Comprehensive Study Guide: 1D Kinematics, Vectors, and Graphical Motion Analysis

Kinematic Terminology and Hierarchy

• Kinematics is the branch of classical mechanics that describes the motion of points, bodies, or systems of bodies without considering the forces that cause the motion.

• Time derivatives form a continuous chain of motion descriptors:

• The sequence of sequential time derivatives ($\frac{d}{dt}$) of position is defined as follows:

  • Position: $\vec{r}$ or $x$
  • Velocity: $\vec{v} = \frac{d\vec{r}}{dt}$
  • Acceleration: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}$
  • Jerk: $\vec{j} = \frac{d\vec{a}}{dt} = \frac{d^3\vec{r}}{dt^3}$
  • Snap: $\vec{s} = \frac{d\vec{j}}{dt} = \frac{d^4\vec{r}}{dt^4}$
  • Crackle: $\vec{c} = \frac{d\vec{s}}{dt} = \frac{d^5\vec{r}}{dt^5}$
  • Pop: $\vec{p} = \frac{d\vec{c}}{dt} = \frac{d^6\vec{r}}{dt^6}$

• Position:

  • Position is a vector quantity that specifies the location of an object relative to a chosen reference point (the origin) along a spatial coordinate axis (e.g., the $x$-axis in one dimension).
  • The sign (positive or negative) denotes the direction relative to the origin.
  • Example: A point located at $x = 3\,\text{m}$ lies $3\,\text{m}$ to the right of the origin; a point at $x = -1\,\text{m}$ lies $1\,\text{m}$ to the left of the origin.

• Displacement:

  • Displacement ($\Delta x$) is a vector quantity representing the straight-line change in position from an initial location ($x_i$) to a final location ($x_f$):

$\Delta x = x_f - x_i$

• It depends only on the initial and final positions, not on the path taken.

  • Distance Travelled:

• Distance travelled is a scalar quantity representing the total length of the physical path traversed by an object during its motion between two points.

• Distance travelled is always non-negative ($\ge 0$).

  • Example Comparison (Mailbox Walk):

• Walking from a front door to a mailbox $10\,\text{m}$ away and returning back to the front door:

• Initial position: $x_i = 0\,\text{m}$

• Final position: $x_f = 0\,\text{m}$

• Displacement: $\Delta x = 0\,\text{m}$

• Distance travelled: $d = 10\,\text{m} + 10\,\text{m} = 20\,\text{m}$

  • Average Velocity:

• Average velocity ($v_{avg}$) is a vector quantity defined as the total displacement divided by the total time interval ($\Delta t$):

$v_{avg} = \frac{\Delta x}{\Delta t}$

• The sign of average velocity matches the sign of displacement.

  • Average Speed:

• Average speed is a scalar quantity defined as the total distance travelled divided by the total time interval ($\Delta t$):

$\text{Average Speed} = \frac{\text{Distance Travelled}}{\Delta t}$

• Example Comparison (Mailbox Walk Calculations):
  • If the round trip to the mailbox ($10\,\text{m}$ away and back, total distance $20\,\text{m}$, total displacement $0\,\text{m}$) takes $4\,\text{s}$ (moving from $x = 0\,\text{m}$ to $x = -3\,\text{m}$ along a path with total length $7\,\text{m}$):
  • Average velocity for displacement $\Delta x = -3\,\text{m}$ over $\Delta t = 4\,\text{s}$:

$v_{avg} = \frac{-3\,\text{m}}{4\,\text{s}} = -0.75\,\text{m/s}$

• Average speed for distance $d = 7\,\text{m}$ over $\Delta t = 4\,\text{s}$:

$\text{Average Speed} = \frac{7\,\text{m}}{4\,\text{s}} = 1.75\,\text{m/s}$

• Example Comparison (Vertical Toss):

  • A ball thrown straight up in the air and caught at the exact same height from which it was released:
  • Displacement: $\Delta y = 0\,\text{m}$
  • Average velocity: $v_{avg} = 0\,\text{m/s}$
  • Distance travelled: Non-zero (twice the peak height reached)
  • Average speed: Non-zero

• Instantaneous Velocity and Instantaneous Speed:

  • Instantaneous velocity ($v$) is the velocity of an object at a specific single point in time, evaluated as the mathematical limit of average velocity as the time interval approaches zero:

$v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$

• Instantaneous speed is the scalar magnitude of the instantaneous velocity ($|v|$).

  • Acceleration:

• Acceleration ($\vec{a}$) is a vector quantity defined as the rate of change of velocity over time:

$\vec{a} = \frac{\Delta \vec{v}}{\Delta t}$

• The SI base unit of acceleration is meters per second squared ($\text{m/s}^2$ or $\text{m}\cdot\text{s}^{-2}$).
• Example calculation: Accelerating a car from $30\,\text{km/h}$ to $50\,\text{km/h}$ over a time period of $60\,\text{s}$:
  • Convert speeds to standard SI units ($\text{m/s}$):

$v_i = \frac{30\,\text{km/h}}{3.6} = 8.333\,\text{m/s}$

$v_f = \frac{50\,\text{km/h}}{3.6} = 13.889\,\text{m/s}$

- Compute change in velocity ($\Delta v$):

$\Delta v = 13.889\,\text{m/s} - 8.333\,\text{m/s} = 5.556\,\text{m/s}$

- Compute acceleration ($a$):

$a = \frac{5.556\,\text{m/s}}{60\,\text{s}} = 0.0926\,\text{m/s}^2 \approx 0.09\,\text{m/s}^2$

Vectors and Trigonometric Foundations

• Vector Attributes:

  • A vector is a mathematical quantity possessing both magnitude (length/size) and direction.
  • Key kinematic physical quantities that are vectors include displacement, velocity, acceleration, and force.

• Forms of Vector Representation:

• Component Notation: Expressing a vector in terms of unit vectors along Cartesian axes:

$\vec{r} = 6\hat{x} + 8\hat{y}$

• Resultant Notation: Expressing a vector in terms of its overall magnitude and directional angle relative to a geographical/coordinate reference:

$\vec{r} = 10 \text{ at } 37^\circ \text{ North of East}$

• Trigonometric Relations (SOH-CAH-TOA):

• For a right-angled triangle with angle $A$, opposite side $a$, adjacent side $b$, and hypotenuse $c$:

$\sin(A) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{a}{c}$

$\cos(A) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{b}{c}$

$\tan(A) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{a}{b}$

• Procedure for Determining Vector Angles:

  1. Define a coordinate origin and choose positive directional axes (conventionally, up and right are defined as positive).
  2. Draw the orthogonal components ($x$ and $y$) of the vector tail-to-head starting at the origin.
  3. Complete the right-angled triangle where the vector itself forms the hypotenuse.
  4. Solve for the interior angle using inverse trigonometric functions (typically relative to the horizontal $x$-axis).

• Worked Example (Vector Direction Angle):

  • Given a vector starting at the origin with horizontal component $x = 9.0$ and vertical component $y = 7.0$:
  • Angle $\theta$ with respect to the horizontal $x$-axis is calculated as:

$\theta = \arctan\left(\frac{\text{Opposite}}{\text{Adjacent}}\right) = \arctan\left(\frac{7.0}{9.0}\right) \approx 38^\circ$

Graphical Analysis of Motion

• Position-Time ($x-t$) Graphs and Velocity:
  • The slope of a tangent line at any point on a position-time graph gives the instantaneous velocity of the object:

$\text{Slope} = \frac{\Delta x}{\Delta t} = v$

• When velocity is constant, the position-time plot is linear, adhering to $y = mx + b$ where $y = x(t)$, slope $m = v$, run $x = t$, and vertical intercept $b = x_i$.

  • Worked Example (Ranking Average Velocity from Position-Time Graph):

• Evaluating slopes across four distinct time intervals:
  • Interval A ($t = 0\,\text{s}$ to $t = 3.5\,\text{s}$): Moves from $x = -2\,\text{m}$ to $x = 5\,\text{m}$. Slope = $\frac{5 - (-2)}{3.5 - 0} = \frac{7}{3.5} = +2\,\text{m/s}$.
  • Interval B ($t = 3.5\,\text{s}$ to $t = 4.5\,\text{s}$): Remains at $x = 5\,\text{m}$. Slope = $\frac{5 - 5}{1} = 0\,\text{m/s}$.
  • Interval C ($t = 4.5\,\text{s}$ to $t = 6.0\,\text{s}$): Moves from $x = 5\,\text{m}$ to $x = 6\,\text{m}$. Slope = $\frac{6 - 5}{6.0 - 4.5} = \frac{1}{1.5} = +0.67\,\text{m/s}$.
  • Interval D ($t = 6.0\,\text{s}$ to $t = 10.0\,\text{s}$): Moves from $x = 6\,\text{m}$ to $x = -2\,\text{m}$. Slope = $\frac{-2 - 6}{10.0 - 6.0} = \frac{-8}{4} = -2\,\text{m/s}$.
• Ranking velocities from most negative to most positive:

$v_D < v_B < v_C < v_A$

• Velocity-Time ($v-t$) Graphs and Acceleration:

• The slope of a velocity-time graph gives the acceleration:

$\text{Slope} = \frac{\Delta v}{\Delta t} = a$

• Region A (Positive Slope): Object accelerates in the positive direction.

• Region B (Zero Slope): Object maintains constant velocity (zero acceleration).

• Region C (Negative Slope, Positive Velocity): Object decelerates while moving in the positive direction.

• Region D (Negative Slope, Negative Velocity): Object accelerates in the negative direction (speeding up in reverse).

  • Worked Examples (Velocity-Time Analysis):

• Finding Acceleration at a Point: On a $v-t$ graph where velocity decreases linearly from $2\,\text{m/s}$ at $t = 1\,\text{s}$ to $-4\,\text{m/s}$ at $t = 4\,\text{s}$:

$a = \frac{\Delta v}{\Delta t} = \frac{-4\,\text{m/s} - 2\,\text{m/s}}{4\,\text{s} - 1\,\text{s}} = \frac{-6\,\text{m/s}}{3\,\text{s}} = -2\,\text{m/s}^2$

• Negative Acceleration Interval: The acceleration is negative over time intervals where the line slopes downwards (e.g., from $t = 1\,\text{s}$ to $t = 4\,\text{s}$).

• Matching Acceleration-Time Plot: A $v-t$ graph consisting of an upward linear line, a flat horizontal line, and a downward linear line corresponds step-for-step to an $a-t$ graph containing a constant positive step, a zero step, and a constant negative step.

  • Area Under Curves:

• Area under Acceleration-Time ($a-t$) Graph:

  • Units of area: $[a] \times [t] = \frac{\text{m}}{\text{s}^2} \times \text{s} = \frac{\text{m}}{\text{s}}$.
  • The area bounded between the acceleration curve and the time axis yields the change in velocity ($\Delta \vec{v}$):

$\Delta \vec{v} = \vec{a} \Delta t$

• Area under Velocity-Time ($v-t$) Graph:
  • Units of area: $[v] \times [t] = \frac{\text{m}}{\text{s}} \times \text{s} = \text{m}$.
  • The area bounded between the velocity curve and the time axis yields the displacement ($\Delta x$).

• Boundary Rule: The area evaluated must always be computed between the function curve and the horizontal time axis ($x$-axis). Areas above the time axis represent positive displacement, while areas below the time axis represent negative displacement.

  • Worked Example (Displacement Calculation from $v-t$ Graph):

• Given a $v-t$ plot where from $t = 1\,\text{s}$ to $t = 2\,\text{s}$, velocity goes from $2\,\text{m/s}$ to $0\,\text{m/s}$ (triangle above axis), and from $t = 2\,\text{s}$ to $t = 4\,\text{s}$, velocity goes from $0\,\text{m/s}$ to $-4\,\text{m/s}$ (triangle below axis):

• Positive Area ($t = 1\,\text{s}$ to $2\,\text{s}$):

$\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2\,\text{s} - 1\,\text{s}) \times 2\,\text{m/s} = +1\,\text{m}$

• Negative Area ($t = 2\,\text{s}$ to $4\,\text{s}$):

$\text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (4\,\text{s} - 2\,\text{s}) \times (-4\,\text{m/s}) = -4\,\text{m}$

• Total Displacement ($\Delta x$):

$\Delta x = \text{Area}_1 + \text{Area}_2 = +1\,\text{m} + (-4\,\text{m}) = -3\,\text{m}$

Derivation of 1D Kinematic Equations

• For one-dimensional motion under constant acceleration, the four fundamental kinematic equations are:

  1. $x_f = x_i + v_i t + \frac{1}{2}a t^2$
  2. $x_f = x_i + \left(\frac{v_i + v_f}{2}\right)t$
  3. $v_f = v_i + a t$
  4. $v_f^2 = v_i^2 + 2a (x_f - x_i)$

• Geometric Derivation from $v-t$ Motion Plots:

  • Consider a velocity-time graph with initial velocity $v_i$ at $t = 0$ and final velocity $v_f$ at time $t$.
  • The total area under the linear curve can be partitioned into a lower rectangle and an upper triangle:
  • Area of rectangle = $v_i t$
  • Area of triangle = $\frac{1}{2} t (v_f - v_i) = \frac{1}{2} t \Delta v$
  • Substituting $\Delta v = a t$ into the triangle area:

$\Delta x = v_i t + \frac{1}{2} t (a t) = v_i t + \frac{1}{2} a t^2$

• Alternatively, substituting $\Delta v = v_f - v_i$ into the total area expression:

$\Delta x = v_i t + \frac{1}{2} t v_f - \frac{1}{2} t v_i = \frac{1}{2} t v_f + \frac{1}{2} t v_i = \left(\frac{v_i + v_f}{2}\right)t$

1D Kinematics Worked Examples (Horizontal Motion)

• Example 1: Cheetah Acceleration:
  • Problem: A cheetah accelerates from rest to $25\,\text{m/s}$ in $6.2\,\text{s}$. How far has the cheetah travelled?
  • Inventory:
  • $v_i = 0\,\text{m/s}$
  • $v_f = 25\,\text{m/s}$
  • $t = 6.2\,\text{s}$
  • Calculation:

$\Delta x = \left(\frac{v_i + v_f}{2}\right)t = \left(\frac{0 + 25\,\text{m/s}}{2}\right)(6.2\,\text{s}) = (12.5\,\text{m/s})(6.2\,\text{s}) = 78\,\text{m}$

• Example 2: Parking at Saba Coffee Shop:
  • Problem: Driving to a parking spot $53.0\,\text{m}$ away, a driver decelerates at $a = -2.48\,\text{m/s}^2$ to come to a complete stop. What was the initial speed in $km/h$?
  • Inventory:
  • $v_f = 0\,\text{m/s}$
  • $a = -2.48\,\text{m/s}^2$
  • $\Delta x = 53.0\,\text{m}$
  • Calculation:

$v_f^2 = v_i^2 + 2a \Delta x \implies 0 = v_i^2 + 2(-2.48\,\text{m/s}^2)(53.0\,\text{m})$

$v_i^2 = 262.88\,\text{m}^2/\text{s}^2 \implies v_i = \sqrt{262.88} \approx 16.2136\,\text{m/s}$

- Convert to $\text{km/h}$:

$v_i = 16.2136\,\text{m/s} \times 3.6 = 58.4\,\text{km/h}$

• Example 3: Car Entering a Freeway:
  • Problem: A car accelerates from rest at $2.40\,\text{m/s}^2$ for $12.0\,\text{s}$. Find total distance travelled and final velocity.
  • Inventory:
  • $v_i = 0\,\text{m/s}$
  • $a = 2.40\,\text{m/s}^2$
  • $t = 12.0\,\text{s}$
  • Calculation 1 (Distance):

$\Delta x = v_i t + \frac{1}{2} a t^2 = 0 + \frac{1}{2}(2.40\,\text{m/s}^2)(12.0\,\text{s})^2 = 0.5 \times 2.40 \times 144 = 173\,\text{m}$

• Calculation 2 (Final Velocity):

$v_f = v_i + a t = 0 + (2.40\,\text{m/s}^2)(12.0\,\text{s}) = 28.8\,\text{m/s}$

• Example 4: Deer Stopping Distance (Zero Reaction Time):
  • Problem: Driving at $30\,\text{m/s}$, a driver spots a deer $75\,\text{m}$ away. What acceleration is needed to stop right before the deer?
  • Inventory:
  • $v_i = 30\,\text{m/s}$
  • $v_f = 0\,\text{m/s}$
  • $\Delta x = 75\,\text{m}$
  • Calculation:

$v_f^2 = v_i^2 + 2a \Delta x \implies 0 = (30\,\text{m/s})^2 + 2 a (75\,\text{m})$

$0 = 900 + 150 a \implies a = -\frac{900}{150} = -6\,\text{m/s}^2$

• Example 5: Deer Stopping Distance with Reaction Time:
  • Problem: Same scenario as Example 4 ($v_i = 30\,\text{m/s}$, initial distance $75\,\text{m}$), but incorporating a driver reaction time of $t_{react} = 0.5\,\text{s}$. Find the required deceleration.
  • Phase 1 (Reaction Phase):
  • Speed is constant ($30\,\text{m/s}$) for $0.5\,\text{s}$.

$d_{react} = v_i \times t_{react} = 30\,\text{m/s} \times 0.5\,\text{s} = 15\,\text{m}$

• Phase 2 (Braking Phase):
  • Remaining distance available for braking: $\Delta x_{brake} = 75\,\text{m} - 15\,\text{m} = 60\,\text{m}$.

$v_f^2 = v_i^2 + 2a \Delta x_{brake} \implies 0 = (30\,\text{m/s})^2 + 2 a (60\,\text{m})$

$0 = 900 + 120 a \implies a = -\frac{900}{120} = -7.5\,\text{m/s}^2$

• Example 6: Two-Phase Motion (Squirrel Distraction):
  • Problem: Starting from rest, a car accelerates at $3.50\,\text{m/s}^2$ for $7.00\,\text{s}$. A squirrel jumps on the road, causing the driver to slam on the brakes, slowing down at $2.25\,\text{m/s}^2$ until stopped. Find total distance and total elapsed time.
  • Phase 1 (Acceleration):
  • $v_{i1} = 0\,\text{m/s}$, $a_1 = 3.50\,\text{m/s}^2$, $t_1 = 7.00\,\text{s}$

$v_{f1} = v_{i1} + a_1 t_1 = 0 + (3.50)(7.00) = 24.5\,\text{m/s}$

$d_1 = \frac{1}{2} a_1 t_1^2 = \frac{1}{2}(3.50)(7.00)^2 = 85.75\,\text{m}$

• Phase 2 (Braking):
  • $v_{i2} = 24.5\,\text{m/s}$, $v_{f2} = 0\,\text{m/s}$, $a_2 = -2.25\,\text{m/s}^2$

$v_{f2} = v_{i2} + a_2 t_2 \implies 0 = 24.5 + (-2.25)t_2 \implies t_2 = \frac{24.5}{2.25} \approx 10.889\,\text{s}$

$d_2 = \frac{v_{f2}^2 - v_{i2}^2}{2 a_2} = \frac{0 - (24.5)^2}{2(-2.25)} = \frac{-600.25}{-4.5} \approx 133.389\,\text{m}$

• Total Motion:

  • Total distance = $d_1 + d_2 = 85.75\,\text{m} + 133.389\,\text{m} = 219.139\,\text{m} \approx 219\,\text{m}$

  • Total time = $t_1 + t_2 = 7.00\,\text{s} + 10.889\,\text{s} = 17.889\,\text{s} \approx 17.9\,\text{s}$

  • Example 7: Direction Reversal (Distance vs. Displacement):

• Problem: An object moves initially at $v_i = -5\,\text{m/s}$ (leftward). It undergoes constant acceleration for $10\,\text{s}$, reaching $v_f = +5\,\text{m/s}$ (rightward). Find distance travelled and total displacement.

• Acceleration Calculation:

$a = \frac{v_f - v_i}{t} = \frac{5\,\text{m/s} - (-5\,\text{m/s})}{10\,\text{s}} = \frac{10}{10} = 1.0\,\text{m/s}^2$

• Time to Come to Rest ($v = 0$):

$v = v_i + a t_{stop} \implies 0 = -5 + 1.0 t_{stop} \implies t_{stop} = 5\,\text{s}$

• Segment 1 ($t = 0$ to $5\,\text{s}$):

$\Delta x_1 = v_i t_{stop} + \frac{1}{2} a t_{stop}^2 = (-5)(5) + \frac{1}{2}(1.0)(5)^2 = -25 + 12.5 = -12.5\,\text{m}$

• Segment 2 ($t = 5$ to $10\,\text{s}$):

$\Delta x_2 = (0)(5) + \frac{1}{2}(1.0)(5)^2 = +12.5\,\text{m}$

• Results:

  • Distance travelled = $|\Delta x_1| + |\Delta x_2| = 12.5\,\text{m} + 12.5\,\text{m} = 25\,\text{m}$

  • Total displacement = $\Delta x_1 + \Delta x_2 = -12.5\,\text{m} + 12.5\,\text{m} = 0\,\text{m}$

  • Example 8: Two-Vehicle Pursuit / Catch-Up:

• Problem: A yellow car travels rightward at constant speed $v_y = 12\,\text{m/s}$. When it is $45\,\text{m}$ behind a stationary blue car, the blue car begins accelerating from rest. The yellow car catches the blue car $5.5\,\text{s}$ later. Find the blue car's acceleration.
• Position Equations:
  • Yellow car position: $x_y(t) = 12 t$
  • Blue car position: $x_b(t) = 45 + \frac{1}{2} a_b t^2$
• Catch-up Condition at $t = 5.5\,\text{s}$:

$x_y(5.5) = x_b(5.5)$

$12(5.5) = 45 + \frac{1}{2} a_b (5.5)^2$

$66 = 45 + \frac{1}{2} a_b (30.25)$

$21 = 15.125 a_b \implies a_b = \frac{21}{15.125} \approx 1.39\,\text{m/s}^2$

Vertical 1D Kinematics (Free Fall Motion)

• Definition of Free Fall:
  • Free fall is one-dimensional vertical motion where the only acting acceleration is due to gravity ($g$).
  • Air resistance and atmospheric drag forces are neglected.
  • Acceleration due to gravity is constant near the surface of the Earth:

$g = 9.81\,\text{m/s}^2$

• The sign assigned to $g$ depends entirely on the defined coordinate system choice (e.g., $a = -9.81\,\text{m/s}^2$ if upward is chosen as positive, or $a = +9.81\,\text{m/s}^2$ if downward is chosen as positive).

  • Worked Example: Cliff Height Measurement:

• Problem: A rock dropped from rest from the top of a cliff takes $6.50\,\text{s}$ to hit the ground. Determine the height of the cliff.

• Coordinate System Definition: Define downward direction as positive.

• Inventory:

  • $v_i = 0\,\text{m/s}$
  • $t = 6.50\,\text{s}$
  • $a = 9.81\,\text{m/s}^2$

• Calculation:

$\Delta y = v_i t + \frac{1}{2} a t^2$

$\Delta y = (0)(6.50\,\text{s}) + \frac{1}{2}(9.81\,\text{m/s}^2)(6.50\,\text{s})^2$

$\Delta y = 0.5 \times 9.81 \times 42.25 = 207.246\,\text{m} \approx 207\,\text{m}$

• The total height of the cliff is $207\,\text{m}$.