Lecture 3: Linear Motion and Kinematics

Introduction to Kinematics

• Kinematics is defined as the study of the motion of objects.

• Linear motion serves as the foundation for exploring kinematics, providing a framework to analyze how far an object travels, how fast it moves, and how quickly its speed changes.

• These principles apply to objects moving at a constant velocity as well as those moving under the influence of external forces, such as gravity.

• Studying linear motion establishes the necessary background for more advanced physics topics, including forces, momentum, and energy.

• The core concepts categorized within this study include path and displacement, speed and velocity, acceleration, and free-fall.

Path and Displacement

• Path and displacement are concepts used to analyze the distance and movement of an object as it travels between points.

• Path describes the total motion of an object from its starting point to its finishing point.

  • It is a scalar quantity, meaning it has magnitude but no specific direction.

  • Symbols used for path include the lowercase letters $d$, $x$, or $y$.

  • Standard practice utilizes $d$ for general distance, $x$ for horizontal motion, and $y$ for vertical motion.

  • The standard unit of measurement for path is meters ($\text{m}$).

• Displacement describes the straight-line distance from the starting point to the finishing point.

  • It is a vector quantity and must include a direction.

  • Symbols used for displacement include $\Delta d$, $\Delta x$, or $\Delta y$.

  • The delta symbol ($\Delta$) specifically indicates interest in the "change of position."

  • The standard unit of measurement for displacement is meters ($\text{m}$).

• Example: Classroom Walk

  • Walking at a constant pace from the $0$ meter mark to the $4$ meter mark results in a path (distance) of $4\,\text{m}$.

• Example: Driving from Naperville to the College of DuPage (COD)

  • Driving a car requires following roads rather than traveling in a perfectly straight line.

  • In this specific scenario, the path (total route driven) is $15,600\,\text{m}$ or $15.6\,\text{km}$.

  • The displacement (change in position from start to finish) is $11,100\,\text{m}$ or $11.1\,\text{km}$ Northeast.

  • Regardless of the specific path taken between these two points, the displacement remains constant as the vector pointing from start to finish.

• Measuring Movement

  • Path can be determined using a car's odometer.

  • Displacement can be calculated if the coordinates of the starting and ending locations are known by drawing a straight line and splitting it into horizontal ($\Delta x$) and vertical ($\Delta y$) components using the Pythagorean theorem.

• Calculation Example: Vector Magnitude

  • Starting location: $(300, 400)\,\text{meters}$.

  • Ending location: $(1200, 900)\,\text{meters}$.

  • Horizontal component ($\Delta x$): $1200\,\text{m} - 300\,\text{m} = 900\,\text{m}$.

  • Vertical component ($\Delta y$): $900\,\text{m} - 400\,\text{m} = 500\,\text{m}$.

  • Magnitude of displacement: $\sqrt{(900\,\text{m})^2 + (500\,\text{m})^2} = 1030\,\text{m}$.

• Example: Coiled Plastic Tubing

  • A BB travels through a coiled tube with a total length of $1.5\,\text{m}$.

  • Path of the BB = $1.5\,\text{m}$.

  • Displacement of the BB = $-0.73\,\text{m}$. The negative sign is used to indicate the vector points downward.

• Displacement in Round Trips

  • If a person walks from the $0$ meter mark to the $4$ meter mark and back to the $2$ meter mark:

    • Path = $4\,\text{m} + 2\,\text{m} = 6\,\text{m}$.

    • Displacement = $2\,\text{m} - 0\,\text{m} = 2\,\text{m}$.

  • If a person walks from the $0$ meter mark to the $4$ meter mark and returns to the $0$ meter mark:

    • Path = $4\,\text{m} + 4\,\text{m} = 8\,\text{m}$.

    • Displacement = $0\,\text{m} - 0\,\text{m} = 0\,\text{m}$.

  • Rule: Whenever the starting and ending locations are identical, the displacement is zero, regardless of path size.

Speed and Velocity

• Speed and velocity describe the rate at which an object changes its position.

• Speed is a scalar quantity.

  • Symbols: lowercase $s$ or $|v|$ (magnitude of velocity).

  • Unit: meters per second ($\text{m/s}$).

• Velocity is a vector quantity.

  • Symbols: lowercase $v$ written in bold or with an arrow over the top.

  • Unit: meters per second ($\text{m/s}$).

• Average vs. Instantaneous Values

  • Average Values: Describe motion from the beginning to the end of a trip.

    • $\text{Average Speed} = \frac{\text{Path}}{\text{Time}}$

    • $\text{Average Velocity} = \frac{\text{Displacement}}{\text{Time}}$

  • Instantaneous Values: Describe how fast an object is moving at a specific snapshot in time.

• Data from Driving Example (Naperville to COD)

  • Path = $15.6\,\text{km}$, Time = $20\,\text{minutes}$ ($\frac{1}{3}\,\text{hour}$).

  • Average Speed: $15.6\,\text{km} / (1/3)\,\text{h} = 46.8\,\text{km/h}$ (approximately $29\,\text{mph}$).

  • This includes time spent at red lights, stop signs, and variations in speed.

  • Instantaneous speed is read on the speedometer and may fluctuate between $0\,\text{km/h}$ and $72\,\text{km/h}$.

  • Average Velocity: $11.1\,\text{km Northeast} / (1/3)\,\text{h} = 33.3\,\text{km/h}$ Northeast.

• Comparison of Speed and Velocity

  • Since the shortest distance between two points is a straight line, displacement is always $\le$ path.

  • Therefore, average velocity is always $\le$ average speed.

  • To determine instantaneous velocity in a car, use the speedometer for speed and a compass for direction.

• Data from BB in Tubing Example

  • Path = $1.5\,\text{m}$, Displacement = $-0.73\,\text{m}$, Time = $1.7\,s$.

  • Average Speed: $1.5\,\text{m} / 1.7\,s = 0.88\,\text{m/s}$.

  • Average Velocity: $-0.73\,\text{m} / 1.7\,s = -0.43\,\text{m/s}$.

• Data from Constant Pace Classroom Walk

  • Path = $8\,\text{m}$, Displacement = $0\,\text{m}$, Time = $14.2\,s$.

  • Average Speed = $0.56\,\text{m/s}$. Since pace was constant, instantaneous speed also = $0.56\,\text{m/s}$.

  • Average Velocity = $0\,\text{m/s}$.

Acceleration

• Acceleration quantifies the rate at which an object changes its velocity.

• Symbol: lowercase $a$.

• Unit: meters per second-squared ($\text{m/s}^2$).

• Definition and Formulas

  • $\text{Acceleration} = \frac{\Delta v}{t} = \frac{v_f - v_i}{t}$

  • In physics, acceleration refers to three distinct changes: speeding up, slowing down, or changing direction.

  • It is incorrect in a physics context to use "acceleration" only for speeding up.

• Constant Acceleration

  • In cases of constant acceleration, the instantaneous acceleration equals the average acceleration.

  • Situations like the BB in the tube or a car driving through city streets involve non-constant acceleration and are more complex to analyze.

• General Motion Equations (Required when acceleration is non-zero)

  • $d = \frac{1}{2}at^2 + v_it$

  • $v_f = v_i + at$

Motion Detector Experiments

• Experiment 1: Flat Track

  • A cart moves along a flat, low-friction track with no forces acting to change its speed.

  • Position vs. Time graph: A straight line pointing upward (steady increase in position).

  • Velocity vs. Time graph: A straight horizontal line (constant velocity).

  • Initial speed = $0.5\,\text{m/s}$, Final speed = $0.5\,\text{m/s}$.

  • Acceleration = $0\,\text{m/s}^2$.

• Experiment 2: Tilted Track (Downward)

  • The cart starts from rest ($v_i = 0$) and rolls down a tilted track.

  • Position vs. Time graph: A curved line sloping upward (increasing rate of change).

  • Velocity vs. Time graph: A straight line pointing upward (constant increase in speed).

  • Measured data: $v_i = 0.11\,\text{m/s}$, $v_f = 1.09\,\text{m/s}$ after $1.9\,s$.

  • Calculated Acceleration: $(1.09 - 0.11) / 1.9 = 0.52\,\text{m/s}^2$.

  • Calculated Distance: $\frac{1}{2}(0.52\,\text{m/s}^2)(1.9\,s)^2 + (0.11\,\text{m/s})(1.9\,s) = 1.15\,\text{m}$.

  • Verification: The detector recorded an initial position of $0.2\,\text{m}$ and a final position of $1.35\,\text{m}$ ($\text{change} = 1.15\,\text{m}$).

• Experiment 3: Tilted Track (Up and Down)

  • The cart is pushed up a ramp with an initial velocity of $-0.96\,\text{m/s}$.

  • Phase 1: Moves up, slows down. Velocity is negative, acceleration is positive (opposing vectors).

  • Phase 2: Briefly stops at the top ($v = 0$) to change direction. Acceleration is still $0.57\,\text{m/s}^2$.

  • Phase 3: Rolls down, speeds up. Velocity is positive, acceleration is positive (vectors in same direction).

  • Position vs. Time graph: A parabola (bowl shape).

  • Calculated Acceleration: $(0.93 - (-0.96)) / 3.3 = 0.57\,\text{m/s}^2$.

Free-Fall and Gravity

• Free-fall is defined as a situation where gravity is the only force acting on an object (ignoring friction and air drag).

• Gravity Particulars

  • Acceleration due to gravity ($g$) near Earth's surface is approximately $9.8\,\text{m/s}^2$ downward.

  • In calculations, this value is often expressed as $-9.8\,\text{m/s}^2$.

  • Gravity is constant and never "turns off," even when an object stops momentarily at the peak of its flight.

• Ballistic Cart Demonstration

  • A ball is launched upward at $v_i = 3.6\,\text{m/s}$.

  • Known values: Launch velocity = $3.6\,\text{m/s}$, Peak velocity = $0\,\text{m/s}$, Return velocity = $-3.6\,\text{m/s}$.

• Trajectory Analysis Calculations

  • Time to reach peak: $t = (v_f - v_i) / a = (0 - 3.6) / -9.8 = 0.37\,s$.

  • Maximum Height: $d = \frac{1}{2}(-9.8)(0.37)^2 + (3.6)(0.37) = 0.66\,\text{m}$.

  • Total round-trip time: $t = (-3.6 - 3.6) / -9.8 = 0.73\,s$.

  • Symmetry: It takes an equal amount of time for the ball to go up as it does to return.

• Calculation at specific time (t = 0.5s)

  • Height above base: $d = \frac{1}{2}(-9.8)(0.5)^2 + (3.6)(0.5) = 0.575\,\text{m}$.

  • Instantaneous Velocity: $v_f = 3.6 + (-9.8)(0.5) = -1.3\,\text{m/s}$.

  • The negative sign confirms the ball is moving downward at that time.