Comprehensive Kinematics Study Notes: Formulas, Grading Rules, and Practice Problems

Grading Criteria and Credit Breakdown

• $\frac{1}{4}$ credit ($+25\%$): Awarded for listing variables properly at the start of a solution (e.g., $x = 8\,m$, $v_0 = 56\,m/s$).

• $\frac{1}{4}$ credit ($+25\%$): Awarded for identifying and writing down the correct formula.

• $-\frac{1}{4}$ credit (Deduction of $25\%$): Applied for arithmetic or math errors (roughly $1$ to $4$ errors).

• $-1$ point penalty: Applied for significant figure (SIGFIGS) errors.

• $-50\%$ penalty: Applied for using wrong formulas to calculate an answer.

Kinematic Equations Summary and Application Guidelines

• Equation 1:

  • Formula for speed: $v = \frac{d}{t}$

  • Formula for total speed: $v = \frac{d_{tot}}{t_{tot}}$

  • Usage Guidelines: Use only when no changes in direction occur during the motion. In most scenarios, position/displacement $x$ may be substituted for distance $d$.

• Equation 2:

  • Formula for velocity (and occasionally speed): $v = \frac{\Delta x}{\Delta t} = \frac{x - x_0}{t}$

  • Alternative algebraic form: $x = x_0 + v t$

  • Usage Guidelines: Primary use is finding distance or displacement. In most problem scenarios, this formula must be combined with Equation 3.

• Equation 3:

  • Formula for acceleration: $a = \frac{\Delta v}{\Delta t} = \frac{v - v_0}{t}$

  • Alternative algebraic form: $v = v_0 + a t$

  • Usage Guidelines: Used to solve for acceleration ($a$), final velocity ($v$), initial velocity ($v_0$), or time ($t$).

• Equation 4:

  • Formula for displacement: $x = x_0 + v_0 t + \frac{1}{2} a t^2$

  • Usage Guidelines: Applied when final velocity ($v$) is neither given nor required. Capable of solving for any of the 5 kinematic variables ($x$, $x_0$, $v_0$, $a$, $t$) provided the other 4 variables are known.

• Equation 5:

  • Formula independent of time: $v^2 = v_0^2 + 2 a (x - x_0)$

  • Usage Guidelines: Applied specifically when time ($t$) is not provided in the problem statement.

• Multi-Step Formula Combination Strategy:

  • When missing 2 variables simultaneously, Equations 1 through 4 are typically used to calculate one missing variable, which is then plugged directly into Equation 5.

Kinematics Vocabulary and Variable Mapping

• HOW FAR: Refers to position/displacement ($x$) or distance ($d$).

• WHERE: Refers to position/displacement ($x$).

• HOW LONG: Refers to either duration of time ($t$) or physical spatial length ($x$).

• HOW FAST: Refers to speed or velocity ($v$ or $\bar{v}$).

• WHEN: Refers to time ($t$).

Relative Velocity Principles

• When an object is launched/projected from a moving vehicle or when moving within a fluid medium (e.g., swimming in a river), relative velocities and speeds add or subtract depending on the relative direction of motion.

Detailed Practice Problems and Step-by-Step Solutions

• Practice Problem 1: Taxi Journey (Average Speed vs. Average Velocity)

  • Scenario: A taxi driver clocks in at the cab company at 9:00 AM. The driver travels to pick up a first fare $15\,km$ North on Main Street at 9:25 AM. The passenger requests to be dropped off South; after driving for $40\,min$ at $40\,km/h$, the fare is dropped off. After receiving a $100 tip, the boss instructs the driver to return and drop off money at the taxi company by 10:20 AM. Determine the average speed and average velocity upon returning to shift origin (in $m/s$).

  • Time Calculation:

    • Total elapsed shift duration: 9:00 AM to 10:20 AM = $1\,\text{hour}\,20\,\text{minutes} = 80\,\text{minutes}$.

    • Total elapsed time in seconds: $t_{tot} = 80 \times 60 = 4800\,s$.

  • Distance Calculation:

    • Segment 1 distance: $d_1 = 15\,km = 15,000\,m$ North.

    • Segment 2 distance: $d_2 = 40\,km/h \times \left(\frac{40}{60}\,h\right) = 26.67\,km = 26,666.67\,m$ South.

    • Position relative to cab company after Segment 2: $+15\,km - 26.67\,km = -11.67\,km$ ($11.67\,km$ South of company).

    • Segment 3 distance (return to cab company): $d_3 = 11.67\,km = 11,666.67\,m$ North.

    • Total distance traveled: $d_{tot} = 15\,km + 26.67\,km + 11.67\,km = 53.33\,km = 53,333.33\,m$

  • Average Speed Solution:

    • $v_{avg} = \frac{d_{tot}}{t_{tot}} = \frac{53,333.33\,m}{4800\,s} \approx 11.11\,m/s$

  • Average Velocity Solution:

    • Since initial position and final position are both the taxi company, net displacement $\Delta x = 0\,m$

    • $\bar{v} = \frac{\Delta x}{t_{tot}} = \frac{0\,m}{4800\,s} = 0\,m/s$

• Practice Problem 2: Engine Acceleration and Emergency Braking

  • Scenario: Starting from rest, a car guns its engine at $7.0\,m/s^2$ for $30\,s$. It then blows a tire, slams on the brakes, and comes to a complete stop over $120\,m$. Calculate the acceleration exerted by the brakes.

  • Phase 1: Acceleration Phase

    • $v_0 = 0\,m/s$

    • $a_1 = 7.0\,m/s^2$

    • $t_1 = 30\,s$

    • Velocity reached at end of acceleration: $v_1 = v_0 + a_1 t_1 = 0 + (7.0\,m/s^2)(30\,s) = 210\,m/s$

  • Phase 2: Braking Phase

    • Initial velocity for braking phase: $v_{0,2} = 210\,m/s$

    • Final velocity: $v_2 = 0\,m/s$

    • Displacement during braking: $\Delta x_2 = 120\,m$

    • Applying Equation 5: $v_2^2 = v_{0,2}^2 + 2 a_2 \Delta x_2$

    • $0^2 = (210\,m/s)^2 + 2 a_2 (120\,m)$

    • $0 = 44100 + 240 a_2$

    • $240 a_2 = -44100$

    • $a_2 = -\frac{44100}{240} = -183.75\,m/s^2$

• Practice Problem 3: Multi-Segment Journey Average Speed Calculation

  • Scenario: A driver cruises at $95\,km/h$ for $5\,min$, then slows down and drives at $65\,km/h$ for $7\,min$. At that point, the driver accelerates at $3\,m/s^2$ for $30\,s$. Calculate the average speed over the entire set of trips.

  • Trip 1 Breakdown:

    • $v_1 = 95\,km/h = \frac{95}{3.6} = 26.39\,m/s$

    • $t_1 = 5 \times 60 = 300\,s$

    • $d_1 = (26.39\,m/s)(300\,s) = 7917\,m$

  • Trip 2 Breakdown:

    • $v_2 = 65\,km/h = \frac{65}{3.6} = 18.06\,m/s$

    • $t_2 = 7 \times 60 = 420\,s$

    • $d_2 = (18.06\,m/s)(420\,s) = 7585.2\,m$

  • Trip 3 Breakdown:

    • $a_3 = 3\,m/s^2$

    • $x_0 = 0\,m$

    • $v_0 = 18.06\,m/s$

    • $t_3 = 30\,s$

    • Formula: $x_3 = x_0 + v_0 t_3 + \frac{1}{2} a_3 t_3^2$

    • $x_3 = 0 + (18.06\,m/s)(30\,s) + \frac{1}{2}(3\,m/s^2)(30\,s)^2$

    • $x_3 = 541.8\,m + 1350\,m = 1891.8\,m$

  • Total Journey Summary & Final Calculation:

    • Total distance: $d_{tot} = 7917\,m + 7585.2\,m + 1891.8\,m = 17394\,m$

    • Total time: $t_{tot} = 300\,s + 420\,s + 30\,s = 750\,s$

    • Average speed calculation: $v = \frac{d_{tot}}{t_{tot}} = \frac{17394\,m}{750\,s} = 23.192\,m/s \approx 23\,m/s$