Kinematics: Displacement, Velocity, and Acceleration – Lecture Notes

Average Velocity

• Definition: average velocity is the displacement divided by the time interval over which the motion occurs.
  $\vec{v}<em>{\text{avg}} = \frac{\Delta \vec{x}}{\Delta t} = \frac{\vec{x}(t</em>2) - \vec{x}(t<em>1)}{t</em>2 - t_1}$
• In 1D, displacement is along the x-axis; the numerator is a vector, time is a scalar, so the average velocity is a vector.
• The path vs displacement idea: the plotted path (set of displacement vectors at successive times) is not the same as the true continuous path; as the time step shrinks, the set of segments better approximates the true path.
• Distance vs displacement: in 1D, total distance can give the same magnitude as displacement, but in more complex motion (2D/3D) they can differ.
• The coordinate system matters: the sign of the displacement or velocity depends on the chosen axis direction.

Displacement vs Time Graph and Instantaneous Velocity

• Graph setup (example): displacement x is plotted as a function of time t (e.g., drag racer example); axes definitions matter for interpretation.
• Instantaneous velocity is the slope of the displacement-versus-time curve as \Delta t → 0:
  $\vec{v}(t) = \frac{d \vec{x}}{dt}$
• In 1D, this reduces to $v = \frac{dx}{dt}$, a vector pointing in the instantaneous direction of motion.
• Graph interpretation:
  • The slope of the curve between two points gives the average velocity over that interval.
  • The tangent slope at a point gives the instantaneous velocity at that moment.
• Motion interpretation along the curve:
  • If the tangent becomes steeper with time, speed increases (positive acceleration in the forward direction).
  • A horizontal tangent implies velocity is zero (instantaneous stop).
  • A negative slope implies negative velocity (motion in the opposite direction).
  • When the curve passes through the origin or changes sign, the direction can flip; the velocity sign encodes that direction along the x-axis.
• Points on the displacement-vs-time plot (labeled a, b, c, d, e in the lecture) illustrate:
  • Point a: positive slope → positive velocity.
  • Point b: slope more vertical → higher speed.
  • Point c: horizontal slope → zero velocity (stopped).
  • Point d: slope indicating reversal of direction (negative velocity).
  • Point e: slope reducing but still positive (slower positive velocity).

Instantaneous Velocity and Acceleration

• Instantaneous velocity is the first derivative of displacement with respect to time:
  $\vec{v}(t) = \frac{d \vec{x}}{dt}$
• Acceleration is the rate of change of velocity:
  $\vec{a} = \frac{d \vec{v}}{dt} = \frac{d^2 \vec{x}}{dt^2}$
• Units (SI):
  • Displacement: meters (m)
  • Velocity: meters per second (m/s)
  • Acceleration: meters per second squared (m/s^2)
• Sign conventions and interpretation:
  • Acceleration is a vector pointing in the direction of the change in velocity.
  • If velocity changes in time, there is acceleration; if acceleration is zero, velocity is constant.
  • Deceleration is acceleration in the opposite direction to velocity.
• Real-world context:
  • Gravitational acceleration is approximately constant: $g \approx 9.81 \ \mathrm{m\,s^{-2}}$
  • The field of forces and acceleration will be revisited in the next chapters.
• Narrative example from the lecture: Colonel John Paul Stapp experienced extreme accelerations (approximately 46.2 g), illustrating that humans can experience very large accelerations under controlled, engineered circumstances.

Constant Acceleration and Kinematic Equations

• When acceleration is constant, we can derive three key equations from calculus: 1) Velocity as a function of time: $\vec{!v}(t) = \vec{v}<em>0 + \vec{a} \; t$ 2) Displacement as a function of time: $\vec{x}(t) = \vec{x}</em>0 + \vec{v}_0 \; t + \tfrac{1}{2} \vec{a} \; t^2$ 3) Velocity–position relation (eliminating time):
  • In 1D:
    $v^2 = v<em>0^2 + 2 a (x - x</em>0)$
  • In general vector form, using dot products:
    $\vec{v} \cdot \vec{v} = \vec{v}<em>0 \cdot \vec{v}</em>0 + 2 \vec{a} \cdot (\vec{x} - \vec{x}_0)$
• Derivation outline (calculus-based):
  • Start with acceleration as the derivative of velocity: $\vec{a} = \frac{d \vec{v}}{dt}$ and integrate to get velocity: $\vec{v}(t) = \vec{v}_0 + \vec{a} t$
  • Then express displacement as the integral of velocity: $\vec{x}(t) = \vec{x}<em>0 + \int</em>0^t \vec{v}(t') dt' = \vec{x}<em>0 + \vec{v}</em>0 t + \tfrac{1}{2} \vec{a} t^2$
  • To obtain the third equation, eliminate time by solving for t from the first equation and substitute into the second; after algebra, you arrive at the velocity–position relation above.
• Key takeaways about the equations:
  • They assume constant acceleration; they are extremely useful in solving problems quickly.
  • They apply in vector form for multi-dimensional motion; in 1D they reduce to scalar equations with sign conventions.
  • The third equation is often the one students memorize for quick problem solving: $v^2 = v<em>0^2 + 2 a (x - x</em>0).$

Sign Conventions, Deceleration, and Examples

• Deceleration is simply acceleration opposite in direction to the velocity; its sign depends on the coordinate choice.
  • Example: car moving to the right (positive x) with initial velocity 15 m/s; after 5 s, velocity is 5 m/s with acceleration -2.0 m/s^2. This is a deceleration in the positive direction.
  • Conversely, a car moving left (negative velocity) can experience a positive acceleration and still be slowing down if the velocity and acceleration are opposite in sign.
• Ball-wall collision thought experiment:
  • Before impact: velocity has a certain direction; after impact: velocity direction reverses.
  • The interval of contact involves a nonzero acceleration; the question’s correct interpretation is that acceleration occurs during the change of velocity, not only before or after the impact.
  • This reinforces that velocity is a vector and direction changes imply acceleration.
• Big-picture takeaway: If velocity changes, there must be acceleration; if acceleration is zero, velocity is constant. This mental model is a powerful problem-solving heuristic.

Constant Acceleration in Context

• Recognize a constant-acceleration model as an idealization that works well for gravity near the Earth, free fall on a short timescale, and many introductory problems.
• The lecture plans to connect constant-acceleration equations to circular motion in a later chapter, where acceleration points toward the center of the circle and changes direction continuously.
• The professor emphasizes the importance of having the three core equations readily available on a formula sheet for exams.

Graphical and Conceptual Takeaways

• Velocity is the slope of the displacement–time graph; speed is the magnitude of velocity (nonnegative by definition).
• Acceleration is the slope of the velocity–time graph; it represents how quickly velocity is changing in time.
• In 1D problems, keep track of sign conventions for v, a, and x; in 2D/3D, use vector forms and consider projections along the direction of motion.
• Always connect the math to the physics: a change in velocity implies acceleration; a nonzero acceleration implies a changing velocity; zero acceleration implies constant velocity.

Practical Notes for Studying and Exam Preparation

• Memorize the three kinematic equations for constant acceleration and know how to derive them from the definitions.
• Be able to switch between vector form and 1D scalar form, including dot products for general 2D/3D motion.
• Practice interpreting displacement–time graphs: identify where velocity is positive/negative, zero, or changing.
• Practice differentiating between displacement and distance, especially in non-straight-line motion.
• Remember SI units and dimensional analysis as a sanity check for equations and derivatives.

Quick Reference: Core Formulas (Constant Acceleration)

• Velocity:
  $\vec{v}(t) = \vec{v}_0 + \vec{a} t$
• Position:
  $\vec{x}(t) = \vec{x}<em>0 + \vec{v}</em>0 t + \tfrac{1}{2} \vec{a} t^2$
• Velocity–Position (scalar form):
  $v^2 = v<em>0^2 + 2 a (x - x</em>0)$
• General dot-product form (multi-dimensional):
  $\vec{v} \cdot \vec{v} = \vec{v}<em>0 \cdot \vec{v}</em>0 + 2 \vec{a} \cdot (\vec{x} - \vec{x}_0)$
• Instantaneous quantities:
  • $\vec{v}(t) = \dfrac{d \vec{x}}{dt}$
  • $\vec{a}(t) = \dfrac{d \vec{v}}{dt} = \dfrac{d^2 \vec{x}}{dt^2}$