Comprehensive Notes on One-Dimensional Kinematics

Fundamentals of One-Dimensional Motion

• One-Dimensional Motion Definition: Motion along a straight line represents the simplest form of kinematics, wherein an object's position changes along a single spatial dimension (commonly designated as the $x$-axis for horizontal motion or the $y$-axis for vertical motion).
• Position and Reference Systems:
  • The position of an object is defined as its precise location relative to a chosen reference point.
  • This chosen reference point is defined as the origin of the coordinate system ($x = 0$).
  • Position is a coordinate value that identifies where the particle resides at any specific instant in time $t$.
• Translational Motion and the Particle Model:
  • When analyzing the motion of macroscopic objects (such as an automobile), internal degrees of freedom, rotations, and deformations can often be neglected.
  • The object is modeled as a mathematical point particle possessing mass but zero volume, focusing exclusively on translational motion.

• Tracking Position Over Time:
  • Motion can be observed by recording the position of the particle on a coordinate axis at successive time intervals.
  • Positions along the coordinate line may take positive values, negative values, or zero relative to the origin:

• Position-Time Graph Analysis:
  • A position-time graph plots time ($t$, in seconds) on the horizontal axis and position ($x$, in meters) on the vertical axis.
  • The curve on a position-time graph describes the continuous trajectory of the particle across time.
  • Trajectory points from the observed particle motion:
  • Point $A$: Located at $t = 0\,s$, where $x = 30\,m$.
  • Point $B$: Located at $t = 10\,s$, where the particle reaches a maximum coordinate of approximately $x = 53\,m$.
  • Point $C$: Located at $t = 20\,s$, where the particle moves back to $x = 38\,m$.
  • Point $D$: Located at $t = 30\,s$, where the particle crosses the origin ($x = 0\,m$).
  • Point $E$: Located at $t = 40\,s$, where the position becomes negative at $x = -38\,m$.
  • Point $F$: Located at $t = 50\,s$, reaching $x = -53\,m$.
  • The straight secant line connecting two points (such as between point $A$ and point $B$) has a horizontal run of $\Delta t = t_B - t_A$ and a vertical rise of $\Delta x = x_B - x_A$.

Displacement Versus Distance

• Displacement ($\Delta x$):
  • Displacement is a vector quantity defined as the change in position of a particle during a specific time interval.
  • Mathematical formula:     $\Delta x = x_f - x_i$
  • In this equation, $x_f$ represents the final position coordinate, and $x_i$ represents the initial position coordinate.
  • Displacement has both magnitude and direction.
  • In one dimension, direction is indicated by the algebraic sign: $\Delta x$ can be strictly positive, negative, or zero.
• Distance ($D$):
  • Distance is a scalar quantity defined as the complete length of the path followed by the particle during its travel.
  • Distance depends entirely on the full path traversed and is completely independent of direction.
  • Distance is always non-negative: $D \ge 0$.
• Direct Comparison Example:
  • Consider a particle that starts at a given position, moves $20\,m$ to the right, and subsequently returns $20\,m$ to the left back to its exact starting point:

• Total displacement calculation:   $\Delta x = x_f - x_i = 0\,m$
• Total distance calculation:   $D = 20\,m + 20\,m = 40\,m$
• This demonstrates that an object can travel a large total distance while having a net displacement of zero.

Velocity and Speed

• Average Velocity ($v_{\text{avg}}$ or $\bar{v}$):
  • Average velocity is a vector quantity defined as the rate at which displacement occurs over an elapsed time interval $\Delta t$.
  • Mathematical formula:     $v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$
  • Geometrically, average velocity represents the slope of the secant line connecting the initial and final points on a position-time graph.
  • The sign of $v_{\text{avg}}$ matches the sign of $\Delta x$; it can be positive, negative, or zero.
• Average Speed:
  • Average speed is a scalar quantity defined as the total distance traveled divided by the total time elapsed:     $\text{Average speed} = \frac{D}{\Delta t}$
  • Average speed possesses magnitude only and has no directional sign; it is always expressed as a positive number (or zero when stationary).
  • Average speed is not necessarily equal to the magnitude of the average velocity.
• Instantaneous Velocity ($v_{\text{inst}}$ or $v$):
  • Instantaneous velocity defines the velocity of a particle at an exact, infinitesimal instant in time.
  • It is defined mathematically as the limit of the average velocity as the time interval $\Delta t$ approaches zero:     $v_{\text{inst}} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$
  • Instantaneous velocity is the first time derivative of the position function with respect to time.
  • Geometrically, instantaneous velocity represents the slope of the tangent line to the position-time curve at any given time coordinate $t$.
  • Instantaneous velocity can be positive (moving toward increasing $x$), negative (moving toward decreasing $x$), or zero (instantaneously at rest).

Acceleration

• Average Acceleration ($a_{\text{avg}}$ or $\bar{a}$):
  • Average acceleration is a vector quantity defined as the rate of change of velocity over an elapsed time interval:     $a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}$
  • The SI unit of acceleration is meters per second squared ($\text{m/s}^2$ or $\text{m/sec}^2$).
• Instantaneous Acceleration ($a_{\text{inst}}$ or $a$):
  • Instantaneous acceleration defines the acceleration of an object at a precise moment in time.
  • It is defined mathematically as the limit of average acceleration as the time interval $\Delta t$ approaches zero:     $a_{\text{inst}} = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}$
  • Because velocity is the first derivative of position, acceleration is the second time derivative of the position function:     $a = \frac{d^2x}{dt^2}$
  • Acceleration describes how rapidly the velocity vector changes in magnitude, direction, or both.

Calculus-Based Kinematic Examples

• Example 1: Kinematics of $x(t) = 3t - 4t^2 + 2$

  • A particle moves along the $x$-axis according to the position function:     $x(t) = 3t - 4t^2 + 2$
  • Part (a): Find position at $t = 1\,s$:
  • Substitute $t = 1$ directly into the position equation:       $x(1) = 3(1) - 4(1)^2 + 2$$x(1) = 3 - 4 + 2 = 1\,m$
  • Part (b): Find displacement from $t = 0\,s$ to $t = 1\,s$:
  • Formula for displacement:       $\Delta x = x_2 - x_1$
  • Evaluate position at $t_1 = 0\,s$:       $x_1(0) = 3(0) - 4(0)^2 + 2 = 2\,m$
  • Evaluate position at $t_2 = 1\,s$:       $x_2(1) = 3(1) - 4(1)^2 + 2 = 1\,m$
  • Calculate $\Delta x$:       $\Delta x = x_2 - x_1 = 1 - 2 = -1\,m$
  • Part (c): Find average velocity from $t = 1\,s$ to $t = 2\,s$:
  • Evaluate position at $t_1 = 1\,s$:       $x_1(1) = 3(1) - 4(1)^2 + 2 = 1\,m$
  • Evaluate position at $t_2 = 2\,s$:       $x_2(2) = 3(2) - 4(2)^2 + 2 = 6 - 16 + 2 = -8\,m$
  • Apply the average velocity definition:       $\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}$$\bar{v} = \frac{-8 - 1}{2 - 1} = \frac{-9}{1} = -9\,m/s$
  • Part (d): Find average acceleration from $t = 0\,s$ to $t = 1\,s$:
  • Derive the instantaneous velocity equation by differentiating position:       $v(t) = \frac{dx}{dt} = 3 - 8t + 0 = 3 - 8t$
  • Compute velocity at $t_1 = 0\,s$:       $v_1(0) = 3 - 8(0) = 3\,m/s$
  • Compute velocity at $t_2 = 1\,s$:       $v_2(1) = 3 - 8(1) = -5\,m/s$
  • Compute average acceleration:       $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}$$\bar{a} = \frac{-5 - 3}{1 - 0} = \frac{-8}{1} = -8\,m/s^2$
  • Part (e): Find the acceleration at $t = 3\,s$:
  • Derive instantaneous acceleration by taking the derivative of velocity:       $a = \frac{dv}{dt} = \frac{d^2x}{dt^2} = \frac{d}{dt}(3 - 8t) = -8\,m/s^2$
  • Because the derivative yields a constant, the acceleration at $t = 3\,s$ is:       $a_{t=3} = -8\,m/s^2$

• Example 2: Kinematics of $x(t) = 2 + 3t - t^2$

  • The position of a particle moving along the $x$-axis is given by:     $x(t) = 2 + 3t - t^2$
  • Part (a): Calculate average velocity from $t = 1\,s$ to $t = 3\,s$:
  • Determine position at $t_1 = 1\,s$:       $x_1(1) = 2 + 3(1) - (1)^2 = 2 + 3 - 1 = 4\,m$
  • Determine position at $t_2 = 3\,s$:       $x_2(3) = 2 + 3(3) - (3)^2 = 2 + 9 - 9 = 2\,m$
  • Calculate average velocity:       $\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}$$\bar{v} = \frac{2 - 4}{3 - 1} = \frac{-2}{2} = -1\,m/s$
  • Part (b): Calculate the velocity at $t = 4\,s$:
  • Compute the first derivative of position with respect to time:       $v(t) = \frac{dx}{dt} = 0 + 3 - 2t = 3 - 2t$
  • Evaluate at $t = 4\,s$:       $v_{t=4} = 0 + 3 - 2(4) = 3 - 8 = -5\,m/s$
  • Part (c): Calculate the acceleration at $t = 3\,s$:
  • Compute the derivative of velocity:       $a = \frac{dv}{dt} = \frac{d}{dt}(3 - 2t) = -2\,m/s^2$
  • Since acceleration is independent of $t$, the acceleration at $t = 3\,s$ is:       $a = -2\,m/s^2$
  • Part (d): Determine the position when velocity $v = 0$:
  • Set the velocity equation equal to zero and solve for time $t$:       $v = \frac{dx}{dt} = 3 - 2t = 0$$3 = 2t$$t = \frac{3}{2}\,s = 1.5\,s$
  • Substitute $t = \frac{3}{2}\,s$ into the position function:       $x\left(\frac{3}{2}\right) = 2 + 3\left(\frac{3}{2}\right) - \left(\frac{3}{2}\right)^2$$x = 2 + \frac{9}{2} - \frac{9}{4}$$x = 2 + 4.5 - 2.25 = 4.25\,m$
  • Part (e): Find average acceleration from $t = 0\,s$ to $t = 1\,s$:
  • Using the velocity function $v(t) = 3 - 2t$:       $v_1(0) = 3 - 2(0) = 3\,m/s$$v_2(1) = 3 - 2(1) = 1\,m/s$
  • Calculate average acceleration:       $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}$$\bar{a} = \frac{1 - 3}{1 - 0} = \frac{-2}{1} = -2\,m/s^2$

Kinematic Equations for Constant Acceleration

• Uniform Acceleration Principle:
  • Uniform acceleration is mathematically equivalent to constant acceleration ($a = \text{constant}$).
  • When acceleration is constant, the instantaneous acceleration equals the average acceleration over any time interval.
• Kinematic Variables and Nomenclature:
  • $v_f \equiv v$: Final velocity of the particle.
  • $v_i \equiv v_0$: Initial velocity of the particle.
  • $a$: Constant acceleration.
  • $\Delta x$: Horizontal displacement along the $x$-axis ($\Delta x = x_f - x_i$).
  • $\Delta y$: Vertical displacement along the $y$-axis ($\Delta y = y_f - y_i$).
  • $g$: Magnitude of the acceleration due to gravity ($g = 9.8\,m/s^2$).
• Equations for Motion Along the Horizontal ($x$-axis):
  1. Velocity as a function of time:      $v_f = v_i + a t$
  2. Displacement as a function of velocity and time:      $\Delta x = v_i t + \frac{1}{2} a t^2$
  3. Velocity as a function of displacement (time-independent relation):      $v_f^2 = v_i^2 + 2 a \Delta x$
• Equations for Motion Along the Vertical ($y$-axis / Free Fall):
  • For vertical motion under gravity near Earth's surface where downward is defined as negative, the acceleration is $a = -g = -9.8\,m/s^2$:

1. Velocity as a function of time:    $v_f = v_i - g t$
2. Displacement as a function of time:    $\Delta y = v_i t - \frac{1}{2} g t^2$
3. Velocity as a function of displacement:    $v_f^2 = v_i^2 - 2 g \Delta y$

Constant Acceleration Numerical Problems

• Example 3: Braking Car Deceleration Analysis

  • Problem Statement: A car moves with constant acceleration and an initial velocity of $20\,m/s$. It traverses a displacement of $100\,m$ over a duration of $10\,s$.
  • Part (a): Find the acceleration of the car.
  • Part (b): Find the velocity at $t = 10\,s$.
  • Given Values:
  • Initial velocity: $v_0 = 20\,m/s$
  • Displacement: $\Delta x = 100\,m$
  • Time elapsed: $t = 10\,s$
  • Solution for Part (a):
  • Select the kinematic relation containing $\Delta x$, $v_0$, $t$, and $a$:       $\Delta x = v_0 t + \frac{1}{2} a t^2$
  • Substitute known values:       $100 = 20(10) + \frac{1}{2} a (10)^2$$100 = 200 + \frac{1}{2} a (100)$$100 = 200 + 50 a$
  • Isolate the acceleration variable $a$:       $100 - 200 = 50 a$$-100 = 50 a$$a = \frac{-100}{50} = -2\,m/s^2$
  • Solution for Part (b):
  • Select the linear velocity relation:       $v = v_0 + a t$
  • Substitute $v_0 = 20\,m/s$, $a = -2\,m/s^2$, and $t = 10\,s$:       $v = 20 + (-2)(10)$$v = 20 - 20 = 0\,m/s$
  • The car comes to a complete rest at the end of the $10\,s$ duration.

• Example 4: Displacement Under Constant Positive Acceleration

  • Problem Statement: A car starts motion with an initial speed of $5\,m/s$ and accelerates uniformly to a final speed of $10\,m/s$ with an acceleration of $2\,m/s^2$. Find the displacement.
  • Given Values:
  • Initial velocity: $v_0 = 5\,m/s$
  • Final velocity: $v = 10\,m/s$
  • Acceleration: $a = 2\,m/s^2$
  • Step 1: Calculate the time elapsed ($t$):
  • Use the first kinematic equation:       $v = v_0 + a t$$10 = 5 + 2 t$$10 - 5 = 2 t$$5 = 2 t$$t = \frac{5}{2} = 2.5\,s$
  • Step 2: Calculate the displacement ($\Delta x$):
  • Apply the displacement-time formula:       $\Delta x = v_0 t + \frac{1}{2} a t^2$
  • Substitute the calculated time $t = 2.5\,s$:       $\Delta x = 5(2.5) + \frac{1}{2}(2)(2.5)^2$$\Delta x = 12.5 + (1)(6.25)$$\Delta x = 18.75\,m$

• Example 5: Multi-Interval Acceleration from Rest

  • Problem Statement: A car starts from rest and reaches a speed of $5\,m/s$ in $3\,s$. Assuming uniform acceleration, find the speed of the car after $5\,s$.
  • Interval 1: Determining Acceleration:
  • Initial velocity: $v_0 = 0\,m/s$
  • Final velocity at $t = 3\,s$: $v = 5\,m/s$
  • Elapsed time: $t = 3\,s$
  • Calculate constant acceleration using $v = v_0 + a t$:       $5 = 0 + a (3)$$5 = 3 a$$a = \frac{5}{3}\,m/s^2$
  • Interval 2: Determining Velocity at Target Time:
  • Initial state from origin: $v_0 = 0\,m/s$
  • Total time elapsed: $t = 5\,s$
  • Apply the velocity equation with the uniform acceleration $a = \frac{5}{3}\,m/s^2$:       $v = v_0 + a t$$v_{t=5} = 0 + \left(\frac{5}{3}\right)(5)$$v_{t=5} = \frac{25}{3}\,m/s \approx 8.33\,m/s$