Calculus Study Guide: Constant Velocity, Average Velocity, and Instantaneous Velocity

Linear Motion and Constant Velocity

• Bicycle Motion Model:

  • At time $t = 0\,\text{hours}$, position $s = 15 \times 0 = 0\,\text{miles}$.

  • At time $t = 1\,\text{hour}$, position $s = 15\,\text{miles}$.

  • Position function: $s_{\text{bicycle}}(t) = 15t$

  • Curve geometry: Equation of a straight line defined uniquely by two distinct points.

  • Slope evaluation: $\text{Slope} = \frac{15}{1} = 15$

• Car Motion Model:

  • At time $t = 1\,\text{hour}$, position $s = 40\,\text{miles}$.

  • Position function: $s_{\text{car}}(t) = 40t$

  • Slope comparison: The slope of the car's position line ($40$) is steeper than the bicycle's position line ($15$).

• Fundamental Physical Rule:

  • Velocity represents the slope of the position-versus-time line.

  • Higher velocity corresponds directly to a steeper slope on a position graph.

• General Case for Constant Velocity:

  • Motion evaluated between two distinct times, $t_0$ (read as $t\text{-zero}$ or $t\text{-naught}$) and $t_1$.

  • Position at time $t_0$: $s(t_0)$

  • Position at time $t_1$: $s(t_1)$

  • Right triangle geometric construction:

  • Run (change in time): $\Delta t = t_1 - t_0$

  • Rise (change in position): $\Delta s = s(t_1) - s(t_0)$

  • Slope formula for constant velocity:     $\text{Velocity} = \text{Slope} = \frac{\Delta s}{\Delta t} = \frac{\text{Rise}}{\text{Run}} = \frac{s(t_1) - s(t_0)}{t_1 - t_0}$

Mathematical Constraint on Division by Zero

• Strict Condition for Slope Calculation:

  • The condition $t_1 \neq t_0$ must strictly hold for all velocity calculations.

  • If $t_1 = t_0$, then $\Delta t = t_1 - t_0 = 0$, resulting in division by zero.

  • Division by zero yields an mathematically undefined expression.

  • Drawing a unique line requires two distinct points; a single point cannot specify a line.

• Pedagogical Warning:

  • Dividing by zero must be avoided in all mathematical steps.

Non-Constant Velocity and Parabolic Trajectories

• Accelerating Motorcycle Scenario:

  • Models a motorcycle accelerating down a freeway on-ramp.

  • Position function: $s(t) = 3t^2$ for $t \ge 0$

  • Units of measurement:

  • Time $t$ in seconds ($\text{s}$).

  • Position $s(t)$ in feet ($\text{ft}$).

  • Velocity units in feet per second ($\text{ft/s}$).

  • Curve geometry: Parabola (exhibits non-linear curvature rather than constant steepness).

• Table of Position Values for $s(t) = 3t^2$:

  • At $t = 0\,\text{s}$: $s(0) = 3(0)^2 = 0\,\text{ft}$

  • At $t = 1\,\text{s}$: $s(1) = 3(1)^2 = 3\,\text{ft}$

  • At $t = 2\,\text{s}$: $s(2) = 3(2)^2 = 3(4) = 12\,\text{ft}$

  • At $t = 3\,\text{s}$: $s(3) = 3(3)^2 = 3(9) = 27\,\text{ft}$

  • At $t = 4\,\text{s}$: $s(4) = 3(4)^2 = 3(16) = 48\,\text{ft}$

  • At $t = 5\,\text{s}$: $s(5) = 3(5)^2 = 3(25) = 75\,\text{ft}$

  • At $t = 6\,\text{s}$: $s(6) = 3(6)^2 = 3(36) = 108\,\text{ft}$

• Properties of Changing Velocity:

  • Non-linear behavior: The graph forms a parabola, not a straight line.

  • Speedometer reading: Speedometer values continuously increase over time as the vehicle accelerates.

  • Slope behavior: The slope of the curve becomes progressively steeper as time $t$ increases.

• Graphing Conventions:

  • Horizontal axis: Time $t$ (marked from $0$ to $6\,\text{s}$).

  • Vertical axis: Position $s$ (scaled with increments at $25$, $50$, $75$, and $100\,\text{ft}$, up to $108\,\text{ft}$).

  • Sketches must accurately reflect smooth parabolic curvature rather than piecewise linear segments.

Average Velocity

• Definition:

  • Average velocity ($v_{\text{avg}}$) is measured over a closed time interval $t \in [t_0, t_1]$.

• Formula:   $v_{\text{avg}} = \frac{s(t_1) - s(t_0)}{t_1 - t_0} = \frac{\Delta s}{\Delta t}$   where $t_1 \neq t_0$

• Geometric Definition:

  • Average velocity represents the slope of a secant line connecting two distinct points $(t_0, s(t_0))$ and $(t_1, s(t_1))$ on the position graph.

• Sample Calculations for $s(t) = 3t^2$:

  • Interval $t \in [2, 6]$:

  • $t_0 = 2\,\text{s}$, $t_1 = 6\,\text{s}$

  • $s(2) = 12\,\text{ft}$, $s(6) = 108\,\text{ft}$

  • $\Delta t = 6 - 2 = 4\,\text{s}$

  • $\Delta s = 108 - 12 = 96\,\text{ft}$

  • $v_{\text{avg}} = \frac{96\,\text{ft}}{4\,\text{s}} = 24\,\text{ft/s}$

  • The secant line passing through $(2, 12)$ and $(6, 108)$ has a slope of $24$

  • Interval $t \in [1, 5]$:

  • $t_0 = 1\,\text{s}$, $t_1 = 5\,\text{s}$

  • $s(1) = 3\,\text{ft}$, $s(5) = 75\,\text{ft}$

  • $\Delta t = 5 - 1 = 4\,\text{s}$

  • $\Delta s = 75 - 3 = 72\,\text{ft}$

  • $v_{\text{avg}} = \frac{72\,\text{ft}}{4\,\text{s}} = 18\,\text{ft/s}$

  • The secant line passing through $(1, 3)$ and $(5, 75)$ has a slope of $18$

Instantaneous Velocity and Limits

• Core Question:

  • What is the exact velocity of the motorcycle at the single instant $t = 4\,\text{s}$?

• Limiting Process Concept:

  • Instantaneous velocity cannot be calculated directly by substituting $t_1 = t_0 = 4$, as this yields $\frac{0}{0}$ (undefined).

  • Solution: Construct a sequence of average velocities over shrinking time intervals $[t_0, t_1]$, fixing $t_0 = 4\,\text{s}$ and letting $t_1 \rightarrow 4\,\text{s}$.

• Numerical / Table Approach ($t_0 = 4\,\text{s}$, $s(4) = 48\,\text{ft}$):

  • Interval $[4, 5]$:

  • $t_1 = 5\,\text{s}$, $s(5) = 75\,\text{ft}$

  • $\Delta t = 5 - 4 = 1\,\text{s}$

  • $\Delta s = 75 - 48 = 27\,\text{ft}$

  • $v_{\text{avg}} = \frac{27}{1} = 27\,\text{ft/s}$ (secant slope = $27$)

  • Interval $[4, 4.5]$:

  • $t_1 = 4.5\,\text{s}$, $s(4.5) = 3(4.5)^2 = 60.75\,\text{ft}$

  • $\Delta t = 4.5 - 4 = 0.5\,\text{s}$

  • $\Delta s = 60.75 - 48 = 12.75\,\text{ft}$

  • $v_{\text{avg}} = \frac{12.75}{0.5} = 25.5\,\text{ft/s}$ (secant slope = $25.5$)

  • Interval $[4, 4.1]$:

  • $t_1 = 4.1\,\text{s}$, $s(4.1) = 3(4.1)^2 = 50.43\,\text{ft}$

  • $\Delta t = 4.1 - 4 = 0.1\,\text{s}$

  • $\Delta s = 50.43 - 48 = 2.43\,\text{ft}$

  • $v_{\text{avg}} = \frac{2.43}{0.1} = 24.3\,\text{ft/s}$

  • Interval $[4, 4.01]$:

  • $t_1 = 4.01\,\text{s}$, $s(4.01) = 3(4.01)^2 = 48.2403\,\text{ft}$

  • $\Delta t = 4.01 - 4 = 0.01\,\text{s}$

  • $\Delta s = 48.2403 - 48 = 0.2403\,\text{ft}$

  • $v_{\text{avg}} = \frac{0.2403}{0.01} = 24.03\,\text{ft/s}$

  • Continuous Refinement ($t_1 = 4.001\,\text{s}$, $t_1 = 4.0001\,\text{s}$):

  • As $t_1 \rightarrow 4$, the average velocity sequence converges toward $24\,\text{ft/s}$.

• Graphical Approach and Tangent Lines:

  • As $t_1$ approaches $t_0$, the series of secant lines converges to a single line called the tangent line.

  • Tangent line definition: A line touching the curve at exactly one localized point (t_0, s(t_0))$.\n * Key Connection: The slope of the tangent line at t = 4\,\text{s}$equals the instantaneous velocity ($v_{\text{inst}}).\n * Value at t = 4\,\text{s}$:$v_{\text{inst}} = 24\,\text{ft/s}.\n\n\n# Questions & Classroom Discussion\n\n* **Question**: What kind of curve is represented by the equation 15t?\n * **Response**: It is a straight line.\n\n* **Question**: What is the slope of the bicycle position equation s_{\text{bicycle}}(t) = 15t?\n * **Response**: The slope is \frac{15}{1} = 15\n\n* **Question**: What happens to the average velocity expression if t_1 = t_0?\n * **Response**: It is undefined due to division by zero.\n\n* **Question**: What is 2^2$, and what is the resulting position value when multiplied by$3$in$s(t) = 3t^2?\n * **Response**: 2^2 = 4$, and$3 \times 4 = 12\n\n* **Question**: What happens to the speedometer reading as the motorcycle accelerates along the freeway on-ramp?\n * **Response**: The speed is increasing, meaning the slope of the position curve becomes steeper.\n\n* **Question**: What values are even closer to 4$than$4.01?\n * **Response**: 4.001$,$4.0001, and so forth.\n\n* **Question**: What value is the sequence of average velocities approaching as t_1 \rightarrow 4?\n * **Response**: It approaches 24$$