Comprehensive Study Guide to Uniformly Accelerated Motion

Defining Average Acceleration

• Acceleration measures the time rate-of-change of velocity. It is defined as the change in the velocity vector divided by the time taken for that change to occur.
• Average Acceleration Formula:   $\mathbf{a}_{av} = \frac{\mathbf{v}_f - \mathbf{v}_i}{t}$   Where:
  • $\mathbf{v}_i$ is the initial velocity.
  • $\mathbf{v}_f$ is the final velocity.
  • $t$ is the time interval over which the change occurred.
• Units of Acceleration: Units are those of velocity divided by time. Common examples include:
  • $\text{(m/s)/s}$ or $\text{m/s}^2$
  • $\text{(km/h)/s}$ or $\text{km/h}\cdot\text{s}$
• Vector Nature: Acceleration is a vector quantity. Its direction is the same as the direction of the velocity change ($\mathbf{v}_f - \mathbf{v}_i$).
• Scalar Simplification: It is common to refer to the magnitude of acceleration simply as "acceleration" if there is no ambiguity. When considering only accelerations tangent to the path of travel, the direction is known, and the equation can be written in scalar form:   $a_{av} = \frac{v_f - v_i}{t}$

Uniformly Accelerated Motion along a Straight Line

• In this specific case, the acceleration vector is constant and lies along the line of the displacement vector. Directions for velocity ($v$) and acceleration ($a$) can be indicated using plus ($+$) and minus ($-$) signs.
• Displacement is represented by $s$, being positive in the designated positive direction and negative in the opposite direction.
• Five Foundational Equations of Uniformly Accelerated Motion:
  1. $s = v_{av} t$
  2. $v_{av} = \frac{v_i + v_f}{2}$
  3. $v_f = v_i + at$
  4. $v_f^2 = v_i^2 + 2as$
  5. $s = v_i t + \frac{1}{2}at^2$
• Notation Variations:
  • Displacement ($s$) is often replaced with $x$ or $y$.
  • Initial velocity ($v_i$) is sometimes written as $v_0$.
  • Final velocity ($v_f$) is sometimes written as $v$.
• Sign Convention: A positive direction must be chosen at the start of analysis. If any displacement, velocity, or acceleration is in the opposite direction of the chosen positive axis, it must be treated as a negative value.

Graphical Interpretations of Motion

• Distance vs. Time Graphs:
  • The plot is always positive (stays above the time axis).
  • The curve never decreases because distance cannot be negative and speed (the slope) is either zero or positive.
• Displacement vs. Time Graphs (Straight Line):
  • Can be positive (plotted above the time axis) when the object is to the right of the origin ($x = 0$).
  • Can be negative (plotted below the time axis) when the object is to the left of the origin.
  • A positive slope indicates positive velocity (moving in the positive direction), whether the graph itself is currently positive or negative.
  • A negative slope indicates negative velocity (moving in the negative direction).
  • Instantaneous Velocity: This is the slope of the displacement-versus-time graph at a specific point in time. It can be positive, negative, or zero.
• Velocity vs. Time Graphs:
  • Instantaneous Acceleration: This is the slope of the velocity-versus-time graph at a specific point in time.
  • For constant-velocity motion, the $x$-versus-$t$ graph is a tilted straight line.
  • For constant-acceleration motion, the $v$-versus-$t$ graph is a straight line.

Acceleration due to Gravity ($g$)

• Gravity is the only force acting on a body in free-fall. The gravitational acceleration ($g$) is always directed vertically downward.
• Standard Values:
  • Earth (Average at surface): $g = 9.81\,\text{m/s}^2$ (or $32.2\,\text{ft/s}^2$). Note that this value varies slightly by location.
  • Moon (Average at surface): $g = 1.6\,\text{m/s}^2$.

Velocity Components and Signs

• When an object moves at an angle $\theta$ from the $x$-axis, its velocity ($v$) has vector components characterized by scalar values:
  • $v_x = v \cos(\theta)$
  • $v_y = v \sin(\theta)$
• Quadrant Logic for Signs:
  • 1st Quadrant: $v_x > 0$, $v_y > 0$
  • 2nd Quadrant: $v_x < 0$, $v_y > 0$
  • 3rd Quadrant: $v_x < 0$, $v_y < 0$
  • 4th Quadrant: $v_x > 0$, $v_y < 0$
• Strictly speaking, "velocity" refers to the vector quantity with an explicit direction, while speed ($v$) is always positive. For an object with velocity $\mathbf{v} = 100\,\text{m/s-WEST}$, the scalar velocity along the $x$-axis is $v_x = -100\,\text{m/s}$, but the speed is $v = 100\,\text{m/s}$.

Projectile Motion and Ballistics

• Independence of Motion: In the absence of air friction, projectile motion is split into two independent components:
  1. Horizontal Motion: Acceleration ($a_x = 0$), horizontal velocity ($v_x$) is constant.
  2. Vertical Motion: Acceleration ($a_y = g = 9.81\,\text{m/s}^2$ downward).
  • A ball dropped vertically and a ball fired horizontally from the same height will hit the ground simultaneously.
• Free-Fall Conventions:
  • Assign signs based on the chosen positive direction.
  • When an object is dropped ($v_i = 0$) or thrown down, choosing "down" as positive makes $v_i$, $g$, and $t$ positive. Displacement ($y$) and final velocity ($v_f$) will also be positive.
  • When an object is fired upward, choosing "up" as positive makes vertical displacement ($y$) and initial velocity ($v_i$) positive, but $g$ becomes negative ($-9.81\,\text{m/s}^2$).
• Key Projectile Formulas:
  • Peak Altitude ($y_p$): Occurs when $v_f = 0$.       $y_p = -\frac{v_i^2}{2g}$
  • Time to Reach Peak Altitude ($t_p$):       $t_p = -\frac{v_i}{g}$
  • General Launch Components:       $v_{ix} = v_i \cos(\theta)$ and $v_{iy} = v_i \sin(\theta)$The trajectory is a parabola.
  • Total Time of Flight ($t_T$): Assuming a symmetrical trajectory (returning to launch height), $t_T = 2t_p$.
  • Range ($R$): The horizontal distance covered when the projectile returns to the launch height.       $R = v_{ix} t_T = (v_i \cos(\theta)) \times \left( 2 \times \left( -\frac{v_{iy}}{g} \right) \right)$       Simplified: $R = \frac{v_i^2 \sin(2\theta)}{g}$ (where $g = 9.81\,\text{m/s}^2$).

Dimensional Analysis

• All mechanical quantities can be expressed using three fundamental dimensions:
  • Length ($L$)
  • Mass ($M$)
  • Time ($T$)
• Dimensions of Common Quantities:
  • Acceleration: $[LT^{-2}]$
  • Volume: $[L^3]$
  • Velocity: $[LT^{-1}]$
  • Force: $[MLT^{-2}]$
• Dimension Equality: Every term in a physical equation must have the same dimensions.
  • Example: In $s = v_i t + \frac{1}{2}at^2$, the dimensions are:       $[L] = [LT^{-1}][T] + [LT^{-2}][T^2] \rightarrow [L] = [L] + [L]$.
  • Terms with different dimensions (e.g., adding a volume to an area) cannot be added or subtracted.

Problem-Solving Methodology

1. Read Carefully: Identify given quantities and what needs to be found.
2. Write Quantities with Symbols: Use appropriate symbols and units immediately. Maintain significant figures throughout.
3. Check Values: Ensure constants like $g$ are applied correctly (e.g., $0.000070$ is not $0.00070$).
4. Component Independence: Treat horizontal and vertical motions separately in ballistics problems.
5. Verify Dimensions: Check that the units of the final answer match the expected quantity.

Case Studies and Solved Problems

Acceleration of a Robot (Case 2.1)

• Scenario: Robot Fred moves at $2.20\,\text{m/s}$ and accelerates to $4.80\,\text{m/s}$ in $0.20\,\text{s}$.
• Calculation: $a_{av} = \frac{4.80\,\text{m/s} - 2.20\,\text{m/s}}{0.20\,\text{s}} = 13\,\text{m/s}^2$.
• Note: The answer has 2 significant figures due to the time ($0.20\,\text{s}$).

Braking Automobile (Case 2.2)

• Scenario: A car travels at $20.0\,\text{m/s}$ and comes to a stop ($v_f = 0$) in $4.2\,\text{s}$.
• Calculation: $a_{av} = \frac{0.0\,\text{m/s} - 20.0\,\text{m/s}}{4.2\,\text{s}} \approx -4.76\,\text{m/s}^2$.
• Result: $-4.8\,\text{m/s}^2$ (rounded to 2 significant figures).

Motion from Rest (Case 2.3)

• Scenario: Object starts from rest ($v_i = 0$) with $a = 8.00\,\text{m/s}^2$ for $5.00\,\text{s}$.
• Findings:
  • (a) Speed at end: $v_f = v_i + at = 0 + (8.00\,\text{m/s}^2)(5.00\,\text{s}) = 40.0\,\text{m/s}$.
  • (b) Average Speed: $v_{av} = \frac{0 + 40.0}{2} = 20.0\,\text{m/s}$.
  • (c) Distance: $x = v_i t + \frac{1}{2}at^2 = 0 + \frac{1}{2}(8.00\,\text{m/s}^2)(5.00\,\text{s})^2 = 100\,\text{m}$.

Truck Speed Increase (Case 2.4)

• Scenario: Truck increases speed from $15\,\text{km/h}$ ($4.17\,\text{m/s}$) to $60\,\text{km/h}$ ($16.7\,\text{m/s}$) in $20\,\text{s}$.
• Findings:
  • (a) Average Speed: $10.4\,\text{m/s}$.
  • (b) Acceleration: $0.63\,\text{m/s}^2$.
  • (c) Distance: $208\,\text{m}$ (or $0.21\,\text{km}$).

Segmented Motion Graph Analysis (Case 2.5)

• Scenario: Analysis of a displacement-time graph (Fig 2-1).
  • 0-2.0 s: Slope is zero; object is at rest.
  • 2.0-4.0 s: Constant positive slope. $v_{av} = \frac{3.0\,\text{m} - 0\,\text{m}}{4.0\,\text{s} - 2.0\,\text{s}} = 1.5\,\text{m/s}$ in the positive $x$-direction.
  • 4.0-6.0 s: Slope is zero; object is at rest.
  • 6.0-10.0 s: Constant negative slope. $v_{av} = \frac{-2.0\,\text{m} - 3.0\,\text{m}}{10.0\,\text{s} - 6.0\,\text{s}} = -1.3\,\text{m/s}$ (negative $x$-direction).

Vertical Motion Graph Analysis (Case 2.6)

• Scenario: Vertical position vs. time (Fig 2-2).
  • Point A: Slope is positive ($v_A = 2.3\,\text{m/s-UP}$).
  • Point B: Peak altitude, slope is zero ($v_B = 0\,\text{m/s}$).
  • Point C: Slope is negative ($v_C = -1.2\,\text{m/s}$ or $1.2\,\text{m/s-DOWN}$).

Dropped Ball from 50 m (Case 2.7)

• Scenario: Ball falls from rest ($v_i = 0, y = 50.0\,\text{m}$, taking down as positive, $a = 9.81\,\text{m/s}^2$).
• Speed just before impact: $v_f = \sqrt{v_i^2 + 2ay} = \sqrt{0 + 2(9.81)(50.0)} = 31.3\,\text{m/s}$.
• Time taken: $t = \frac{v_{fy} - v_{iy}}{a} = \frac{31.3 - 0}{9.81} = 3.19\,\text{s}$.

Skier Acceleration (Case 2.8)

• Scenario Phase 1: Skier travels $9.0\,\text{m}$ in $3.0\,\text{s}$ from rest.
  • $a = \frac{2x}{t^2} = \frac{18\,\text{m}}{(3.0\,\text{s})^2} = 2.0\,\text{m/s}^2$.
• Scenario Phase 2: Time to reach $24\,\text{m/s}$ from rest.
  • $t = \frac{v_{fx} - v_{ix}}{a} = \frac{24\,\text{m/s}}{2.0\,\text{m/s}^2} = 12\,\text{s}$.

Bus Deceleration (Case 2.9)

• Scenario: Bus travels at $20\,\text{m/s}$, slows down at $3.0\,\text{m/s}^2$ ($a = -3.0\,\text{m/s}^2$).
• Stopping Distance: $0 = v_i^2 + 2ax \rightarrow x = \frac{-(20)^2}{2(-3.0)} = 67\,\text{m}$.

Segmented Distance (Case 2.10)

• Scenario: Car at $30\,\text{m/s}$ slows to $10\,\text{m/s}$ in $5.0\,\text{s}$.
• Acceleration: $a = \frac{10 - 30}{5.0} = -4.0\,\text{m/s}^2$.
• Distance in the 3rd Second:
  • $x_{3rd} = \text{Distance}_{3.0\text{s}} - \text{Distance}_{2.0\text{s}}$.
  • Using $x = v_{ix}t + \frac{1}{2}at^2$:
  • $x = [30(3) + 0.5(-4)(3)^2] - [30(2) + 0.5(-4)(2)^2] = 20\,\text{m}$.

Stone Caught on Downward Path (Case 2.13)

• Scenario: Stone thrown up at $20\,\text{m/s}$, caught $5.0\,\text{m}$ above launch on the way down.
• Velocity when caught: $v_y^2 = (20)^2 + 2(-9.81)(5.0) = 302 \rightarrow v_y = -17\,\text{m/s}$ (negative due to downward motion).
• Time of trip: $t = \frac{-17.4 - 20}{-9.81} = 3.8\,\text{s}$.

Projectile from Cliff (Case 2.17)

• Scenario: Horizontal speed $30\,\text{m/s}$, cliff height $80\,\text{m}$.
• Time to strike: Using $y = \frac{1}{2}gt^2$: $-80 = 0 + \frac{1}{2}(-9.81)t^2 \rightarrow t = 4.04\,\text{s}$.
• Horizontal Distance: $x = v_x t = 30 \times 4.04 = 121\,\text{m}$.
• Velocity at Impact:
  • $v_y = 0 + (-9.81)(4.04) = -39.6\,\text{m/s}$.
  • $v = \sqrt{(30)^2 + (-39.6)^2} = 50\,\text{m/s}$.
  • Direction: $\tan^{-1}\left( \frac{39.6}{30} \right) = 53^{\circ}$ below $x$-axis.

Building to Building Projectile (Case 2.20)

• Scenario: Target building $50\,\text{m}$ away. Launch $v_i = 20\,\text{m/s}$ at $40^{\circ}$.
• Components: $v_{ix} = 15.3\,\text{m/s}$, $v_{iy} = 12.9\,\text{m/s}$.
• Time to reach building: $t = \frac{50}{15.3} = 3.27\,\text{s}$.
• Vertical Displacement: Taking down as positive ($v_{iy}$ is negative, $g$ is positive):
  • $y = v_{iy}t + \frac{1}{2}gt^2 = (-12.9)(3.27) + 0.5(9.81)(3.27)^2 = 10.3\,\text{m}$.
• Result: Hits $10\,\text{m}$ below original level.