Equations and Graphical Representations of Uniformly Accelerated Motion

Fundamental Kinematic Quantities of Rectilinear Motion

• Uniform rectilinear acceleration describes a physical state where an object travels along a straight line in a single, unchanging direction while undergoing a constant rate of change in velocity.

• The mathematical study of constant acceleration coordinates five primary physical quantities:

  • Displacement ($\Delta x$): Defined as the direct distance traversed along a straight line in a single direction, measured in meters ($\text{m}$).
  • Initial Velocity ($v_0$): Defined as the instantaneous velocity of the body at the onset of motion ($t = 0$), measured in meters per second ($\text{m/s}$).
  • Final Velocity ($v$): Defined as the velocity of the body at the end of the specified time interval, measured in meters per second ($\text{m/s}$).
  • Acceleration ($a$): Defined as the constant, steady rate of change of velocity over time, measured in meters per second squared ($\text{m/s}^2$).
  • Time ($t$): Defined as the total duration or elapsed time period during which the body undergoes motion, measured in seconds ($\text{s}$).

• Foundational philosophy: As stated by Galileo Galilei, "Mathematics is the basic language for understanding the universe."

The Three Primary Equations of Uniformly Accelerated Motion

• The five fundamental kinematic quantities are mathematically interrelated through three standard equations for rectilinear motion with uniform acceleration:

• First Equation of Motion:

  • Mathematical formula:     $v = v_0 + at$
  • Operational criterion: Utilized when displacement ($\Delta x$) is neither provided as a known variable nor required to be determined.

• Second Equation of Motion:

  • Mathematical formula:     $\Delta x = v_0 t + \frac{1}{2} a t^2$
  • Operational criterion: Utilized when final velocity ($v$) is neither provided as a known variable nor required to be determined.

• Third Equation of Motion:

  • Mathematical formula:     $v^2 = v_0^2 + 2a(\Delta x)$
  • Operational criterion: Utilized when elapsed time ($t$) is neither provided as a known variable nor required to be determined.

• Forensic and accident reconstruction utility: When analyzing vehicular collision scenes to determine a driver's initial velocity prior to braking, the appropriate kinematic equation is selected based entirely on which parameters (skid marks for displacement, impact speed, deceleration rate, or duration) are measurable.

Step-by-Step Worked Problems and Computational Methods

• Example 1 (Truck Deceleration Under Braking):

  • Scenario: A truck travels with a speed of $22.2\,\text{m/s}$. The driver applies the brakes, causing the truck's speed to decrease uniformly at a deceleration magnitude of $2\,\text{m/s}^2$. Calculate the final velocity of the truck after a duration of $8\,\text{s}$.
  • Step 1: Identify known and unknown variables:
  • Initial velocity ($v_0$): $v_0 = 22.2\,\text{m/s}$
  • Acceleration ($a$): $a = -2\,\text{m/s}^2$ (negative sign denotes deceleration)
  • Elapsed time ($t$): $t = 8\,\text{s}$
  • Final velocity ($v$): $v = ?$
  • Step 2: Select and apply the appropriate kinematic equation:
  • Formula: $v = v_0 + at$
  • Substitution:       $v = 22.2 + (-2 \times 8)$
  • Solution:       $v = 6.2\,\text{m/s}$
  • Physical verification: The final velocity of $6.2\,\text{m/s}$ after $8\,\text{s}$ confirms that the truck loses exactly $2\,\text{m/s}$ of its speed during each passing second ($2\,\text{m/s} \times 8\,\text{s} = 16\,\text{m/s}$ total loss, $22.2 - 16 = 6.2\,\text{m/s}$).

• Example 2 (Biological Transmission of Nerve Impulses):

  • Scenario: Assume the propagation speed of a nerve impulse throughout the human body is constant and equals $100\,\text{m/s}$. Calculate the time required for a nerve signal to travel from the foot to the brain of a man whose height is $1.7\,\text{m}$.
  • Step 1: Identify known and unknown variables:
  • Initial velocity ($v_0$): $v_0 = 100\,\text{m/s}$
  • Acceleration ($a$): $a = 0\,\text{m/s}^2$ (propagation occurs at constant velocity)
  • Displacement ($\Delta x$): $\Delta x = 1.7\,\text{m}$
  • Elapsed time ($t$): $t = ?$
  • Step 2: Select and apply the appropriate kinematic equation:
  • Formula:       $\Delta x = v_0 t + \frac{1}{2} a t^2$
  • Substitution:       $1.7 = (100 \times t) + \left(\frac{1}{2} \times 0 \times t^2\right)$
  • Simplified relation:       $1.7 = 100 \times t$
  • Solution:       $t = 0.017\,\text{s}$

• Example 3 (Accelerating Vehicle Over Known Displacement):

  • Scenario: A moving vehicle accelerates constantly over a fixed linear distance. Calculate the final velocity given the displacement and acceleration parameters.
  • Step 1: Identify known and unknown variables:
  • Initial velocity ($v_0$): $v_0 = 27.7\,\text{m/s}$
  • Acceleration ($a$): $a = 2\,\text{m/s}^2$
  • Displacement ($\Delta x$): $\Delta x = 30\,\text{m}$
  • Final velocity ($v$): $v = ?$
  • Step 2: Select and apply the appropriate kinematic equation:
  • Formula:       $v^2 = v_0^2 + 2a(\Delta x)$
  • Substitution:       $v^2 = (27.7)^2 + (2 \times 2 \times 30)$
  • Intermediate squared sum:       $v^2 = 767.29 + 120 = 887.29$
  • Extraction of root:       $v = \sqrt{887.29}$
  • Final computed velocity:       $v = 29.78\,\text{m/s}$

• Example 5 (Graphical Slope Analysis for Displacement and Acceleration):

  • Scenario: An object starts moving from rest with constant uniform acceleration. The slope of the straight-line graph representing the relationship between displacement and the square of time ($x - t^2$) equals $3$. Calculate:
  1. The acceleration of the object ($a$).
  2. The displacement ($\Delta x$) covered after an elapsed time of $5\,\text{s}$ from the start of motion.
  • Step 1: Identify known and unknown variables:
  • Slope value of the graph:       $\frac{\Delta x}{\Delta t^2} = 3$
  • Initial velocity: $v_0 = 0$
  • Target time for displacement: $t = 5\,\text{s}$
  • Step 2: Select and apply the appropriate mathematical relations:
  • Part 1: Finding acceleration ($a$):
    • The mathematical slope of an object starting from rest on a $\Delta x - t^2$ graph is defined as:         $\text{slope} = \frac{1}{2} a$
    • Setting up the equality:         $\frac{1}{2} a = 3$
    • Solving for acceleration:         $a = 2 \times 3 = 6\,\text{m/s}^2$
  • Part 2: Finding displacement ($\Delta x$) at $t = 5\,\text{s}$:
    • Using the second equation of motion:         $\Delta x = v_0 t + \frac{1}{2} a t^2$
    • Substituting known values ($v_0 = 0$, $a = 6\,\text{m/s}^2$, $t = 5\,\text{s}$):         $\Delta x = 0 + \left(\frac{1}{2} \times 6 \times (5)^2\right)$
    • Computation:         $\Delta x = 3 \times 25 = 75\,\text{m}$

Graphical Representation of Uniform Acceleration

• Analyzing the motion of uniformly accelerating bodies via graphic coordinate systems yields direct physical interpretations of slopes and geometric boundaries:

• First Relationship: Velocity versus Time ($v - t$ Graph):

  • Governed by the linear equation:     $v = v_0 + at$
  • Case A (Starting from rest, $v_0 = 0$):
  • The curve represents a straight line passing through the coordinate origin $(0,0)$.
  • Displayed in Figure (7).
  • Case B (Starting with non-zero velocity, $v_0 \neq 0 \rightarrow v_0 > 0$):
  • The curve is an ascending straight line with a non-zero vertical intercept at $(0, v_0)$.
  • Displayed in Figure (8).
  • Physical meaning of the slope:     $\text{slope} = \frac{\Delta y}{\Delta x} = a$
  • The slope of any line on a velocity-time coordinate graph directly represents the acceleration ($a$) of the body.

• Second Relationship: Displacement versus Square of Time ($\Delta x - t^2$ Graph):

  • Governed by the quadratic displacement equation:     $\Delta x = v_0 t + \frac{1}{2} a t^2$
  • Case A (Starting from rest, $v_0 = 0$):
  • When $v_0 = 0$, the relation reduces to:       $\Delta x = \frac{1}{2} a t^2$
  • Plotting $\Delta x$ on the vertical axis against $t^2$ on the horizontal axis yields a direct proportional straight line through the origin $(0,0)$.
  • Displayed in Figure (9).
  • Mathematical slope interpretation:       $\text{slope} = \frac{\Delta y}{\Delta x} = \frac{1}{2} a$
  • The slope of the straight-line graph of displacement versus time squared when motion starts from rest equals exactly half of the acceleration ($\frac{1}{2} a$).
  • Case B (Starting with non-zero velocity, $v_0 \neq 0 \rightarrow v_0 > 0$):
  • The plot of $\Delta x$ against $t^2$ appears as an upward-curving parabolic curve rather than a linear slope, caused by the non-zero linear first-order term ($v_0 t$).
  • Displayed in Figure (10).

• Third Relationship: Square of Velocity versus Displacement ($v^2 - \Delta x$ Graph):

  • Governed by the third equation:     $v^2 = v_0^2 + 2a(\Delta x)$
  • When motion initiates from rest ($v_0 = 0$), the line passes through the origin $(0,0)$ with a constant slope equal to $2a$.
  • When motion initiates with an initial speed ($v_0 > 0$), the graph intercepts the vertical axis at $v_0^2$.

Geometric Derivation of the Second Equation of Motion

• Interdisciplinary link between physics and geometry:

  • The mathematical proof and derivation of the second kinematic equation, $\Delta x = v_0 t + \frac{1}{2} a t^2$, is obtained geometrically through area integration under a velocity-time graph.

• Theorem: The total linear displacement ($\Delta x$) equals the total area enclosed under the curve on a velocity-time ($v - t$) graph.

• Geometric decomposition of the region beneath the line from time $0$ to $t$:

  • The enclosed geometric shape bounded between $v_0$ and $v$ over time $t$ is partitioned into two distinct geometric figures: a bottom rectangle and a top right triangle.

• Derivation steps:

  • Area of the lower rectangle:     $\text{Area}_{\text{rectangle}} = \text{base} \times \text{height} = t \times v_0 = v_0 t$
  • Area of the upper right triangle:     $\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times (v - v_0)$
  • Applying the first equation of motion to substitute for $(v - v_0)$:     $v - v_0 = at$
  • Substituting $at$ into the triangle area formula:     $\text{Area}_{\text{triangle}} = \frac{1}{2} \times t \times (at) = \frac{1}{2} a t^2$
  • Summing the geometric components:     $\text{Displacement} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{triangle}}$
  • Final derived relationship:     $\Delta x = v_0 t + \frac{1}{2} a t^2$

Real-World Animal Kinematics: The Cheetah

• The hunting cheetah (الفهد الصياد) ranks as the fastest terrestrial animal on Earth.

• Dynamic performance characteristics:

  • Top speed: Reaches velocities up to $120\,\text{km/h}$.
  • Acceleration profile: Attains its peak velocity within a span of only a few seconds.

• Physiological limitation:

  • A cheetah cannot sustain this extreme speed for more than a brief duration due to catastrophic rates of metabolic energy expenditure and physiological fatigue.

Applied Exercises and Problem-Solving Activities

• Explore on Your Own (1):
  • Problem statement: A school bus moves at an initial speed of $20\,\text{m/s}$ and accelerates uniformly at $1.5\,\text{m/s}^2$. Determine the velocity of the bus after an elapsed time of $6\,\text{s}$.
  • Analytical setup:
  • Known values: $v_0 = 20\,\text{m/s}$, $a = 1.5\,\text{m/s}^2$, $t = 6\,\text{s}$
  • Target: $v = ?$
  • Governing formula: $v = v_0 + at$

• Explore on Your Own (2):

  • Problem statement: A sprinter in a footrace gradually accelerates from rest with a uniform acceleration of $4\,\text{m/s}^2$. Calculate the distance covered over a specified initial acceleration interval.
  • Analytical setup:
  • Known values: $v_0 = 0\,\text{m/s}$, $a = 4\,\text{m/s}^2$
  • Governing formula: $\Delta x = v_0 t + \frac{1}{2} a t^2$

• Explore on Your Own (3):

  • Problem statement: A car moves at speed $v$. The driver depresses the brakes, resulting in a uniform deceleration that brings the vehicle to a complete stop over a distance of $\Delta x$. If the car were instead traveling at an initial speed of $2v$, what stopping distance (expressed in terms of $\Delta x$) would be required to bring the car to rest under identical braking deceleration?
  • Analytical derivation:
  • From the third equation with $v_{\text{final}} = 0$:       $0 = v_0^2 + 2a(\Delta x) \rightarrow \Delta x = \frac{-v_0^2}{2a}$
  • Because stopping distance is proportional to the square of initial velocity ($\Delta x \propto v_0^2$):       $\Delta x' = \frac{-(2v)^2}{2a} = 4 \left(\frac{-v^2}{2a}\right) = 4\Delta x$
  • Doubling the initial speed quadruples ($4\Delta x$) the required stopping distance.

• Explore on Your Own (4) (Al-Jahra Road Accident Investigation):

  • Problem statement: During an accident investigation on Al-Jahra Road (طريق الجهراء), speed enforcement traffic cameras recorded two identical vehicles possessing identical specifications:
  • Car A (السيارة أ): Was traveling at a speed of $80\,\text{km/h}$. The driver applied the brakes, bringing the vehicle to a complete stop over a distance of $10\,\text{m}$.
  • Car B (السيارة ب): Was traveling at twice the speed of Car A ($160\,\text{km/h}$). The driver applied the brakes with the exact same braking force (producing identical deceleration).
  • Question: Determine the displacement covered by Car B before it comes to a complete halt.
  • Proportionality solution:
  • Since both vehicles experience identical braking acceleration ($a$) and decelerate to rest ($v = 0$):       $v^2 = v_0^2 + 2a(\Delta x) \rightarrow 0 = v_0^2 + 2a(\Delta x) \rightarrow \Delta x = \frac{-v_0^2}{2a}$
  • The stopping displacement is directly proportional to the square of the initial velocity ($\Delta x \propto v_0^2$).
  • The ratio of stopping displacements equals the ratio of the squares of initial velocities:       $\frac{\Delta x_B}{\Delta x_A} = \left(\frac{v_{0B}}{v_{0A}}\right)^2 = \left(\frac{160\,\text{km/h}}{80\,\text{km/h}}\right)^2 = (2)^2 = 4$
  • Evaluating the final stopping displacement for Car B:       $\Delta x_B = 4 \times \Delta x_A = 4 \times 10\,\text{m} = 40\,\text{m}$

• Explore on Your Own (5):
  • Concluding conceptual and computational synthesis exercise on rectilinear uniformly accelerated motion.