{\rho}\right] e^{-\rho t} - \frac{g}{\rho} \] \[ x'(t) = v(t) = \left[\frac{\rho v_0 + g}{\rho}\right] e^{-\rho t} - \frac{g}{\rho} \]

integrate

\[ x(t) = -\left[\frac{\rho v_0 + g}{\rho^2}\right] e^{-\rho t} - \frac{g t}{\rho} + C \]

\[ x(t) = -\left[\frac{\rho v_0 + g}{\rho^2}\right] e^{-\rho t} - \frac{g}{\rho} t + C \]

plug in \( x(0) = x_0 \) to find \( C \)

\[ x_0 = x(0) = -\left[\frac{\rho v_0 + g}{\rho^2}\right] + C \Rightarrow C = x_0 + \left[\frac{\rho v_0 + g}{\rho^2}\right] \] \[ x(t) = -\left[\frac{\rho v_0 + g}{\rho^2}\right] e^{-\rho t} - \frac{g}{\rho} t + x_0 + \left[\frac{\rho v_0 + g}{\rho^2}\right] \] \[ = x_0 + \left(-\frac{g}{\rho}\right)t + \left[\left(\frac{\rho v_0 + g}{\rho^2}\right)\right](1 - e^{-\rho t}) \]

Go back to \( v(t) \):

\[ v(t) = \left[ \frac{\rho v_0 + g}{\rho} \right] e^{-\rho t} - \frac{g}{\rho} \]

\[ \lim_{t \to \infty} v(t) = \lim_{t \to \infty} \left[ \frac{\rho v_0 + g}{\rho} \right] e^{-\rho t} - \frac{g}{\rho} = -\frac{g}{\rho} \]

\( v_t = -\frac{g}{\rho} \) is called limiting velocity OR Terminal velocity

\( |v_t| = \frac{g}{\rho} \) is terminal speed

Rewrite:

\[ v(t) = \left[ v_0 + \frac{g}{\rho} \right] e^{-\rho t} - \frac{g}{\rho} = (v_0 - v_t) e^{-\rho t} + v_t \]

\[ x(t) = x_0 + v_t t + \frac{1}{\rho} [v_0 - v_t] (1 - e^{-\rho t}) \]

Going back to Arrow problem

\( v(0) = 49 \), \( x(0) = 0 \), \( a = -9.8 \), \( g = 9.8 \)

\[ \rho = \frac{1}{25} \quad [\text{assume}] \]

\[ v_t = -\frac{g}{\rho} = -\frac{9.8}{(1/25)} = -245\text{ m/s} \]

max height: \( v(t) = 0 \Rightarrow (v_0 - v_t)\bar{e}^{-\rho t} + v_t = 0 \)

\[ (-196)\bar{e}^{-t/25} - 245 = 0 \]

Calculator: \[ t = 4.86 \] \[ \times 2 = 9.72 \]

max height: \( x(4.86) \approx 108\text{ m} \)

\[ x(t) = 0 \leadsto t = \underline{9.41}\text{ sec} \]