Velocity, Position, Displacement, and Integration Applications

One-Dimensional Motion: Position, Velocity, and Displacement

• Position Function $s(t)$

  • Represents the position of an object in one-dimensional space at time $t$.

• Displacement Definition over Interval $[a, b]$

  • Displacement is defined as the net change in position $s$ over the time interval $[a, b]$:     $\text{Displacement} = s(b) - s(a)$
  • $s(b)$ denotes the position at terminal time $t = b$
  • $s(a)$ denotes the position at initial time $t = a$

• Directionality and Sign of Displacement

  • Positive Displacement ($s(b) - s(a) > 0$): Occurs when terminal position $s(b)$ is strictly greater than initial position $s(a)$.
  • Negative Displacement ($s(b) - s(a) < 0$): Occurs when terminal position $s(b)$ is strictly less than initial position $s(a)$.
  • Path Independence: Displacement depends exclusively on the initial position $s(a)$ at time $t = a$ and the terminal position $s(b)$ at time $t = b$. It does not account for intermediate back-and-forth trajectory or turning behavior between $a$ and $b$.

• Fundamental Theorem of Calculus Relation

  • Velocity $v(t)$ is the time derivative of position:     $v(t) = s'(t)$
  • The antiderivative of velocity $v(t)$ is position $s(t)$.
  • Integrating velocity from $a$ to $b$ yields net displacement:     $\int_{a}^{b} v(t)\,dt = \int_{a}^{b} s'(t)\,dt = s(b) - s(a)$

Relationship Between Velocity, Displacement, and Distance Traveled

• Net Displacement via Integration

  • Calculated by directly integrating velocity over the time interval $[a, b]$:     $\text{Displacement} = \int_{a}^{b} v(t)\,dt$
  • Gives only the net difference between initial position $s(a)$ and terminal position $s(b)$.

• Total Distance Traveled via Integration

  • Calculated by integrating the absolute value of velocity over the time interval $[a, b]$:     $\text{Total Distance Traveled} = \int_{a}^{b} |v(t)|\,dt$
  • Accumulates all path distance regardless of direction, preventing positive and negative motion from canceling out.

• Graphical Analysis of Velocity Graphs

  • Subintervals where $v(t) > 0$:
  • Curve lies above the horizontal axis.
  • Particle moves rightward (positive direction).
  • Integral over this interval yields positive displacement.
  • Subintervals where $v(t) < 0$:
  • Curve lies below the horizontal axis.
  • Particle moves leftward (negative direction).
  • Integral over this interval yields negative displacement.
  • Turning Point $t = c$:
  • Point where velocity changes sign ($v(c) = 0$).
  • The object turns around and reverses direction.
  • Absolute Value Transformation $|v(t)|$:
  • Reflects the negative velocity portions above the horizontal axis (abscissa).
  • Makes the integrand strictly non-negative across the entire domain, yielding total distance traveled.

Motion Formulas with Initial Time ($t = 0$)

• General Position Function from Initial Conditions

  • Applying the Fundamental Theorem of Calculus with initial time $a = 0$ and terminal time $t$:     $s(t) = s(0) + \int_{0}^{t} v(\tau)\,d\tau$
  • $s(0)$: Initial position at time $t = 0$
  • $\int_{0}^{t} v(\tau)\,d\tau$: Displacement accumulated over the time interval $[0, t]$
  • Position at time $t$ equals initial position plus displacement over $[0, t]$.

• General Velocity Function from Acceleration

  • Acceleration $a(t)$ is the time derivative of velocity:     $a(t) = v'(t)$
  • Applying the Fundamental Theorem of Calculus with initial time $a = 0$ and terminal time $t$:     $v(t) = v(0) + \int_{0}^{t} a(\tau)\,d\tau$
  • $v(0)$: Initial velocity at time $t = 0$
  • $\int_{0}^{t} a(\tau)\,d\tau$: Change in velocity over the time interval $[0, t]$

Motion Analysis Example: Jogger along a Straight Road

• Given Velocity Function

  • A jogger (or H ogre) runs along a straight road with velocity in miles per hour:     $v(t) = 2t^2 - 8t + 6$
  • Time domain: $t \in [0, 3]$

• Factoring and $t$--Intercepts

  • Factoring out constant factor $2$:     $v(t) = 2(t^2 - 4t + 3)$
  • Factoring quadratic expression:     $v(t) = 2(t - 1)(t - 3)$
  • $t$--intercepts occur at $t = 1$ and $t = 3$
  • Graph geometry: Parabola opening upward intersecting the $t$--axis at $1$ and $3$.

• Direction of Motion Analysis

  • Interval $[0, 1]$:
  • Velocity $v(t) > 0$
  • Jogger moves rightward (positive direction)
  • Interval $[1, 3]$:
  • Velocity $v(t) < 0$
  • Jogger moves leftward (negative direction)

• Step-by-Step Displacement Calculations

  • Displacement over Interval $[0, 1]$:     $\int_{0}^{1} (2t^2 - 8t + 6)\,dt = \left[ \frac{2}{3}t^3 - 4t^2 + 6t \right]_{0}^{1}$$\text{Evaluating at } t = 1: \frac{2}{3}(1)^3 - 4(1)^2 + 6(1) = \frac{2}{3} - 4 + 6 = \frac{2}{3} + 2 = \frac{8}{3}\,\text{miles}$$\text{Evaluating at } t = 0: 0$$\text{Displacement over } [0, 1] = \frac{8}{3}\,\text{miles}$

  • Physical Interpretation: Assuming initial position $s(0) = 0$, at $t = 1$ the jogger is at position $\frac{8}{3}\,\text{miles}$ to the right.

  • Displacement over Interval $[1, 3]$:     $\int_{1}^{3} (2t^2 - 8t + 6)\,dt = \left[ \frac{2}{3}t^3 - 4t^2 + 6t \right]_{1}^{3}$$\text{Evaluating at } t = 3: \frac{2}{3}(3)^3 - 4(3)^2 + 6(3) = \frac{2}{3}(27) - 36 + 18 = 18 - 36 + 18 = 0$$\text{Evaluating at } t = 1: \frac{8}{3}$$\text{Displacement over } [1, 3] = 0 - \frac{8}{3} = -\frac{8}{3}\,\text{miles}$

  • Physical Interpretation: Starting from position $\frac{8}{3}\,\text{miles}$ at $t = 1$, moving leftward by $\frac{8}{3}\,\text{miles}$ brings the terminal position back to $0$ at $t = 3$

  • Displacement over Full Interval $[0, 3]$:     $\int_{0}^{3} (2t^2 - 8t + 6)\,dt = \int_{0}^{1} v(t)\,dt + \int_{1}^{3} v(t)\,dt = \frac{8}{3} + \left( -\frac{8}{3} \right) = 0\,\text{miles}$

  • Physical Interpretation: Net displacement over $[0, 3]$ is zero because initial position and terminal position are identical.

• Total Distance Traveled Calculation over Interval $[0, 3]$

  • Integral Setup:     $\text{Total Distance} = \int_{0}^{3} |v(t)|\,dt = \int_{0}^{1} v(t)\,dt - \int_{1}^{3} v(t)\,dt$
  • Evaluation:     $\text{Total Distance} = \frac{8}{3} - \left( -\frac{8}{3} \right) = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}\,\text{miles}$
  • Physical Interpretation: Sum of rightward distance ($\frac{8}{3}\,\text{miles}$) and leftward distance ($\frac{8}{3}\,\text{miles}$).

Application Example: Upward Motion of an Artillery Shell

• Given Parameters

  • Initial velocity at time $t = 0$: $v(0) = 300\,\text{m/s}$
  • Initial position above ground at time $t = 0$: $s(0) = 30\,\text{m}$ (also noted as 3 meters above ground / 300 meters / 30 meters)
  • Acceleration due to gravity: $a(t) = -9.8\,\text{m/s}^2$
  • Negative Sign Explanation: Gravity pulls downward toward the center of the Earth, operating in the direction opposite to upward motion.

• Derivation of Velocity Function $v(t)$

  • Integration Formula:     $v(t) = v(0) + \int_{0}^{t} a(\tau)\,d\tau$
  • Substitution and Evaluation:     $v(t) = 300 + \int_{0}^{t} (-9.8)\,d\tau$$v(t) = 300 - \left[ 9.8\tau \right]_{0}^{t}$$v(t) = 300 - 9.8t$

• Derivation of Position Function $s(t)$

  • Integration Formula:     $s(t) = s(0) + \int_{0}^{t} v(\tau)\,d\tau$
  • Substitution and Evaluation:     $s(t) = 30 + \int_{0}^{t} (300 - 9.8\tau)\,d\tau$$s(t) = 30 + \left[ 300\tau - \frac{9.8}{2}\tau^2 \right]_{0}^{t}$$s(t) = 30 + 300t - 4.9t^2$

Application Example: Cell Population Growth

• Given Parameters

  • Population function: $N(t)$ denotes the number of cells at time $t$
  • Initial population at time $t = 0$: $N(0) = 100\,\text{cells}$
  • Population growth rate: $N'(t) = 90 e^{-0.1t}$

• Fundamental Theorem of Calculus Formulation

  • Population Equation:     $N(t) = N(0) + \int_{0}^{t} N'(\tau)\,d\tau$$N(t) = 100 + \int_{0}^{t} 90 e^{-0.1\tau}\,d\tau$

• Integration of Exponential Component

  • Exponential integration rule:     $\int e^{a\tau}\,d\tau = \frac{1}{a} e^{a\tau}$
  • Application with $a = -0.1$, constant factor $90$:     $\int 90 e^{-0.1\tau}\,d\tau = \frac{90}{-0.1} e^{-0.1\tau} = -900 e^{-0.1\tau}$

• Definite Integration Evaluation from $0$ to $t$

  • Evaluating antiderivative at limits:     $\left[ -900 e^{-0.1\tau} \right]_{0}^{t} = -900 e^{-0.1t} - (-900 e^{0})$
  • Since $e^{0} = 1$:     $-900 e^{-0.1t} + 900$

• Final Population Expression $N(t)$

  • Combining terms:     $N(t) = 100 - 900 e^{-0.1t} + 900$$N(t) = 1000 - 900 e^{-0.1t}$

Application Example: Marginal Cost in Book Publishing

• Definition of Cost Concepts

  • $C(x)$: Total cost of producing $x$ units (books)
  • Marginal Cost $C'(x)$: Approximate cost of producing one additional unit after $x$ units have already been produced
  • Given Marginal Cost Function:     $C'(x) = 4 - 0.00002x$

• Problem Objective

  • Calculate the cost of producing units $12,001$ through $15,000$ ($12,000^{\text{th}}$ first through $15,000^{\text{th}}$ book).

• Definite Integral Setup via FTC

  • Difference in total cost:     $\Delta C = C(15000) - C(12000) = \int_{12000}^{15000} C'(x)\,dx$

• Antiderivative and Calculation Steps

  • Antiderivative expression:     $\int (4 - 0.00002x)\,dx = 12x - 0.00001x^2$
  • Upper Limit Evaluation ($x = 15000$):     $12 \times 15000 - 0.00001 \times (15000)^2$
  • Lower Limit Evaluation ($x = 12000$):     $12 \times 12000 - 0.00001 \times (4000)^2 \text{ (or } 12000^2 \text{)}$
  • Subtracting lower limit evaluation from upper limit evaluation gives the total cost for producing the $12,001^{\text{st}}$ through $15,000^{\text{th}}$ books.