Notes on Motion in a Straight Line (Chapter 2)

2.1 Introduction

• Motion is the change in position of an object with time; study focuses on describing motion using velocity and acceleration.
• Focussed on rectilinear (one-dimensional) motion along a straight line and, for uniform acceleration, a set of simple kinematic equations can be derived.
• Treat objects as point objects when their size is negligible compared with the distances involved (valid in many real-life situations).
• The aim is to describe motion without addressing causes of motion; causes are addressed in Chapter 4.
• Relative velocity introduces the idea that motion is frame-dependent and depends on the observer’s frame of reference.
• The material links to foundational principles (sign conventions, limits, calculus) and has real-world relevance (vehicles, projectiles, falling bodies).

2.2 Instantaneous velocity and speed

• Instantaneous velocity (velocity) at an instant t is defined as the limit of the average velocity as the time interval ∆t → 0:
  v =
  \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}.

• Interpretation: velocity is the rate of change of position with respect to time at that instant; it is the slope of the position–time (x–t) curve at that instant.

• Graphical method (conceptual): use smaller and smaller time intervals around t to approximate the tangent slope; the velocity at t is the slope of the tangent to the x–t curve at that point.

• Numerical demonstration (Fig./Table reference): for a position function x(t) = 0.08 t^3, as ∆t decreases (2.0 s, 1.0 s, 0.5 s, 0.1 s, 0.01 s) around t = 4.0 s, the average velocity ∆x/∆t approaches 3.84 m s⁻¹, the velocity at t = 4.0 s.

• Key relation: instantaneous velocity is the derivative dx/dt; it can be obtained graphically, numerically, or analytically if x(t) is known.

• Instantaneous speed: the magnitude of instantaneous velocity, i.e. |v|. For uniform motion, instantaneous speed equals the constant velocity magnitude and the average velocity over any interval equals the instantaneous velocity at any instant within that interval.

• Example (Example 2.1): If x(t) = a + b t^2, then

  v(t) = \frac{dx}{dt} = 2 b t.
  With a = 8.5 m and b = 2.5 m s⁻², at t = 0 s, v = 0; at t = 2.0 s, v = 10 m s⁻¹.

  • The average velocity over a time interval can be computed as the change in x divided by the interval.

2.3 Acceleration

• Acceleration describes the rate of change of velocity with time. The average acceleration over a time interval is
  \bar{a} = \frac{v2 - v1}{t2 - t1}.
• Instantaneous acceleration at time t is defined similarly to velocity:
  a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}.
• The acceleration at an instant is the slope of the tangent to the v–t (velocity–time) curve at that instant.
• Because velocity has both magnitude and direction, acceleration can result from changes in speed, changes in direction, or both. Consequently, acceleration can be positive, negative, or zero.
• Graphical notes: with positive acceleration, the x–t graph curves upward; with negative acceleration, it curves downward; with zero acceleration, the x–t graph is a straight line. The v–t graph is a straight line when acceleration is constant.
• For motion with constant acceleration, the average acceleration equals the constant acceleration during the interval.
• If the velocity at t = 0 is v₀ and at time t is v, then
  v = v_0 + a t.
• The area under the velocity–time (v–t) curve between two instants equals the displacement over that interval (a general result from calculus; here illustrated for constant acceleration).
• Velocity–time and position–time diagrams: constant acceleration yields a linear v–t graph and a parabolic x–t graph; the area under the v–t curve from t₁ to t₂ gives the displacement: \Delta x = \int{t1}^{t_2} v\, dt.
• Non-smooth graphs are idealizations; in realistic situations, acceleration and velocity change smoothly (differentiable functions).

2.4 Kinematic equations for uniformly accelerated motion

• For motion with constant acceleration a, the following relationships hold among displacement x, time t, initial velocity v₀, final velocity v, and acceleration a:

  1) v = v_0 + a t.

  2) x = x0 + v0 t + \tfrac{1}{2} a t^2.

  These were derived by recognizing that the instantaneous velocity is the slope of the v–t relation and that displacement equals the area under the v–t curve over the interval.

• An equivalent form (eliminating t between v and x) is the velocity–squared relation:

  3) v^2 = v0^2 + 2 a (x - x0).

• Special case when x0 = 0 (origin at t = 0):

  • v = v_0 + a t
  • x = v_0 t + \tfrac{1}{2} a t^2
  • v^2 = v_0^2 + 2 a x.

• These five quantities (x, t, v, v0, a) are linked by the set of kinematic equations for one-dimensional motion with constant acceleration.

• Generalization for x0 ≠ 0: replace x by x − x0 in the equations to obtain the modified forms.

• Example approaches (calculus-based derivation):

  • Start from the definition a = dv/dt with a constant, integrate to get v = v0 + a t.
  • Use dx/dt = v and integrate to obtain x = x0 + v0 t + (1/2) a t^2.
  • Eliminate t to obtain v^2 = v0^2 + 2 a (x − x0).

• Example 2.3 (upward throw) and Example 2.4 (free fall) illustrate applications of these equations:

  • Example 2.3 (upward throw): with v0 = +20 m s⁻¹, a = −g = −10 m s⁻², and y0 = 25 m, the maximum height relative to launch is 20 m; total time to hit ground is 5 s (two methods shown: (i) split into ascent and descent; (ii) use a single quadratic in y with y = 0 as final position).
  • Example 2.4 (free fall): for an object released from rest near Earth, with a = −g (g ≈ 9.8 m s⁻²), the standard relations are
  • v = - g t,
  • y = -\tfrac{1}{2} g t^2,
  • v^2 = -2 g y.

• Example 2.2 (calculus method for deriving the equations) demonstrates that the same results can be obtained via integrals:

  • Start with dv/dt = a (constant), integrate to obtain v(t).
  • Then use dx/dt = v(t) and integrate to obtain x(t).

• Example 2.5 (Galileo’s law of odd numbers) shows that, for free fall from rest, the distances covered in successive equal time intervals τ are in the ratio 1:3:5:7:9:11:…

• Example 2.6 (stopping distance): for a moving vehicle braking with constant deceleration −a, the stopping distance ds is
  ds = -\frac{v_0^2}{2 a}.
  Since a is negative, ds is positive. Doubling the initial speed increases stopping distance by a factor of 4 (since ds ∝ v₀^2).

• Example 2.7 (reaction time): the reaction time tᵣ can be inferred from a ruler-drop experiment where d = (1/2) g tᵣ^2 (with vo = 0 and a = −g). Thus
  tᵣ = \sqrt{\frac{2 d}{g}}.

2.5 Relative velocity

• The transcript mentions relative velocity as part of the chapter outline, emphasizing that motion is frame-dependent. (Details are not included in the provided content; conceptually, relative velocity involves comparing velocities of objects as seen from different reference frames.)

Summary of key results and concepts

• Motion is defined by changes in position with time; rectilinear motion uses a single spatial axis with sign conventions.

• Instantaneous velocity is the limit of average velocity as the time interval goes to zero:
  v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}.

• Instantaneous speed is the magnitude of instantaneous velocity: |v|.

• Acceleration is the rate of change of velocity:

  • Average: \bar{a} = \frac{v2 - v1}{t2 - t1}.
  • Instantaneous: a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}.

• The area under the v–t curve between t₁ and t₂ equals the displacement: \Delta x = \int{t1}^{t_2} v \; dt.

• For motion with constant acceleration, the kinematic equations link x, t, v, v0, and a:

  v = v0 + a t, x = x0 + v0 t + \tfrac{1}{2} a t^2, v^2 = v0^2 + 2 a (x - x_0).

• These equations reduce to simpler forms when x0 = 0: x = v0 t + \tfrac{1}{2} a t^2\,,\quad v^2 = v0^2 + 2 a x.

• Physical interpretations and caveats:

  • The sign of acceleration depends on the chosen positive direction; it does not alone indicate whether speed is increasing or decreasing.
  • A particle at rest at a given instant can still have non-zero acceleration.
  • The equations assume one-dimensional motion with constant acceleration over the interval considered.

Exercises (topics and illustrative prompts)

• 2.1: When can a body be treated as a point object? (Orders of magnitude, jerky motion vs smooth motion, etc.)
• 2.2: Interpret x–t graphs of two walkers; deduce who lives closer, who starts earlier, who walks faster, and whether they overtake each other and when.
• 2.3: Plot the x–t graph for a woman walking 2.5 km to work at 5 km/h, staying until 5:00 pm, and returning by auto at 25 km/h; determine scales and behavior.
• 2.4: Drunkard’s walk: 5 steps forward, 3 back, repeated; length of a step = 1 m, 1 s per step; plot x–t and determine time to fall into a 13 m pit.
• 2.5: Stopping distance: a car stops within 200 m from initial speed 126 km/h; determine the uniform deceleration and stopping time.
• 2.6: A ball thrown upward with initial speed 29.4 m/s; (a) direction of acceleration during upward motion? (b) velocity and acceleration at the highest point? (c) define the origin at the highest point with downward as positive; signs of x, v, a during upward and downward motion; (d) maximum height and total time to return (g = 9.8 m/s²).
• 2.7: True/false statements about 1D motion with zero speed, zero velocity, constant speed vs acceleration, etc., with reasons.
• 2.8: A ball dropped from 90 m with a loss of one-tenth of speed on each floor collision; plot speed–time from t = 0 to 12 s.
• 2.9: Distinguish between (a) magnitude of displacement over an interval and total path length; (b) magnitude of average velocity and average speed; show the latter is always greater than or equal to the former, with equality conditions.
• 2.10: A man walks 2.5 km to market at 5 km/h, returns at 7.5 km/h; compute the magnitude of average velocity and the average speed over intervals (i) 0–30 min, (ii) 0–50 min, (iii) 0–40 min; illustrates why average speed is total path length divided by time, not the magnitude of average velocity.
• 2.11: Explain why instantaneous speed equals the magnitude of instantaneous velocity.
• 2.12–2.18: Graph analysis questions: which graphs cannot represent 1D motion, SHM plots, and interpretation of speed-time graphs; sign conventions and accelerations in different time intervals.

Connections and implications

• The instantaneous concepts bridge algebraic equations with calculus, enabling precise descriptions of motion.
• The kinematic equations provide practical tools for solving projectile-like and car braking problems without needing forces or energy arguments.
• Understanding the difference between average velocity vs average speed informs correct interpretation of motion over intervals, especially when direction changes occur.
• Relative velocity highlights that motion is frame-dependent; measurements can differ between observers depending on their reference frames.

Notation and conventions used in this chapter

• Positive direction along the x-axis is a matter of choice and must be stated; signs of x, v, a follow this convention.
• Displacements, velocities, and accelerations can be negative depending on the chosen axis.
• The calculus-based definitions ensure exactness (dx/dt, dv/dt) while the kinematic equations assume constant acceleration over the interval considered.

Key formulas to memorize

• Instantaneous velocity:
  v = \frac{dx}{dt}
• Average acceleration:
  \bar{a} = \frac{\Delta v}{\Delta t}
• Instantaneous acceleration:
  a = \frac{dv}{dt}
• Displacement from velocity:
  \Delta x = \int{t1}^{t_2} v\, dt
• Constant-acceleration kinematics:
  v = v0 + a t x = x0 + v0 t + \tfrac{1}{2} a t^2 v^2 = v0^2 + 2 a (x - x_0)
• Special case x0 = 0:
  x = v0 t + \tfrac{1}{2} a t^2, \quad v^2 = v0^2 + 2 a x.