Chapter 2 PDLC Notes: Step 1 & Step 2 for Problems on Specific Speed and Travel Time

Step 1: PDLC — Problem Definition and Knowns

• Context: Chapter 2, Assignment requires doing Step 1 and Step 2 of the PDLC for two problems involving speed and time.
• PDLC focus here: Plan (define the problem, identify given data and unknowns) and Prepare for calculation.
• Core relationship (kinematics for constant speed):
  • $d = v\,t$
  • $v = \frac{d}{t}$
  • $t = \frac{d}{v}$
• Problem 1: Specific speed to arrive 14 km ahead in a given amount of time
  • Given data (knowns): distance to destination ahead, $d = 14\ \text{km}$; time available is a given value, denote as $t_{target}$ (units should be hours or minutes, with conversion as needed).
  • Unknown: required speed $v$ to meet the target time, i.e., the speed that results in arriving 14 km sooner than or exactly at the destination given by the time constraint.
  • Key relation to apply: $v = \frac{d}{t_{target}}$ (assuming the driver maintains constant speed over the 14 km).
• Problem 2: Time to reach a distance at a constant speed of 60 km/h
  • Given data (knowns): speed $v = 60\ \text{km/h}$; distance to cover is $d$ (unknown in this statement, but treated as a variable for time calculation).
  • Unknown: time $t$ required to travel distance $d$ at this constant speed.
  • Key relation to apply: $t = \frac{d}{v}$. With $v = 60\ \text{km/h}$, this becomes $t = \frac{d}{60}$ hours.
• Important unit notes
  • Distances are in kilometers (km).
  • Time can be in hours (h) or minutes (min); when using $v = d/t$, ensure consistent time units.
  • To convert minutes to hours: $t<em>{\text{hours}} = \frac{t</em>{\text{minutes}}}{60}$; to convert hours to minutes: $t<em>{\text{minutes}} = t</em>{\text{hours}} \times 60$.
  • 60 km/h is equivalent to 1 km/min, since
  • $60\ \text{km/h} = \frac{60\ \text{km}}{60\ \text{min}} = 1\ \text{km/min}.$
• Plan/Strategic considerations for Step 1
  • Identify knowns and unknowns for each problem.
  • Decide which primary equation to use (d = v t) based on given data.
  • Check units and consider necessary conversions before solving.
  • Note special cases: if time is not given in hours, convert; if distance is in km, speed should be in km/h for consistency.
• Connections to foundational principles
  • The inverse relationship between speed and time for a fixed distance (holding distance constant, higher speed yields smaller time and vice versa).
  • Dimensional analysis checks: units must resolve to a unit of velocity (km/h) for v, time (h or min) for t, distance (km) for d.
• Practical and ethical implications
  • Understanding how speed planning affects arrival times can inform trip planning, fuel efficiency, and safety.
  • Always consider safety, traffic laws, and realistic driving conditions when applying constant-speed assumptions.
• Expected outcomes from Step 1
  • A clear statement of the problem with identified knowns and unknowns for each item.
  • A chosen equation set (based on d = v t) and a plan to compute the unknown quantities.

Step 2: PDLC — Solve and Compute (Apply Plan to Problems 1 & 2)

• Core equations to use
  • Distance–speed–time fundamental relations:
  • $d = v t$
  • $v = \frac{d}{t}$
  • $t = \frac{d}{v}$
• Problem 1: Solve for the specific speed to arrive 14 km ahead in a given time
  • Formula to compute speed: $v = \frac{d}{t_{target}}$
  • Substitution for the given data: $v = \frac{14\ \text{km}}{t<em>{target}}$, where $t</em>{target}$ must be in hours for $v$ in km/h. If $t<em>{target}$ is given in minutes, convert first: $t</em>{target} (\text{hours}) = \frac{t_{target} (\text{minutes})}{60}$.
  • Example calculations (illustrative):
  • If $t_{target} = 0.5\ \text{h}$, then $v = \frac{14}{0.5} = 28\ \text{km/h}.$
  • If $t_{target} = 1\ \text{h}$, then $v = \frac{14}{1} = 14\ \text{km/h}.$
  • If $t_{target} = 30\ \text{min} = 0.5\ \text{h}$, then $v = 28\ \text{km/h}$ (same as first example).
  • Key takeaways from solving Problem 1
  • The required speed is inversely proportional to the available time for a fixed distance.
  • Ensure time units match speed units (hours for km/h).
• Problem 2: Solve for the time to reach a distance at 60 km/h
  • Formula to compute time: $t = \frac{d}{v}$ with $v = 60\ \text{km/h}$
  • Substitution: $t = \frac{d}{60}\quad \text{hours}$
  • Examples (illustrative):
  • If $d = 120\ \text{km}$, then $t = \frac{120}{60} = 2\ \text{hours} = 120\ \text{minutes}$.
  • If $d = 30\ \text{km}$, then $t = \frac{30}{60} = 0.5\ \text{hours} = 30\ \text{minutes}$.
  • Key takeaways from solving Problem 2
  • Time scales linearly with distance when speed is constant.
  • Converting between hours and minutes is a routine step; for 60 km/h, 1 km corresponds to 1 minute of travel time.
• Additional notes and checks
  • Dimensional checks: confirm that the units satisfy the target quantity (velocity for v, time for t).
  • If a time or distance is not physically reasonable (e.g., negative, zero time), re-check data or constraints.
  • Real-world caveats: traffic, stops, weather, and road conditions can affect actual travel time; the equations assume ideal constant speed.
• Summary of PDLC outputs
  • For Problem 1: v = d / t_target; example values show how the required speed changes with the allowed time.
  • For Problem 2: t = d / 60 hours; conversion to minutes yields t_min = d minutes when distance is in kilometers and speed is 60 km/h.
• Connections to previous lectures
  • The d = v t relationship is a staple result in introductory physics/engineering math and appears repeatedly in problem-solving strategies.
• Final practical hint
  • When presenting solutions, clearly state the assumed time units, distances, and any conversions performed to ensure clarity and reproducibility.