Motion along a Straight Line: Physics I Study Notes

Learning Outcomes

  • Define position, displacement, velocity, speed, and acceleration.
  • Distinguish between vector and scalar kinematic quantities.
  • Use the definitions of average velocity and acceleration to perform simple computations.
  • State the definition of free fall and identify common characteristics of free-falling objects.
  • Apply the kinematic equations for 1D motion of particles/objects.
  • Interpret displacement-time, velocity-time, and acceleration-time graphs.
  • Apply a simple mathematical model to solve one-dimensional and two-dimensional motion problems (CLO 2).

Study of Motion: Kinematics vs. Dynamics

  • Kinematics: Often referred to as the "geometry of motion," kinematics deals with pure motion without reference to the masses or forces involved in it. It is concerned with describing how objects move.
  • Dynamics: Dynamics is concerned with the study of forces and torques and their effect on motion. It is concerned with explaining why motion occurs.
  • Both Kinematics and Dynamics are branches of Mechanics.

Displacement and Average Velocity

  • Relative Position: When a particle moves along a straight line (e.g., the x-axis), its position is described by a single coordinate xx.
  • Displacement: The change in position of a particle. If at time t1t_1 a point is at position x1x_1 and at time t2t_2 it is at position x2x_2, the displacement is the vector from x1x_1 to x2x_2. The x-component of the displacement is given by:   Δx=x2x1\Delta x = x_2 - x_1
  • Average Velocity (vav,xv_{av,x}): A vector defined in the direction from x1x_1 to x2x_2. The x-component is calculated as:   vav,x=x2x1t2t1=ΔxΔtv_{av,x} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t}
  • Units: The SI unit for velocity is meters per second (m/sm/s).
  • Displacement-Time Graph:   - Position xx is a function of time tt.   - The average velocity between two positions is represented by the slope of a line (secant line) connecting the two corresponding points on a graph of position as a function of time.
Example: Travel from Abu Dhabi to Dubai
  • Scenario: A vehicle leaves Abu Dhabi at 09:45 AM and arrives in Dubai at 11:20 AM. The distance is 135km135\,km.
  • Time Interval Calculation:   - 09:45AM09:45\,AM to 11:20AM=1hour35minutes=95minutes11:20\,AM = 1\,hour\,35\,minutes = 95\,minutes.   - 95min×60s/min=5700s95\,min \times 60\,s/min = 5700\,s.
  • Distance Conversion:   - 135km=135,000m135\,km = 135,000\,m.
  • Average Velocity Calculation:   - vav,x=135,000m5700s=23.7m/sv_{av,x} = \frac{135,000\,m}{5700\,s} = 23.7\,m/s.
Example: Ant Crawling Exercise
  • Scenario: An ant crawls along a wire (x-axis) through points A, B, C, and D. OO is the origin. Measurements: AB=50cmAB = 50\,cm, BC=30cmBC = 30\,cm, and AO=5cmAO = 5\,cm. Point DD overlaps point AA.
  • Positions:   - xA=5cmx_A = -5\,cm   - xB=xA+50cm=45cmx_B = x_A + 50\,cm = 45\,cm   - xC=xB30cm=15cmx_C = x_B - 30\,cm = 15\,cm   - xD=5cmx_D = -5\,cm
  • Interval Analysis:   - A to B: Displacement = +50cm+50\,cm, Distance = 50 cm\text{50 cm}.   - B to C: Displacement = 30cm-30\,cm, Distance = 30 cm\text{30 cm}.   - C to D: Displacement = xDxC=515=20cmx_D - x_C = -5 - 15 = -20\,cm, Distance = 20 cm\text{20 cm}.   - A to D: Displacement = xDxA=5(5)=0cmx_D - x_A = -5 - (-5) = 0\,cm, Distance = 50+30+20=100cm50 + 30 + 20 = 100\,cm.

Instantaneous Velocity and Speed

  • Definition: The instantaneous velocity at any point on a graph of coordinate xx vs. time tt is equal to the slope of the tangent line to the curve at that specific point.
  • Instantaneous Speed: The magnitude of the instantaneous velocity at a specific moment.
  • Technology Example: Instantaneous speed cameras measure speed at a specific point, whereas average speed cameras calculate speed over a distance between two points.

Average and Instantaneous Acceleration

  • Definition: Acceleration is a quantitative description of the rate of change of velocity with time.
  • Average Acceleration (aav,xa_{av,x}): If at time t1t_1 a particle has velocity v1xv_{1x} and at time t2t_2 it has velocity v2xv_{2x}, the average acceleration is:   aav,x=v2xv1xt2t1=ΔvxΔta_{av,x} = \frac{v_{2x} - v_{1x}}{t_2 - t_1} = \frac{\Delta v_x}{\Delta t}
  • Instantaneous Acceleration (axa_x): The limit approached by the average acceleration as the time interval approaches zero (Δt0\Delta t \to 0). It is the slope of the tangent line to the curve on a velocity-time (vtv-t) graph.
  • Positive and Negative Acceleration:   - If velocity and acceleration have the same sign, the object is speeding up.   - If velocity and acceleration have opposite signs, the object is slowing down.

Graphical Representation of Motion

  • Types of Graphs:   1. Displacement-time (xtx-t)   2. Velocity-time (vtv-t)   3. Acceleration-time (ata-t)
  • Slope Relationships:   - The slope of an xtx-t graph equals velocity.   - The slope of a vtv-t graph equals acceleration.
  • Area Relationships:   - The area under a vtv-t graph equals the displacement (or distance) travelled.

Motion with Constant Acceleration

  • If acceleration is constant, velocity changes at the same rate throughout the motion.
  • Kinematic Equations:   1. vx=v0x+axtv_x = v_{0x} + a_x t   2. x=x0+v0xt+12axt2x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2   3. vx2=v0x2+2ax(xx0)v_x^2 = v_{0x}^2 + 2 a_x (x - x_0)   4. xx0=12(v0x+vx)tx - x_0 = \frac{1}{2} (v_{0x} + v_x) t
  • Where:   - v0xv_{0x} = Initial velocity   - vxv_x = Final velocity   - axa_x = Constant acceleration   - x0x_0 = Initial position   - xx = Final position   - tt = Time elapsed
Example: Passing Speed
  • Scenario: A car traveling at 15m/s15\,m/s accelerates at 2.0m/s22.0\,m/s^2 for 5.0s5.0\,s.
  • Calculations:   - Final Velocity: vx=15+(2.0×5.0)=25m/sv_x = 15 + (2.0 \times 5.0) = 25\,m/s.   - Distance Travelled: xx0=(15×5.0)+12(2.0)(5.02)=75+25=100mx - x_0 = (15 \times 5.0) + \frac{1}{2} (2.0) (5.0^2) = 75 + 25 = 100\,m.
Example: Traffic Signal Braking
  • Scenario: A car at 20m/s20\,m/s brakes to a stop in 5.0s5.0\,s.
  • Calculations:   - Acceleration: ax=0205.0=4.0m/s2a_x = \frac{0 - 20}{5.0} = -4.0\,m/s^2.   - Distance to line: xx0=12(20+0)×5.0=50mx - x_0 = \frac{1}{2} (20 + 0) \times 5.0 = 50\,m.
Example: Ball Rebounding off Wall
  • Scenario: A ball hits a wall horizontally at 4.8m/s4.8\,m/s and rebounds at 3.6m/s3.6\,m/s. Contact time is 280ms280\,ms.
  • Calculations:   - Change in velocity: Δv=vfinalvinitial=(3.6)4.8=8.4m/s\Delta v = v_{final} - v_{initial} = (-3.6) - 4.8 = -8.4\,m/s.   - Acceleration: a=8.4m/s0.280s=30m/s2a = \frac{-8.4\,m/s}{0.280\,s} = -30\,m/s^2.

Free Fall Motion

  • Core Definition: Free fall is motion with constant acceleration under the influence of Earth's gravity, where air resistance is neglected.
  • Galileo’s Discovery: In the absence of air resistance, all objects fall with the same constant acceleration regardless of their weight/mass.
  • Acceleration due to Gravity (gg):   - Magnitude near Earth's surface: g9.8m/s2g \approx 9.8\,m/s^2.   - Magnitude on the Moon: g1.6m/s2g \approx 1.6\,m/s^2.
  • Conventions for Equation Application:   - Acceleration in the y-direction (aya_y) is usually taken as g-g (where upward is positive).   - At the highest point of a trajectory, the vertical velocity component is 00.
Example: Tennis Serving Machine
  • Scenario: Ball shot vertically UP at 30m/s30\,m/s.
  • Graphs Analysis:   - Acceleration: Continuous horizontal line at 9.8m/s2-9.8\,m/s^2.   - Velocity: Downward sloping straight line starting at +30+30, crossing zero at the peak, and reaching 30-30 upon return.   - Position: Parabola opening downwards (inverted U-shape).
Example: Ball Thrown from Building Roof
  • Scenario: Ball thrown upward from a roof with v0y=15.0m/sv_{0y} = 15.0\,m/s. Position y0=0y_0 = 0.
  • Calculations for 1.00 s:   - Velocity: vy=15.0+(9.8×1.0)=5.2m/sv_y = 15.0 + (-9.8 \times 1.0) = 5.2\,m/s.   - Position: y=(15.0×1.0)+12(9.8)(1.02)=10.1my = (15.0 \times 1.0) + \frac{1}{2} (-9.8) (1.0^2) = 10.1\,m.
  • Calculations for 4.00 s:   - Velocity: vy=15.0+(9.8×4.0)=24.2m/sv_y = 15.0 + (-9.8 \times 4.0) = -24.2\,m/s.   - Position: y=(15.0×4.0)+12(9.8)(4.02)=6078.4=18.4my = (15.0 \times 4.0) + \frac{1}{2} (-9.8) (4.0^2) = 60 - 78.4 = -18.4\,m.
  • Calculations for Maximum Height:   - Velocity at max height = 00.   - Time to reach top: t=015.09.8=1.53st = \frac{0 - 15.0}{-9.8} = 1.53\,s.   - Max Height: y=(15.0×1.53)+12(9.8)(1.532)=11.5my = (15.0 \times 1.53) + \frac{1}{2} (-9.8) (1.53^2) = 11.5\,m.

Summary Key Points

  • Displacement segments Δy\Delta y and Δx\Delta x are interchangeable based on the axis of motion.
  • In free fall problems, always assume ay=g=9.8m/s2a_y = -g = -9.8\,m/s^2.
  • Launch points are typically defined as t=0t=0, x=0x=0, and y=0y=0.
  • Reading assignment: Section 2.7 regarding Relative Velocity along a straight line (excluding the Theory of Relativity section).