Velocity, Acceleration, and One-Dimensional Kinematics

Fundamentals of Position, Velocity, and Speed

• Vector Nature of Motion Variables:

  • Both velocity and acceleration are vector quantities.
  • Every vector quantity is defined by having both a magnitude and a direction.

• Velocity Defined:

  • Velocity measures how rapidly an object's position changes with respect to elapsed time.
  • Displacement represents the net change in position: final position minus initial position.
  • Velocity represents displacement divided by time.

• Formulas for Velocity:

  • Average Velocity: Calculated over a measurable time interval using the formula:         $v = \frac{x_f - x_i}{t}$         Where $x_f$ is the final position, $x_i$ is the initial position, and $t$ is the elapsed time.
  • Instantaneous Velocity: Obtained from the same fundamental ratio as the elapsed time $t$ approaches zero ($t \rightarrow 0$), measuring velocity at an exact point in time.

• Distinction Between Speed and Velocity:

  • Speed: A scalar quantity indicating how rapidly distance is changing with respect to time.
  • Velocity: A vector quantity indicating how rapidly displacement is changing with respect to time.
  • Speed represents the magnitude of velocity.

• Concrete Rate-of-Change Examples for Velocity and Speed:

  • Vehicle Traveling East at $30\,\text{miles per hour}$:
    • Magnitude is $30\,\text{miles per hour}$; direction is East.
    • After $1\,\text{hour}$, distance traveled is $30\,\text{miles}$.
    • After $2\,\text{hours}$, distance traveled is $60\,\text{miles}$.
    • After $3\,\text{hours}$, distance traveled is $90\,\text{miles}$.
  • Object Moving at a Speed of $8\,\text{m/s}$:
    • Distance changes by $8\,\text{meters}$ every second.
    • After $1\,\text{second}$, distance traveled is $8\,\text{meters}$.
    • After $2\,\text{seconds}$, distance traveled is $16\,\text{meters}$.
    • After $3\,\text{seconds}$, distance traveled is $24\,\text{meters}$.
  • Vehicle Moving at a Speed of $15\,\text{m/s}$:
    • Distance changes by $15\,\text{meters}$ every elapsed second.

Acceleration and the First Kinematic Equation

• Acceleration Defined:

  • Acceleration measures how rapidly velocity is changing with respect to time.
  • Just as speed quantifies how fast distance changes, acceleration quantifies how fast velocity changes.

• Formulas for Acceleration:

  • Average Acceleration: Defined as the change in velocity ($\Delta v$) divided by the elapsed time ($t$) over a significant time interval:         $a = \frac{\Delta v}{t} = \frac{v_f - v_i}{t}$         Where $v_f$ is the final velocity, $v_i$ is the initial velocity, and $t$ is the elapsed time.
  • Instantaneous Acceleration: Defined as the change in velocity over time as the time interval $t$ approaches zero ($t \rightarrow 0$).

• Conceptual Rate of Change for Acceleration ($8\,\text{m/s}^2$):

  • An acceleration of $8\,\text{m/s}^2$ means the velocity increases by $8\,\text{m/s}$ every second.
  • Second-by-second velocity tracking starting from rest:
    • At $t = 1\,\text{s}$, velocity is $8\,\text{m/s}$.
    • At $t = 2\,\text{s}$, velocity is $16\,\text{m/s}$.
    • At $t = 3\,\text{s}$, velocity is $24\,\text{m/s}$.
    • At $t = 4\,\text{s}$, velocity is $32\,\text{m/s}$.

• Derivation of the Final Velocity Formula:

  • Begin with the average acceleration formula:         $a = \frac{v_f - v_i}{t}$
  • Multiply both sides by $t$:         $a \times t = v_f - v_i$
  • Add $v_i$ to both sides:         $v_i + a \times t = v_f$
  • Reordered kinematic equation:         $v_f = v_i + a \times t$
  • Prerequisite Condition: This kinematic equation is strictly valid only when the acceleration is constant.

Problem Set: Acceleration Calculations and Velocity Tabulation

• Problem 1: Average Acceleration Calculation and Tabulation:

  • Given Information:
    • Initial speed ($v_i$): $15\,\text{m/s}$
    • Final speed ($v_f$): $45\,\text{m/s}$
    • Time elapsed ($t$): $5\,\text{s}$
  • Formula:$a = \frac{v_f - v_i}{t}$
  • Calculation:$a = \frac{45\,\text{m/s} - 15\,\text{m/s}}{5\,\text{s}} = \frac{30\,\text{m/s}}{5\,\text{s}} = 6\,\text{m/s}^2$
  • Second-by-Second Velocity Progression Table:
    • When acceleration is positive, velocity is increasing.
    • At $t = 0\,\text{s}$, velocity is $15\,\text{m/s}$.
    • At $t = 1\,\text{s}$, velocity is $15 + 6 = 21\,\text{m/s}$.
    • At $t = 2\,\text{s}$, velocity is $21 + 6 = 27\,\text{m/s}$.
    • At $t = 3\,\text{s}$, velocity is $27 + 6 = 33\,\text{m/s}$.
    • At $t = 4\,\text{s}$, velocity is $33 + 6 = 39\,\text{m/s}$.
    • At $t = 5\,\text{s}$, velocity reaches the final speed of $45\,\text{m/s}$.

• Problem 2: Unit Conversions with Truck Acceleration:

  • Given Information:
    • Initial speed ($v_i$): $25\,\text{km/h}$
    • Final speed ($v_f$): $45\,\text{km/h}$
    • Time elapsed ($t$): $40\,\text{s}$
  • Part A: Acceleration in $\text{km/(h}\cdot\text{s)}$:$a = \frac{45\,\text{km/h} - 25\,\text{km/h}}{40\,\text{s}} = \frac{20\,\text{km/h}}{40\,\text{s}} = 0.5\,\text{km/(h}\cdot\text{s)}$
  • Part B: Conversion to $\text{m/s}^2$ via Dimensional Analysis:
    • Conversion factors: $1\,\text{km} = 1000\,\text{m}$, $1\,\text{h} = 60\,\text{min}$, and $1\,\text{min} = 60\,\text{s}$.
    • Fractional setup:             $\frac{0.5\,\text{km}}{1\,\text{h} \times 1\,\text{s}} \times \frac{1000\,\text{m}}{1\,\text{km}} \times \frac{1\,\text{h}}{60\,\text{min}} \times \frac{1\,\text{min}}{60\,\text{s}}$
    • Cancelling units leaves $\text{m/s}^2$.
    • Stepwise arithmetic:             $0.5 \times 1000 = 500$$\frac{500}{60} \approx 8.33$$\frac{8.33}{60} \approx 0.13\bar{8}\,\text{m/s}^2$
    • Result: $0.138\,\text{repeating}\,\text{m/s}^2$

Problem Set: Calculating Final Speed under Constant Acceleration

• Problem 3: Acceleration from Rest:

  • Terminology Rule: The phrase "from rest" explicitly signifies that the initial speed is zero ($v_i = 0\,\text{m/s}$).
  • Given Information:
    • Initial speed ($v_i$): $0\,\text{m/s}$
    • Acceleration ($a$): $3.5\,\text{m/s}^2$
    • Time elapsed ($t$): $12\,\text{s}$
  • Formula:$v_f = v_i + a \times t$
  • Calculation:$v_f = 0 + (3.5 \times 12) = 42\,\text{m/s}$

• Problem 4: Acceleration from Non-Zero Initial Velocity:

  • Given Information:
    • Initial speed ($v_i$): $12\,\text{m/s}$
    • Constant acceleration ($a$): $1.2\,\text{m/s}^2$
    • Time elapsed ($t$): $15\,\text{s}$
  • Formula:$v_f = v_i + a \times t$
  • Calculation:$v_f = 12 + (1.2 \times 15)$$1.2 \times 15 = 18$$v_f = 12 + 18 = 30\,\text{m/s}$

Deceleration and Negative Acceleration

• Problem 5: Deceleration to a Complete Stop:

  • Given Information:
    • Initial speed: $95\,\text{miles per hour}$
    • Coming to rest implies final speed ($v_f$): $0\,\text{m/s}$
    • Time elapsed ($t$): $4\,\text{s}$
  • Initial Velocity Conversion ($\text{miles per hour}$ to $\text{m/s}$):
    • Conversion factors:             $1\,\text{h} = 60\,\text{min}$$1\,\text{min} = 60\,\text{s}$$1\,\text{km} = 0.6214\,\text{miles}$$1\,\text{km} = 1000\,\text{m}$
    • Setup using dimensional analysis:             $\frac{95\,\text{miles}}{1\,\text{h}} \times \frac{1\,\text{h}}{60\,\text{min}} \times \frac{1\,\text{min}}{60\,\text{s}} \times \frac{1\,\text{km}}{0.6214\,\text{miles}} \times \frac{1000\,\text{m}}{1\,\text{km}}$
    • Cancelling units leaves $\text{m/s}$.
    • Calculation steps:             $\frac{95}{60 \times 60 \times 0.6214} \times 1000 \approx 42.47\,\text{m/s}$
    • Initial velocity ($v_i$): $42.47\,\text{m/s}$
  • Calculation of Average Acceleration:$a = \frac{v_f - v_i}{t}$$a = \frac{0\,\text{m/s} - 42.47\,\text{m/s}}{4\,\text{s}} = \frac{-42.47\,\text{m/s}}{4\,\text{s}} \approx -10.6\,\text{m/s}^2$

• Physical Meaning of Negative Acceleration:

  • Whenever velocity is decreasing, acceleration is negative.
  • A negative acceleration value signifies that the vehicle is slowing down, decelerating, and coming to a stop in the direction of motion.