Physics Chapter 2: Motion Along a Straight Line

Chapter 2: Motion Along a Straight Line

Introduction to Classical Physics

• This chapter deals with "classical physics" or "Newtonian physics".
• The field was essentially invented by Isaac Newton (1642-1726).

Reference Frames

• A reference frame is crucial for describing motion.
• It typically includes:
  • An origin (a fixed point).
  • A reference direction.
• These are chosen by convention or for convenience.
• We commonly use perpendicular axes through the origin (e.g., North, South, West, East).

Unit Vectors

• A unit vector is a vector with a magnitude of 1 (dimensionless).
• It indicates a specific direction without units.
• Common unit vectors for Cartesian coordinates:
  • \hat{i} = unit vector along the +x-direction.
  • \hat{j} = unit vector along the +y-direction.
  • \hat{k} = unit vector along the +z-direction.
• Any general vector \vec{A} can be expressed in terms of its components and unit vectors as: \vec{A} = Ax \hat{i} + Ay \hat{j} + A_z \hat{k} .

Vector Example: Displacement

• Consider a displacement: Run 20 m East, then Go 15 m North.
• Using a coordinate system where East is +x and North is +y:
  • The x-component of the displacement vector is 20 m.
  • The y-component of the displacement vector is 15 m.
  • The resultant displacement vector D is expressed as: D = 20\hat{i} + 15\hat{j} .

Position, Distance, and Displacement

• One-dimensional or Linear Motion: Motion that occurs along a straight line.
  • In this case, we only need one axis, and direction is specified by positive (+) or negative (-) signs.
• Position
  • The location of an object relative to a fixed origin.
• Distance
  • The total length traveled by an object.
  • It is always a positive scalar quantity.
  • Example: A car's odometer measures distance.
• Displacement (\Delta x)
  • The difference between the ending position (xf) and the starting position (xi).
  • It is a vector quantity, having both magnitude and direction.
  • Formula: \Delta x = xf - xi
  • Direction:
    • If \Delta x > 0, displacement is to the right (positive direction).
    • If \Delta x < 0, displacement is to the left (negative direction).

Position vs. Time Graph

• Mathematically, the position (x) can be described as a function of time (t), i.e., x = f(t). This is represented by a Position versus time Graph.

Speed and Average Velocity

• Speed
  • A positive scalar quantity.
  • Your car's speedometer measures speed.
  • Formula: \text{Speed} = \frac{\text{Total distance}}{\text{Total time}}
• Velocity
  • A vector quantity, possessing both size (magnitude) and direction.
  • In one-dimension, its direction is represented by positive (+) or negative (-) signs.
  • Example: "I travel south at 60 mi/hr" specifies both magnitude and direction.
• Average Velocity (v_{av})
  • Defined as the total displacement divided by the total time taken.
  • Formula: v{av} = \frac{\text{Total displacement}}{\text{Total time}} = \frac{xf - xi}{tf - t_i} = \frac{\Delta x}{\Delta t} .
  • Note: \Delta always signifies final minus initial.
• Derived Formulas from Average Velocity:
  • \Delta x = v_{av} \cdot \Delta t
  • \Delta t = \frac{\Delta x}{v_{av}}

Relation to Position-Time Graphs

• The slope of an x-t (position versus time) graph represents the average velocity between two points.
  • \text{Slope} = \frac{\Delta x}{\Delta t} = v_{av} .
  • When \Delta x > 0, the slope is positive, indicating net motion to the right.
  • When \Delta x < 0, the slope is negative, indicating net motion to the left.

Example: Jogger's Average Speed

• Scenario: A jogger runs at 6 m/s for 2 min (120 s) and then walks at 2 m/s for 400 m.
• Objective: Find his average speed.
• Solution Strategy: Average speed = \frac{\text{Total distance}}{\text{Total time}} .
  1. Calculate distance for running segment: d1 = \text{speed}1 \cdot \text{time}_1 = 6 \text{ m/s} \cdot 120 \text{ s} = 720 \text{ m}.
  2. Total distance: d{ ext{total}} = d1 + d_2 = 720 \text{ m} + 400 \text{ m} = 1120 \text{ m}.
  3. Calculate time for walking segment: t2 = \frac{d2}{\text{speed}_2} = \frac{400 \text{ m}}{2 \text{ m/s}} = 200 \text{ s}.
  4. Total time: t{ ext{total}} = t1 + t_2 = 120 \text{ s} + 200 \text{ s} = 320 \text{ s}.
  5. Average speed: v_{avg} = \frac{1120 \text{ m}}{320 \text{ s}} = 3.5 \text{ m/s}.
• Note: The average speed is not simply the average of the two speeds ((6 + 2)/2 = 4 m/s), because the time intervals or distances for each speed are different.

Instantaneous Velocity

• Concept: While average velocity provides overall motion, instantaneous velocity gives a more detailed picture of motion at a specific moment.
• Definition: Instantaneous velocity is the average velocity as the time interval becomes infinitesimally small.
  • Formula: v = \lim_{\Delta t\to0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} .
• Graphical Interpretation: It is the slope of the tangent line to the 'x' vs 't' (position versus time) plot at a specific time 't'.
• Instantaneous Speed: What a car's speedometer measures; it is the magnitude of the instantaneous velocity.
• Usage: When we simply say 'velocity', we usually mean the instantaneous velocity.

Rules for the Sign of x-Velocity

Acceleration

• Definition: Acceleration is the rate of change of velocity with respect to time.
• Nature: It is a vector quantity.
• Deceleration: Often, when an object's speed decreases with time, we refer to it as deceleration, which is acceleration in the direction opposite to the velocity.
• Conditions for Acceleration: A car accelerates when it:
  • 'Speeds up' (velocity magnitude increases).
  • 'Slows down' (velocity magnitude decreases, i.e., deceleration).
  • Changes direction (even if speed is constant, velocity changes, thus acceleration occurs).
• Average Acceleration (a_{av})
  • Formula: a{av} = \frac{vf - vi}{tf - t_i} = \frac{\Delta v}{\Delta t} .
  • Units: meters per second per second (m/s/s) or meters per second squared (m/s^2).
• Graphical Interpretation:
  • On a v-t (velocity vs. time) graph: Acceleration is the slope of the tangent to the graph.
  • On an x-t (position vs. time) graph: Acceleration is represented by the curvature of the curve.

Velocity vs. Time Graph Interpretation

• Increasing velocity (Speeding): Positive slope.
• Constant velocity: Zero slope (horizontal line).
• Decreasing velocity (Slowing): Negative slope.
• Zero velocity (Turning point): Intersects the time axis.
• Increasing negative velocity (Speeding in negative direction): Negative slope, moving further from zero on the negative velocity axis.

Example: Car Acceleration

• Scenario 1 (Speeding up): A car goes from 0 mi/hr to 60 mi/hr in 8 seconds.
  • Convert 60 mi/hr to m/s: 60 \text{ mi/hr} \approx 26.82 \text{ m/s}.
  • Average acceleration: a_{av} = \frac{26.82 \text{ m/s} - 0 \text{ m/s}}{8 \text{ s}} = 3.35 \text{ m/s}^2 .
  • This means, on average, the car's speed increases by 3.35 m/s every second.
• Scenario 2 (Slowing down): A car goes from 60 mi/hr to 0 mi/hr in 8 seconds.
  • Average acceleration: a_{av} = \frac{0 \text{ m/s} - 26.82 \text{ m/s}}{8 \text{ s}} = -3.35 \text{ m/s}^2 .
  • This negative acceleration means the car is slowing down; its speed decreases by 3.35 m/s every second on average.

Instantaneous Acceleration

• Definition: Instantaneous acceleration is the average acceleration as the time interval approaches zero.
  • Formula: a = \lim_{\Delta t\to0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2x}{dt^2} .
• Graphical Interpretation: It is the slope of the tangent line to the velocity versus time graph at a specific instant.
• Relationship between Velocity and Acceleration Signs:
  • 'Speeding Up': The velocity (vx) and acceleration (ax) must have the same sign (both positive or both negative).
  • 'Slowing Down': The velocity (vx) and acceleration (ax) must have opposite signs (one positive, one negative).

x-t Graph Curvature and Acceleration

• The curvature of an x-t graph at any point indicates the particle's acceleration at that point.
• Curvature upward: Positive acceleration (a_x > 0).
• Curvature downward: Negative acceleration (a_x < 0).
• No curvature (straight line): Zero acceleration (a_x = 0).
• The greater the curvature (positive or negative), the greater the magnitude of the particle's acceleration.

Summary of Definitions So Far

• Displacement: \Delta x
• Average Velocity: v_{av} = \frac{\Delta x}{\Delta t}
• Instantaneous Velocity: v = \lim_{\Delta t\to0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} (slope of x-t plot)
• Average Acceleration: a_{av} = \frac{\Delta v}{\Delta t}
• Instantaneous Acceleration: a = \lim_{\Delta t\to0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2x}{dt^2} (slope of v-t plot, curvature of x-t plot)

Motion with Constant Acceleration

• Importance: Constant acceleration is a common scenario, often corresponding to a constant force (e.g., gravity).
• If acceleration (a) is constant, then a = \frac{vf - vi}{t - 0} = \text{constant}. This leads to the first kinematic equation:
  1. Velocity as a function of time: vf = vi + at

Derivations of Kinematic Equations (for constant acceleration)

• From the definition of average velocity: v{av} = \frac{xf - x_i}{t - 0}.
• This leads to: xf = xi + v{av}t (or x = x0 + v_{av}t ).
• For constant acceleration, the average velocity can also be expressed as the arithmetic mean of the initial and final velocities:
  2. Average velocity: v{av} = \frac{1}{2}(v0 + v)
• Combining (2) with x = x0 + v{av}t: Substitute v_{av} from (2) into the position equation:
  3. Position as a function of time, using average velocity: x = x0 + \frac{1}{2}(v0 + v)t .
• Deriving position without final velocity: Substitute equation (1) (v = v_0 + at) into equation (3):
  • x = x0 + \frac{1}{2}(v0 + (v_0 + at))t
  • x = x0 + \frac{1}{2}(2v0 + at)t
  • x = x0 + v0t + \frac{1}{2}at^2
  1. Position as a function of time: x = x0 + v0t + \frac{1}{2}at^2
     • x = Position at time t
     • x_0 = Position at time 0
     • v_0 = x-velocity at time 0
     • t = Time
     • a = Constant acceleration
• Deriving final velocity without time: Eliminate t between equations (1) and (3):
  • From (1): t = \frac{v - v_0}{a}
  • Substitute this into (3): x = x0 + \frac{1}{2}(v + v0)\left(\frac{v - v_0}{a}\right)
  • x - x0 = \frac{1}{2a}(v^2 - v0^2)
  • 2a(x - x0) = v^2 - v0^2
  • v^2 = v0^2 + 2a(x - x0) or v^2 = v_0^2 + 2a\Delta x
  1. Velocity squared: v^2 = v_0^2 + 2a\Delta x

Summary of Kinematic Equations (Constant Acceleration)

Problem-Solving Steps for Kinematic Problems

1. Draw a picture: Visualize the situation.
2. List knowns and unknowns: Identify all given variables and what needs to be found.
3. Find the equation: Select a kinematic equation that contains only one unknown, allowing you to solve for it.

Example: Boy Sliding Down a Slide

• Scenario: A boy slides down a 12 ft long slide with constant acceleration, reaching the bottom in 2 s.

• Knowns: x0 = 0 \text{ ft}, x = 12 \text{ ft}, t = 2 \text{ s}, v0 = 0 \text{ ft/s} (starts from rest).

• Unknowns: a (acceleration), v (bottom speed).

• a) What was his acceleration?

  • Use equation (4): x = x0 + v0t + \frac{1}{2}at^2
  • 12 \text{ ft} = 0 + (0)(2 \text{ s}) + \frac{1}{2}a(2 \text{ s})^2
  • 12 \text{ ft} = 2a \text{ s}^2
  • a = \frac{12 \text{ ft}}{2 \text{ s}^2} = 6 \text{ ft/s}^2

• b) What is his bottom speed?

  • Method 1 (using avg. velocity): Use equation (3) x = x0 + \frac{1}{2}(v0 + v)t
    • 12 \text{ ft} = 0 + \frac{1}{2}(0 + v)(2 \text{ s})
    • 12 \text{ ft} = v \text{ s}
    • v = 12 \text{ ft/s}
  • Method 2 (using acceleration): Use equation (1) v = v_0 + at
    • v = 0 \text{ ft/s} + (6 \text{ ft/s}^2)(2 \text{ s})
    • v = 12 \text{ ft/s}

Example: Car and Motorcycle Catch Up

• Scenario: At t=0, a car has a constant velocity of 20 m/s. At the same time, a motorcycle starts from rest with an acceleration of 2 m/s^2.

• Objective: How long does it take for the motorcycle to catch up? What distance is traveled?

• 1. Analyze the car's motion (constant velocity):

  • v_c = 20 \text{ m/s} (constant)
  • Since velocity is constant, a_c = 0.
  • Position equation: xc = x{c0} + v_ct
  • If starting from x{c0} = 0, then xc = v_ct = 20t

• 2. Analyze the motorcycle's motion (constant acceleration):

  • x_{m0} = 0 (starts from rest)
  • v_{m0} = 0 (starts from rest)
  • a_m = 2 \text{ m/s}^2
  • Position equation: xm = x{m0} + v{m0}t + \frac{1}{2}amt^2
  • x_m = 0 + (0)t + \frac{1}{2}(2)t^2 = t^2

• 3. Determine when they catch up (where their positions are equal):

  • xc = xm
  • 20t = t^2
  • t^2 - 20t = 0
  • t(t - 20) = 0
  • Possible solutions: t = 0 (starting point) or t = 20 \text{ s}.
  • The motorcycle catches up at t = 20 \text{ s}.

• 4. Determine the distance traveled: Substitute t = 20 \text{ s} into either position equation.

  • Using car's position: x_c = 20 \text{ m/s} \cdot 20 \text{ s} = 400 \text{ m}.
  • Using motorcycle's position: x_m = (20 \text{ s})^2 = 400 \text{ m}.
  • The distance traveled is 400 m.

Gravity and Free Fall

• Fundamental Question: Do different objects fall at the same rate?
  • In a vacuum, yes (e.g., ball and feathers dropped in air vs. vacuum).
  • In air, air resistance plays a role, especially for lighter objects.
• Observations for a Falling Object (ignoring air resistance):
  • Distance increases with time.
  • Velocity increases with time.
  • Velocity increases by the same amount every second (due to constant acceleration).
  • Position increases as the square of the time.

Acceleration due to Gravity

• On the surface of the Earth, gravity causes objects to fall with a nearly constant acceleration.
• This acceleration is denoted by g.
• Value of g:
  • g = 9.81 \text{ m/s}^2
  • g = 32.2 \text{ ft/s}^2
• Variation: 'g' varies slightly with location and elevation.
• Air Resistance: For simplicity, we typically ignore air resistance in introductory physics, assuming ideal free-fall conditions.
• Sign Convention: 'g' itself is a positive magnitude (9.81 m/s^2). When using it in equations, we account for the downward direction of gravity with a negative sign (e.g., a = -g if upward is positive).

Kinematic Equations for Free Fall

(Replacing a with -g and often x with y for vertical motion, assuming upward is positive)

1. v = v_0 - gt
2. v{av} = \frac{v0 + v}{2}
3. x = x0 + \frac{1}{2}(v0 + v)t
4. x = x0 + v0t - \frac{1}{2}gt^2
5. v^2 = v0^2 - 2g(x - x0)

Example: Rock Falling into a Well

• Scenario: It takes a rock 3 s to hit the bottom of a well.
• Knowns: t = 3 \text{ s}, x0 = 0 (starting point), v0 = 0 (dropped from rest), g = 9.81 \text{ m/s}^2.
• Objective: What is the depth of the well?
• Solution: Use equation (4), with a = -g
  • x = x0 + v0t - \frac{1}{2}gt^2
  • x = 0 + (0)(3 \text{ s}) - \frac{1}{2}(9.81 \text{ m/s}^2)(3 \text{ s})^2
  • x = -\frac{1}{2}(9.81)(9) \text{ m}
  • x = -44.145 \text{ m} (approximately -44.1 m)
• The negative sign indicates that the final position is below the initial (ground) surface level.

Symmetrical Motion of a Projectile Tossed Upward

• For an object tossed upward, the path is symmetrical (assuming no air resistance).
• Key Points:
  • Velocity is zero (v=0) at the very top of the path, where the object momentarily stops before turning around and falling back down.
  • Acceleration (a = -g) remains constant throughout the entire flight (upward, at the peak, and downward).

Example: Diver Jumping Off a Cliff

• Scenario: A diver jumps off a cliff 4 m above the water with an initial upward speed of 5 m/s. Let positive x be upward, and water level be x=0.

• Knowns: x0 = 4 \text{ m}, v0 = 5 \text{ m/s}, a = -g = -9.81 \text{ m/s}^2.

• a) How much time elapses before she reaches her highest point?

  • At the highest point, v = 0 \text{ m/s}.
  • Use equation (1): v = v_0 - gt
  • 0 = 5 \text{ m/s} - (9.81 \text{ m/s}^2)t
  • t = \frac{5 \text{ m/s}}{9.81 \text{ m/s}^2} = 0.51 \text{ s}.

• b) How high up did she rise (above the water level)?

  • At the highest point, v = 0 \text{ m/s}.
  • Use equation (5) (since time is not involved): v^2 = v0^2 - 2g(x - x0)
  • 0^2 = (5 \text{ m/s})^2 - 2(9.81 \text{ m/s}^2)(x - 4 \text{ m})
  • 0 = 25 - 19.62(x - 4)
  • 19.62(x - 4) = 25
  • x - 4 = \frac{25}{19.62} \approx 1.274
  • x = 4 + 1.274 = 5.274 \text{ m} (approximately 5.28 m) above the water level.

• c) With what speed did she hit the water?

  • When she hits the water, x = 0 \text{ m}.
  • Use equation (5): v^2 = v0^2 - 2g(x - x0)
  • v^2 = (5 \text{ m/s})^2 - 2(9.81 \text{ m/s}^2)(0 \text{ m} - 4 \text{ m})
  • v^2 = 25 - 2(9.81)(-4)
  • v^2 = 25 + 78.48 = 103.48
  • v = \pm\sqrt{103.48} \approx \pm 10.17 \text{ m/s}
  • Since she is moving downward, her velocity is negative: v = -10.17 \text{ m/s}.

• d) How long did the diver take to hit the water?

  • Method 1 (Quadratic Formula): Use equation (4): x = x0 + v0t - \frac{1}{2}gt^2

    • 0 = 4 \text{ m} + (5 \text{ m/s})t - \frac{1}{2}(9.81 \text{ m/s}^2)t^2
    • Rearrange into standard quadratic form (at^2 + bt + c = 0):
      (-\frac{1}{2}g)t^2 + (v0)t + (x0) = 0
      -4.905t^2 + 5t + 4 = 0
    • Apply quadratic formula: t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
      t = \frac{-5 \pm \sqrt{5^2 - 4(-4.905)(4)}}{2(-4.905)}
      t = \frac{-5 \pm \sqrt{25 + 78.48}}{-9.81}
      t = \frac{-5 \pm \sqrt{103.48}}{-9.81} = \frac{-5 \pm 10.17}{-9.81}
    • Two solutions: t = \frac{-5 + 10.17}{-9.81} = \frac{5.17}{-9.81} \approx -0.53 \text{ s} (extraneous) or t = \frac{-5 - 10.17}{-9.81} = \frac{-15.17}{-9.81} \approx 1.55 \text{ s} .
    • Choose the positive time: t = 1.55 \text{ s}.

  • Method 2 (Using final velocity from part c): Use equation (1): v = v_0 - gt

    • -10.17 \text{ m/s} = 5 \text{ m/s} - (9.81 \text{ m/s}^2)t
    • -15.17 \text{ m/s} = -(9.81 \text{ m/s}^2)t
    • t = \frac{-15.17 \text{ m/s}}{-9.81 \text{ m/s}^2} = 1.55 \text{ s}.

  • Method 3 (Using average velocity from part c): Use equation (3): x = x0 + \frac{1}{2}(v0 + v)t

    • 0 = 4 \text{ m} + \frac{1}{2}(5 \text{ m/s} + (-10.17 \text{ m/s}))t
    • -4 \text{ m} = \frac{1}{2}(-5.17 \text{ m/s})t
    • -4 \text{ m} = (-2.585 \text{ m/s})t
    • t = \frac{-4 \text{ m}}{-2.585 \text{ m/s}} = 1.55 \text{ s}.

Example: Determining 'g' Experimentally

• Scenario: An object falls from rest for 2 m in 0.64 s.
• Knowns: x0 = 0 (fall from origin), x = -2 \text{ m} (ends below origin), t = 0.64 \text{ s}, v0 = 0 (from rest).
• Objective: What is the value of g?
• Solution: Use equation (4): x = x0 + v0t - \frac{1}{2}gt^2
  • -2 \text{ m} = 0 + (0)(0.64 \text{ s}) - \frac{1}{2}g(0.64 \text{ s})^2
  • -2 \text{ m} = -\frac{1}{2}g(0.4096 \text{ s}^2)
  • g = \frac{-2 \text{ m}}{-\frac{1}{2}(0.4096 \text{ s}^2)} = \frac{4 \text{ m}}{0.4096 \text{ s}^2}
  • g = 9.765625 \text{ m/s}^2 (approximately 9.77 m/s^2)