Comprehensive Study Guide: Motion and Acceleration

Kinematic Variables and Systematic Problem-Solving Routine

• Fundamental Kinematic Variables and Standard SI Units:

  • Initial position: $x_i$, measured in meters ($\text{m}$
  • Final position: $x_f$, measured in meters ($\text{m}$
  • Displacement: $\Delta x = x_f - x_i$, defined as the net change in position, measured in meters ($\text{m}$
  • Initial velocity: $v_i$, measured in meters per second ($\text{m/s}$
  • Final velocity: $v_f$, measured in meters per second ($\text{m/s}$
  • Acceleration: $a$, measured in meters per second squared ($\text{m/s}^2$
  • Elapsed time: $t$ or $\Delta t$, measured in seconds ($\text{s}$

• Step-by-Step Problem-Solving Routine:

  • Step 1: Establish a positive reference direction and ensure signs remain consistent throughout all calculations.
  • Step 2: Explicitly list all known physical quantities and identify the target unknown variable.
  • Step 3: Convert all given measurements to standard SI units ($\text{m}$, $\text{s}$, $\text{m/s}$, $\text{m/s}^2$) before performing calculations.
  • Step 4: Select a kinematic equation that connects the known parameters directly to the unknown quantity.
  • Step 5: Substitute values, perform numerical calculations, append appropriate physical units, and verify that the sign and magnitude make physical sense.

Speed and Velocity

• Core Concepts and Distinctions:

  • Speed: Describes how rapidly distance is covered over time. Speed is a scalar quantity possessing magnitude only, with no directional component.
  • Average Speed Formula:$\text{average speed} = \frac{\text{total distance}}{\text{total time}}$
  • Velocity: Describes both the speed and directional orientation of motion. Velocity is a vector quantity possessing both magnitude and direction.
  • Average Velocity Formula: Defined as displacement divided by the elapsed time interval:     $v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{\Delta t}$
  • Standard SI Unit: Both speed and velocity are expressed in meters per second ($\text{m/s}$).

• Worked Examples and Practice:

  • Worked Example 1 (Average Speed Calculation):
  • Problem Statement: A car travels $200\,\text{m}$ in $2.0\,\text{s}$. Calculate its average speed.
  • Known Values: $\text{distance} = 200\,\text{m}$, $\text{time} = 2.0\,\text{s}$
  • Substitution: $\text{average speed} = \frac{200\,\text{m}}{2.0\,\text{s}}$
  • Calculation: $\text{average speed} = 100\,\text{m/s}$
  • Final Answer: $100\,\text{m/s}$
  • Worked Example 2 (Average Velocity Calculation):
  • Problem Statement: A car moves $200\,\text{m}$ east in $2.0\,\text{s}$. Calculate its average velocity.
  • Known Values: $\Delta x = +200\,\text{m}\text{ east}$, $\Delta t = 2.0\,\text{s}$
  • Substitution: $v_{\text{avg}} = \frac{+200\,\text{m}}{2.0\,\text{s}}$
  • Calculation and Physical Interpretation: The positive sign directly represents the designated eastward direction.
  • Final Answer: $+100\,\text{m/s}\text{ east}$
  • Question 1 (Grizzly Bear Average Speed with Unit Conversion):
  • Problem Statement: A grizzly bear runs $1870\,\text{m}$ in $2.0\,\text{min}$. What is its average speed in $\text{m/s}$?
  • Unit Conversion: $2.0\,\text{min} \times 60\,\text{s/min} = 120\,\text{s}$
  • Substitution: $\text{average speed} = \frac{1870\,\text{m}}{120\,\text{s}}$
  • Calculation: $\text{average speed} = 15.58\,\text{m/s}$
  • Final Answer: $15.6\,\text{m/s}$

Average Acceleration

• Definitions and Mathematical Formulation:

  • Acceleration: The rate of change of velocity with respect to time. An object experiences acceleration when it speeds up, slows down, or alters its direction of travel.
  • Direction and Sign: Acceleration is a vector quantity; its algebraic sign ($+\text{ or } -$) depends strictly on the established coordinate direction.
  • Average Acceleration Formula:$a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{\Delta t}$
  • Standard SI Unit: Meters per second squared ($\text{m/s}^2$).

• Worked Examples and Practice Applications:

  • Worked Example 3 (Car Accelerating from Rest):
  • Problem Statement: A car starts from rest and reaches $+10.0\,\text{m/s}$ in $8.0\,\text{s}$. Calculate its average acceleration.
  • Known Values: $v_i = 0\,\text{m/s}$, $v_f = +10.0\,\text{m/s}$, $\Delta t = 8.0\,\text{s}$
  • Substitution: $a = \frac{(+10.0\,\text{m/s}) - (0\,\text{m/s})}{8.0\,\text{s}}$
  • Calculation: $a = +1.25\,\text{m/s}^2$
  • Final Answer: $+1.25\,\text{m/s}^2$
  • Question 1 (Runner Accelerating in Positive Direction):
  • Problem Statement: A runner increases velocity from $+2.0\,\text{m/s}$ to $+10.0\,\text{m/s}$ in the positive direction over $3.0\,\text{s}$. Find the average acceleration.
  • Method: Apply $a = \frac{v_f - v_i}{\Delta t}$
  • Calculation: $a = \frac{(+10.0) - (+2.0)}{3.0} = \frac{8.0}{3.0} = +2.67\,\text{m/s}^2$
  • Final Answer: $+2.67\,\text{m/s}^2$
  • Question 2 (Vehicle Decelerating to a Stop):
  • Problem Statement: A driver applies the brakes for $5.0\,\text{s}$, slowing the car from $+30.0\,\text{m/s}$ to rest. Determine the average acceleration.
  • Method: Apply $a = \frac{v_f - v_i}{\Delta t}$
  • Calculation: $a = \frac{0 - 30.0}{5.0} = \frac{-30.0}{5.0} = -6.0\,\text{m/s}^2$
  • Final Answer: $-6.0\,\text{m/s}^2$
  • Question 3 (Bicyclist Decreasing Speed):
  • Problem Statement: A bicyclist decreases speed from $23\,\text{m/s}$ to $13\,\text{m/s}$ over $5.0\,\text{s}$. Calculate the bicyclist's acceleration.
  • Method: Apply $a = \frac{v_f - v_i}{\Delta t}$
  • Calculation: $a = \frac{13 - 23}{5.0} = \frac{-10}{5.0} = -2\,\text{m/s}^2$
  • Final Answer: $-2\,\text{m/s}^2$
  • Question 4 (Car Increasing Speed):
  • Problem Statement: A car accelerates from $11\,\text{m/s}$ to $25\,\text{m/s}$ over $7.0\,\text{s}$. What is the acceleration of the car?
  • Method: Apply $a = \frac{v_f - v_i}{\Delta t}$
  • Calculation: $a = \frac{25 - 11}{7.0} = \frac{14}{7.0} = 2\,\text{m/s}^2$
  • Final Answer: $2\,\text{m/s}^2$

Kinematic Equations for Constant Acceleration

• Kinematics Principles and Equation Selection:

  • Kinematics describes motion purely mathematically without considering the forces causing that motion.
  • When acceleration is uniform (constant), four primary algebraic relations link displacement, velocity, acceleration, and time.
  • Selection Criteria:
  • Velocity as a function of time: $v_f = v_i + a t$ (applied when time is known or sought, and displacement is omitted).
  • Displacement via average velocity: $x_f = x_i + \frac{1}{2}(v_i + v_f) t$ (applied when initial and final velocities are known and acceleration is omitted).
  • Position as a function of time and acceleration: $x_f = x_i + v_i t + \frac{1}{2} a t^2$ (applied when time, initial velocity, and acceleration are known).
  • Velocity as a function of displacement: $v_f^2 = v_i^2 + 2 a (x_f - x_i)$ or $v_f^2 = v_i^2 + 2 a \Delta x$ (applied when time is absent from the problem statement).

• Comprehensive Worked Problems:

  • Worked Example 4 (Two-Part Acceleration Analysis):
  • Problem Statement: An object starts from rest at $x_i = 0\,\text{m}$ and accelerates at $+1.25\,\text{m/s}^2$. (a) Determine its position after $6.0\,\text{s}$. (b) Determine its velocity when it reaches position $x_f = +20.0\,\text{m}$.
  • Known Values: $x_i = 0\,\text{m}$, $v_i = 0\,\text{m/s}$, $a = +1.25\,\text{m/s}^2$, $t = 6.0\,\text{s}$
  • Part (a) Position Calculation: Using $x_f = x_i + v_i t + \frac{1}{2} a t^2$:       $x_f = 0 + 0(6.0) + \frac{1}{2}(1.25)(6.0)^2 = 22.5\,\text{m}$
  • Part (b) Velocity Calculation: Using $v_f^2 = v_i^2 + 2 a (x_f - x_i)$:       $v_f^2 = 0^2 + 2(1.25)(20.0 - 0) = 50.0\,\text{m}^2/\text{s}^2$$v_f = \sqrt{50.0} = +7.07\,\text{m/s}$
  • Final Answers: (a) $+22.5\,\text{m}$, (b) $+7.07\,\text{m/s}$
  • Question 1 (Runner Starting from Rest):
  • Problem Statement: A runner starts from rest at $x_i = 0\,\text{m}$ and accelerates at $+2.0\,\text{m/s}^2$ for $5.0\,\text{s}$. Find the runner's position after $5.0\,\text{s}$.
  • Method: Use $x_f = x_i + v_i t + \frac{1}{2} a t^2$
  • Calculation: $x_f = 0 + 0(5.0) + \frac{1}{2}(2.0)(5.0)^2 = 25\,\text{m}$
  • Final Answer: $+25\,\text{m}$
  • Question 2 (Cyclist Speeding Up Over a Set Distance):
  • Problem Statement: A cyclist accelerates at $+1.50\,\text{m/s}^2$ over a distance of $10.0\,\text{m}$. If the cyclist's initial velocity is $+10.0\,\text{m/s}$, determine the final velocity.
  • Method: Time is omitted; use $v_f^2 = v_i^2 + 2 a \Delta x$
  • Calculation: $v_f^2 = (10.0)^2 + 2(1.50)(10.0) = 100 + 30.0 = 130$
  • Square Root Evaluation: $v_f = \sqrt{130} = 11.4\,\text{m/s}$
  • Final Answer: $+11.4\,\text{m/s}$
  • Question 3 (Runner Decelerating Over a Set Distance):
  • Problem Statement: A runner slows down with a constant acceleration of $-1.0\,\text{m/s}^2$ over a distance of $5.0\,\text{m}$. Given an initial velocity of $+10.0\,\text{m/s}$, find the final velocity.
  • Method: Use $v_f^2 = v_i^2 + 2 a \Delta x$
  • Calculation: $v_f^2 = (10.0)^2 + 2(-1.0)(5.0) = 100 - 10.0 = 90$
  • Square Root Evaluation: $v_f = \sqrt{90} = 9.49\,\text{m/s}$
  • Final Answer: $+9.5\,\text{m/s}$
  • Question 4 (Bicyclist Accelerating from Rest):
  • Problem Statement: A bicyclist starts from rest and accelerates at $5.50\,\text{m/s}^2$ for $12.0\,\text{s}$. Find the bicyclist's position at $12.0\,\text{s}$.
  • Method: Use $x_f = v_i t + \frac{1}{2} a t^2$
  • Calculation: $x_f = 0(12.0) + \frac{1}{2}(5.50)(12.0)^2 = \frac{1}{2}(5.50)(144) = 396\,\text{m}$
  • Final Answer: $396\,\text{m}$
  • Question 5 (Cheetah Accelerating from Rest):
  • Problem Statement: A cheetah starts from rest and accelerates at $10.0\,\text{m/s}^2$ for $3.0\,\text{s}$. Find its position after $3.0\,\text{s}$.
  • Method: Use $x_f = v_i t + \frac{1}{2} a t^2$
  • Calculation: $x_f = 0(3.0) + \frac{1}{2}(10.0)(3.0)^2 = \frac{1}{2}(10.0)(9.0) = 45\,\text{m}$
  • Final Answer: $45\,\text{m}$
  • Question 6 (Dropped Watermelon Velocity Prior to Impact):
  • Problem Statement: A watermelon is dropped from rest from a roof $15.0\,\text{m}$ above the ground. Ignoring air resistance, calculate its velocity immediately prior to ground impact.
  • Coordinate Convention: Select downward as positive: $v_i = 0$, $a = +9.81\,\text{m/s}^2$, $\Delta x = +15.0\,\text{m}$
  • Method: Use $v_f^2 = v_i^2 + 2 a \Delta x$
  • Calculation: $v_f = \sqrt{2(9.81)(15.0)} = \sqrt{294.3} = 17.2\,\text{m/s}$
  • Final Answer: $17.2\,\text{m/s}\text{ downward}$

Free Fall and Gravitational Acceleration

• Free Fall Dynamics and Conditions:

  • Definition: Free fall occurs when the gravitational force is the sole significant influence acting upon a body (aerodynamic drag and air resistance are negligible).
  • Universal Gravitational Acceleration: Near Earth's surface, all bodies experience identical gravitational acceleration regardless of mass, geometry, or composition:     $g = 9.81\,\text{m/s}^2\text{ downward}$
  • Kinematics at Apex: At the highest point of an upward trajectory, instantaneous vertical velocity equals zero ($v = 0\,\text{m/s}$). However, acceleration remains non-zero and continues acting downward ($a = -9.81\,\text{m/s}^2$).
  • Application of Equations: Constant-acceleration kinematic equations apply directly by replacing the general acceleration $a$ with the signed value of gravitational acceleration ($+9.81\,\text{m/s}^2$ or $-9.81\,\text{m/s}^2$ depending on chosen directional signs).

• Worked Examples and Vertical Trajectory Analysis:

  • Worked Example 5 (Time to Attain Maximum Height):
  • Problem Statement: A toy rocket is launched vertically upward at $+25.0\,\text{m/s}$. Determine the time required to reach its highest point.
  • Known Values: $v_i = +25.0\,\text{m/s}$, $v_f = 0\,\text{m/s}$, $a = -9.81\,\text{m/s}^2$
  • Governing Equation: $v_f = v_i + a t$
  • Derivation: $0 = 25.0 + (-9.81)t \implies 9.81 t = 25.0$
  • Calculation: $t = \frac{25.0}{9.81} = 2.55\,\text{s}$
  • Final Answer: $2.55\,\text{s}$
  • Worked Example 6 (Maximum Height Attained):
  • Problem Statement: Using the rocket from Worked Example 5 reaching apex at $t = 2.55\,\text{s}$, calculate its maximum altitude above the launch point.
  • Known Values: $x_i = 0\,\text{m}$, $v_i = +25.0\,\text{m/s}$, $a = -9.81\,\text{m/s}^2$, $t = 2.55\,\text{s}$
  • Governing Equation: $x_f = x_i + v_i t + \frac{1}{2} a t^2$
  • Substitution: $x_f = 0 + (25.0)(2.55) + \frac{1}{2}(-9.81)(2.55)^2$
  • Calculation: $x_f = 63.75 - 31.895 = 63.8 - 31.9 = +31.9\,\text{m}$
  • Final Answer: $+31.9\,\text{m}$
  • Practice Set A, Question 1 (Thrown Ball Height After Fixed Duration):
  • Problem Statement: A ball is thrown straight upward with an initial velocity of $10.0\,\text{m/s}$. Determine its height after $2.0\,\text{s}$.
  • Coordinate Convention: Establish upward as positive; $a = -9.81\,\text{m/s}^2$
  • Method: Apply $x_f = v_i t + \frac{1}{2} a t^2$
  • Calculation: $x_f = (10.0)(2.0) + \frac{1}{2}(-9.81)(2.0)^2 = 20.0 - 19.62 = +0.38\,\text{m}$
  • Final Answer: $+0.38\,\text{m}$
  • Practice Set A, Question 2 (Rocket Height After Fixed Duration):
  • Problem Statement: A toy rocket is launched straight upward at $35.0\,\text{m/s}$. Find its height after $3.0\,\text{s}$.
  • Coordinate Convention: Upward positive, $a = -9.81\,\text{m/s}^2$
  • Method: Apply $x_f = v_i t + \frac{1}{2} a t^2$
  • Calculation: $x_f = (35.0)(3.0) + \frac{1}{2}(-9.81)(3.0)^2 = 105.0 - 44.145 = 60.855\,\text{m}$
  • Final Answer: $+60.9\,\text{m}$
  • Worked Example 7 (Velocity at a Specified Elevation):
  • Problem Statement: A toy rocket is launched vertically from the ground at $+25.0\,\text{m/s}$. Calculate its velocity when it is $30.0\,\text{m}$ above the ground.
  • Known Values: $x_i = 0\,\text{m}$, $x_f = +30.0\,\text{m}$, $v_i = +25.0\,\text{m/s}$, $a = -9.81\,\text{m/s}^2$
  • Governing Equation: $v_f^2 = v_i^2 + 2 a (x_f - x_i)$
  • Substitution: $v_f^2 = (25.0)^2 + 2(-9.81)(30.0) = 625 - 588.6 = 36.4$
  • Root Extraction and Sign Resolution: $v_f = \pm \sqrt{36.4} = \pm 6.03\,\text{m/s}$
  • Physical Distinction: The positive root represents ascending motion ($+6.03\,\text{m/s}$, upward); the negative root represents descending motion ($-6.03\,\text{m/s}$, downward).
  • Final Answer: $+6.03\,\text{m/s}$
  • Practice Set B, Question 4 (Rocket Velocity at Lower Elevation):
  • Problem Statement: A toy rocket is launched straight upward at $35.0\,\text{m/s}$. Determine its velocity when it reaches a height of $10.0\,\text{m}$.
  • Method: Use $v_f^2 = v_i^2 + 2 a \Delta x$
  • Calculation: $v_f^2 = (35.0)^2 + 2(-9.81)(10.0) = 1225 - 196.2 = 1028.8$
  • Root Extraction: $v_f = \pm \sqrt{1028.8} = \pm 32.1\,\text{m/s}$
  • Final Answer: $+32.1\,\text{m/s}$

Cumulative Free Fall Review and Master Kinematic Reference

• Cumulative Multi-Point Trajectory Analysis:

  • Problem Context: A ball is launched vertically upward from ground level ($y_0 = 0\,\text{m}$) with an initial velocity $v_i = 5.0\,\text{m/s}$. Acceleration due to gravity is downward ($a = -9.81\,\text{m/s}^2$). Its height is recorded at $0.1\,\text{s}$, $0.2\,\text{s}$, $0.5\,\text{s}$, $0.75\,\text{s}$, and $1.0\,\text{s}$.
  • Analytical Formula:$y(t) = v_i t + \frac{1}{2} a t^2 = 5.0 t - 4.905 t^2$
  • Calculated Altitudes:
  • At $t = 0.1\,\text{s}$: $y(0.1) = 5.0(0.1) - 4.905(0.1)^2 = 0.500 - 0.04905 = 0.45095\,\text{m} \approx 0.451\,\text{m}$
  • At $t = 0.2\,\text{s}$: $y(0.2) = 5.0(0.2) - 4.905(0.2)^2 = 1.000 - 0.1962 = 0.8038\,\text{m} \approx 0.804\,\text{m}$
  • At $t = 0.5\,\text{s}$: $y(0.5) = 5.0(0.5) - 4.905(0.5)^2 = 2.500 - 1.22625 = 1.27375\,\text{m} \approx 1.274\,\text{m}$
  • At $t = 0.75\,\text{s}$: $y(0.75) = 5.0(0.75) - 4.905(0.75)^2 = 3.750 - 2.75906 = 0.99094\,\text{m} \approx 0.991\,\text{m}$
  • At $t = 1.0\,\text{s}$: $y(1.0) = 5.0(1.0) - 4.905(1.0)^2 = 5.000 - 4.905 = 0.095\,\text{m}$
  • Height Comparison and Chronological Ordering:
  • Comparing calculated magnitudes: $0.095\,\text{m} < 0.451\,\text{m} < 0.804\,\text{m} < 0.991\,\text{m} < 1.274\,\text{m}$
  • Corresponding time values arranged from least height to greatest height: $1.0\,\text{s}$, $0.1\,\text{s}$, $0.2\,\text{s}$, $0.75\,\text{s}$, $0.5\,\text{s}$

• Master Equation Reference and Application Matrix:

  • Average Speed:$\text{speed} = \frac{\text{distance}}{\text{time}}$Condition: Scalar calculation of distance rate.
  • Average Velocity:$v_{\text{avg}} = \frac{\Delta x}{\Delta t}$Condition: Vector displacement divided by elapsed time.
  • Average Acceleration:$a_{\text{avg}} = \frac{\Delta v}{\Delta t}$Condition: Rate of velocity change over an elapsed interval.
  • Velocity-Time Relation:$v_f = v_i + a t$Condition: Connects velocity and time when position/displacement is omitted.
  • Position via Average Velocity:$x_f = x_i + \frac{1}{2}(v_i + v_f) t$Condition: Position determined when both initial and final velocities are known.
  • Position-Time-Acceleration Relation:$x_f = x_i + v_i t + \frac{1}{2} a t^2$Condition: Connects position, elapsed time, and acceleration when final velocity is omitted.
  • Velocity-Displacement Relation:$v_f^2 = v_i^2 + 2 a \Delta x$Condition: Connects velocities, acceleration, and displacement when time is omitted.
  • Earth Gravitational Acceleration Constant:$g = 9.81\,\text{m/s}^2\text{ downward}$Condition: Near-surface free fall ignoring atmospheric drag.