Motion in 1D Vocabulary

Fundamentals of Kinematics and One-Dimensional Motion

Kinematics focuses on describing the motion of objects using concepts such as distance, displacement, speed, velocity, and acceleration. Defining a coordinate system requires specifying a starting point (the origin) and designating a positive direction. Any change in position relative to this coordinate frame is defined as displacement.

Displacement is a vector quantity representing how far an object is from its starting point along with the direction of travel, whereas distance traveled is a scalar quantity representing the total path length traversed regardless of direction. Quantities in physics are categorized as either scalars or vectors:

  • Displacement: Vector

  • Distance: Scalar

  • Velocity: Vector

  • Speed: Scalar

  • Acceleration: Vector

  • Temperature: Scalar

When standing on first base and running to second base, the motion is characterized by a displacement vector toward second base. Conversely, running one full lap around a track and returning to the exact starting line results in a non-zero scalar distance equal to the track length, but a vector displacement of exactly 0 m0\,\text{m}.

A car speedometer measures instantaneous speed, which is a scalar quantity indicating how fast the vehicle is moving at a specific instant, rather than velocity, because it does not provide directional information. While speed and velocity are often used interchangeably in everyday practical contexts, velocity includes direction whereas speed represents magnitude only.

Directional relationships between velocity and acceleration reveal several operational principles:

  • An object moving North can have a Southward acceleration if it is slowing down (decelerating).

  • An object can have a negative velocity while maintaining a positive acceleration if it is moving in the negative direction and slowing down.

  • An object can increase in speed while its acceleration is decreasing, provided velocity and acceleration remain in the same direction; as long as acceleration is positive, speed continues to rise.

  • An object can experience zero acceleration and non-zero velocity simultaneously if it maintains a constant velocity in a straight line.

Unit Conversions, Calculations, and Time Arithmetic

Simplifying time calculations requires converting minutes and seconds into pure seconds or decimal minutes.

  • Converting 2 min 30 s2\,\text{min } 30\,\text{s} into decimal form yields 2.5 min2.5\,\text{min}.

  • Summing 1 min 25 s1\,\text{min } 25\,\text{s} and 32 min 55 s32\,\text{min } 55\,\text{s} gives 34 min 20 s34\,\text{min } 20\,\text{s}, which equals 34.33 min34.33\,\text{min} in decimal form.

  • Converting 12.66 min12.66\,\text{min} to minutes and seconds yields 12 min 40 s12\,\text{min } 40\,\text{s} (calculated as 0.66×60 s=39.6 s≈40 s0.66 \times 60\,\text{s} = 39.6\,\text{s} \approx 40\,\text{s}).

  • Converting seconds to decimal minutes:

    • 6 s=0.1 min6\,\text{s} = 0.1\,\text{min}

    • 15 s=0.25 min15\,\text{s} = 0.25\,\text{min}

    • 30 s=0.5 min30\,\text{s} = 0.5\,\text{min}

    • 36 s=0.6 min36\,\text{s} = 0.6\,\text{min}

    • 45 s=0.75 min45\,\text{s} = 0.75\,\text{min}

Standard unit conversions for time and length include:

  • 14 years=168 months14\,\text{years} = 168\,\text{months}

  • 14 years=5113.5 days14\,\text{years} = 5113.5\,\text{days}

  • 3 minutes=180 seconds3\,\text{minutes} = 180\,\text{seconds}

  • 200 seconds=3 min 20 s200\,\text{seconds} = 3\,\text{min } 20\,\text{s}

  • 2 days=2880 minutes2\,\text{days} = 2880\,\text{minutes}

  • 1000 minutes=16.67 hours1000\,\text{minutes} = 16.67\,\text{hours}

  • Length ordering from shortest to longest: 3.3 cm3.3\,\text{cm} (0.033 m0.033\,\text{m}), 400 mm400\,\text{mm} (0.4 m0.4\,\text{m}), 170 m170\,\text{m}, 22 km22\,\text{km} (22000 m22000\,\text{m}).

  • Metric conversions:

    • 3 km=3000 m3\,\text{km} = 3000\,\text{m}

    • 1.5 m=150 cm1.5\,\text{m} = 150\,\text{cm}

    • 110 cm=1.1 m110\,\text{cm} = 1.1\,\text{m}

    • 35.7 cm=357 mm35.7\,\text{cm} = 357\,\text{mm}

Basic kinematic scenarios are calculated using standard algebraic relationships:

  • Average speed of a bicycle traveling 40 miles40\,\text{miles} in 2 hours2\,\text{hours}: v=40 miles2 h=20 mphv = \frac{40\,\text{miles}}{2\,\text{h}} = 20\,\text{mph}.

  • Distance traveled by the bicycle in 7 hours7\,\text{hours} at 20 mph20\,\text{mph}: d=20 mph×7 h=140 milesd = 20\,\text{mph} \times 7\,\text{h} = 140\,\text{miles}.

  • Time required to walk 15 km15\,\text{km} at a speed of 5 km/h5\,\text{km/h}: t=15 km5 km/h=3 hourst = \frac{15\,\text{km}}{5\,\text{km/h}} = 3\,\text{hours}.

  • Distance covered by a snail moving at 0.03 miles/hour0.03\,\text{miles/hour} over 20 hours20\,\text{hours}: d=0.03 mph×20 h=0.6 milesd = 0.03\,\text{mph} \times 20\,\text{h} = 0.6\,\text{miles}.

  • Time used by a biker moving at 22 miles/hour22\,\text{miles/hour} to complete 40 miles40\,\text{miles}: t=40 miles22 mph≈1.82 hourst = \frac{40\,\text{miles}}{22\,\text{mph}} \approx 1.82\,\text{hours}.

  • Height of a tree climbed by a squirrel running upward at 1.82 m/s1.82\,\text{m/s} for 6.5 seconds6.5\,\text{seconds}: h=1.82 m/s×6.5 s=11.83 mh = 1.82\,\text{m/s} \times 6.5\,\text{s} = 11.83\,\text{m}.

  • A person lost in a desert walking in a straight line at 2 km/h2\,\text{km/h} for 4 hours4\,\text{hours} establishes a search radius r=2 km/h×4 h=8 kmr = 2\,\text{km/h} \times 4\,\text{h} = 8\,\text{km}. The search party must search a circular area A=π⋅r2=π⋅(8 km)2≈201.06 km2A = \pi \cdot r^2 = \pi \cdot (8\,\text{km})^2 \approx 201.06\,\text{km}^2

  • Running at a pace of 7.15 minutes/mile7.15\,\text{minutes/mile} for 6.1 miles6.1\,\text{miles} (10 km10\,\text{km}) requires a total time of t=7.15 min/mile×6.1 miles=43.615≈43.62 minutest = 7.15\,\text{min/mile} \times 6.1\,\text{miles} = 43.615 \approx 43.62\,\text{minutes}.

  • A car trip covering 60 miles60\,\text{miles} in 1 h 20 minutes1\,\text{h } 20\,\text{minutes} (1.333 hours1.333\,\text{hours}) reflects an average speed v=60 miles1.333 h=45 mphv = \frac{60\,\text{miles}}{1.333\,\text{h}} = 45\,\text{mph}. Time to cover 200 miles200\,\text{miles} at this constant rate: t=200 miles45 mph=4.444≈4.45 hourst = \frac{200\,\text{miles}}{45\,\text{mph}} = 4.444 \approx 4.45\,\text{hours}.

Graphical Analysis of Motion

Motion can be analyzed visually through position-time (p/tp/t), velocity-time (v/tv/t), and acceleration-time (a/ta/t) graphs.

A position-time (p/tp/t) graph plots time in seconds (s\text{s}) on the horizontal xx-axis and position in meters (m\text{m}) on the vertical yy-axis. It displays how position changes over time and represents the relationship between distance, velocity, and time. The linear slope equation is given by p2=v⋅t+p1p_2 = v \cdot t + p_1, where the slope of a p/tp/t graph equals velocity (m/s\text{m/s}). A horizontal line with a slope of 00 signifies that the object is stationary. A curved line indicates changing velocity (acceleration), as varying amounts of distance are covered per unit of time.

Data collected from Alex's bicycle ride provides an example of p/tp/t graph analysis:

  • Data Set: Position values at specified times are 0 m0\,\text{m} at 0 s0\,\text{s}, 105 m105\,\text{m} at 30 s30\,\text{s}, 270 m270\,\text{m} at 60 s60\,\text{s}, 400 m400\,\text{m} at 90 s90\,\text{s}, 540 m540\,\text{m} at 120 s120\,\text{s}, and 600 m600\,\text{m} at 150 s150\,\text{s}.

  • Average Speed for Entire Trip: vavg=600 m−0 m150 s−0 s=4 m/sv_{avg} = \frac{600\,\text{m} - 0\,\text{m}}{150\,\text{s} - 0\,\text{s}} = 4\,\text{m/s}.

  • Speed Between 60 s60\,\text{s} and 90 s90\,\text{s}: v=400 m−270 m90 s−60 s=130 m30 s≈4.33 m/sv = \frac{400\,\text{m} - 270\,\text{m}}{90\,\text{s} - 60\,\text{s}} = \frac{130\,\text{m}}{30\,\text{s}} \approx 4.33\,\text{m/s}.

  • Interval Speeds:

    • 0 s to 30 s:v=105 m30 s=3.5 m/s0\,\text{s} \text{ to } 30\,\text{s}: v = \frac{105\,\text{m}}{30\,\text{s}} = 3.5\,\text{m/s}

    • 30 s to 60 s:v=270 m−105 m30 s=165 m30 s=5.5 m/s30\,\text{s} \text{ to } 60\,\text{s}: v = \frac{270\,\text{m} - 105\,\text{m}}{30\,\text{s}} = \frac{165\,\text{m}}{30\,\text{s}} = 5.5\,\text{m/s}

    • 60 s to 90 s:v=400 m−270 m30 s=130 m30 s≈4.33 m/s60\,\text{s} \text{ to } 90\,\text{s}: v = \frac{400\,\text{m} - 270\,\text{m}}{30\,\text{s}} = \frac{130\,\text{m}}{30\,\text{s}} \approx 4.33\,\text{m/s}

    • 90 s to 120 s:v=540 m−400 m30 s=140 m30 s≈4.67 m/s90\,\text{s} \text{ to } 120\,\text{s}: v = \frac{540\,\text{m} - 400\,\text{m}}{30\,\text{s}} = \frac{140\,\text{m}}{30\,\text{s}} \approx 4.67\,\text{m/s}

    • 120 s to 150 s:v=600 m−540 m30 s=60 m30 s=2 m/s120\,\text{s} \text{ to } 150\,\text{s}: v = \frac{600\,\text{m} - 540\,\text{m}}{30\,\text{s}} = \frac{60\,\text{m}}{30\,\text{s}} = 2\,\text{m/s}

    • The interval with the greatest speed is between 30 s30\,\text{s} and 60 s60\,\text{s}, with a speed of 5.5 m/s5.5\,\text{m/s}.

A velocity-time (v/tv/t) graph plots time in seconds (s\text{s}) on the horizontal xx-axis and velocity in meters per second (m/s\text{m/s}) on the vertical yy-axis. Calculating the slope of a v/tv/t graph gives acceleration (m/s2\text{m/s}^2). A slope equal to 00 represents constant velocity, meaning speed and direction remain unchanged over time.

Data recorded from Jenny's GPS watch during a walk illustrates v/tv/t graph analysis:

  • Data Set: Speed values at specified times are 0 m/s0\,\text{m/s} at 0 s0\,\text{s}, 3 m/s3\,\text{m/s} at 30 s30\,\text{s}, 5 m/s5\,\text{m/s} at 60 s60\,\text{s}, 5 m/s5\,\text{m/s} at 90 s90\,\text{s}, 2 m/s2\,\text{m/s} at 120 s120\,\text{s}, and 0 m/s0\,\text{m/s} at 150 s150\,\text{s}.

  • Behavior Between 0 s0\,\text{s} and 30 s30\,\text{s}: Speed increases linearly from 0 m/s0\,\text{m/s} to 3 m/s3\,\text{m/s}, representing uniform acceleration.

  • Interpolation Assumption: Connecting data points assumes constant acceleration across intervals.

  • Speed at 20 s20\,\text{s}: Assuming constant acceleration a=3 m/s30 s=0.1 m/s2a = \frac{3\,\text{m/s}}{30\,\text{s}} = 0.1\,\text{m/s}^2, the speed at t=20 st = 20\,\text{s} is v=0.1 m/s2×20 s=2 m/sv = 0.1\,\text{m/s}^2 \times 20\,\text{s} = 2\,\text{m/s}.

  • Behavior Between 60 s60\,\text{s} and 90 s90\,\text{s}: Speed stays constant at 5 m/s5\,\text{m/s}, represented on the graph as a horizontal line with a slope of 00.

  • Behavior Between 90 s90\,\text{s} and 120 s120\,\text{s}: The graph line slants downward from 5 m/s5\,\text{m/s} to 2 m/s2\,\text{m/s}, representing negative acceleration (deceleration).

In acceleration-time (a/ta/t) analysis, acceleration is modeled as either constant or zero. When acceleration is constant, velocity changes along a straight sloped line on a v/tv/t graph. When acceleration is zero, velocity remains completely constant.

Advanced Kinematics Word Problems

Problem 10: A group of students runs away from physics class with an average velocity of 0.98 m/s0.98\,\text{m/s} to the East for 34 minutes34\,\text{minutes}. To find their final location:

  • Time Conversion: t=34 minutes×60 s/min=2040 secondst = 34\,\text{minutes} \times 60\,\text{s/min} = 2040\,\text{seconds}.

  • Calculation: d=v⋅t=0.98 m/s×2040 s=1999.2 m Eastd = v \cdot t = 0.98\,\text{m/s} \times 2040\,\text{s} = 1999.2\,\text{m} \text{ East}.

Problem 11: A polar bear stalks a caribou for 9.5 minutes9.5\,\text{minutes} before the caribou escapes. The polar bear maintains an average velocity of 1.2 m/s1.2\,\text{m/s} towards the North. To find the bear's displacement:

  • Time Conversion: t=9.5 minutes×60 s/min=570 secondst = 9.5\,\text{minutes} \times 60\,\text{s/min} = 570\,\text{seconds}.

  • Calculation: d=v⋅t=1.2 m/s×570 s=684 m Northd = v \cdot t = 1.2\,\text{m/s} \times 570\,\text{s} = 684\,\text{m} \text{ North}.

Problem 12: Two friends take a road trip to the beach, traveling East.

  • Part A: They travel at 48 km/h48\,\text{km/h} towards a beach 144 km144\,\text{km} to the East. The time taken is t1=144 km48 km/h=3 hourst_1 = \frac{144\,\text{km}}{48\,\text{km/h}} = 3\,\text{hours}.

  • Part B: One week later, they travel at 56 km/h56\,\text{km/h} to the East along the same route. The time taken is t2=144 km56 km/h≈2.5714 hourst_2 = \frac{144\,\text{km}}{56\,\text{km/h}} \approx 2.5714\,\text{hours} (2 hours 34.29 minutes2\,\text{hours } 34.29\,\text{minutes}). The time saved is Δt=t1−t2=3 h−2.5714 h=0.4286 hours≈25.71 minutes\Delta t = t_1 - t_2 = 3\,\text{h} - 2.5714\,\text{h} = 0.4286\,\text{hours} \approx 25.71\,\text{minutes}.

Problem 13: A car traveling at 9 m/s9\,\text{m/s} comes to a complete stop at a red light with an acceleration of −4.1 m/s2-4.1\,\text{m/s}^2. To find the stopping time:

  • Formula: a=vf−vit  ⟹  t=vf−viaa = \frac{v_f - v_i}{t} \implies t = \frac{v_f - v_i}{a}.

  • Calculation: t=0 m/s−9 m/s−4.1 m/s2=−9−4.1≈2.20 secondst = \frac{0\,\text{m/s} - 9\,\text{m/s}}{-4.1\,\text{m/s}^2} = \frac{-9}{-4.1} \approx 2.20\,\text{seconds}.

Problem 14: A train backing up at −1.2 m/s-1.2\,\text{m/s} hits an incline and accelerates over 25 minutes25\,\text{minutes}, reaching a final velocity of −6.5 m/s-6.5\,\text{m/s}. To find the acceleration:

  • Time Conversion: t=25 minutes×60 s/min=1500 secondst = 25\,\text{minutes} \times 60\,\text{s/min} = 1500\,\text{seconds}.

  • Calculation: a=vf−vit=−6.5 m/s−(−1.2 m/s)1500 s=−5.3 m/s1500 s≈−0.00353 m/s2=−3.53×10−3 m/s2a = \frac{v_f - v_i}{t} = \frac{-6.5\,\text{m/s} - (-1.2\,\text{m/s})}{1500\,\text{s}} = \frac{-5.3\,\text{m/s}}{1500\,\text{s}} \approx -0.00353\,\text{m/s}^2 = -3.53 \times 10^{-3}\,\text{m/s}^2

Problem 15: Two cars approach each other, both moving at a speed of 95 km/h95\,\text{km/h}, initially separated by 8.5 km8.5\,\text{km}. To find when they reach each other:

  • Relative Velocity: vrel=95 km/h+95 km/h=190 km/hv_{rel} = 95\,\text{km/h} + 95\,\text{km/h} = 190\,\text{km/h}.

  • Calculation: t=dvrel=8.5 km190 km/h≈0.04474 hours≈2.68 minutes≈161.05 secondst = \frac{d}{v_{rel}} = \frac{8.5\,\text{km}}{190\,\text{km/h}} \approx 0.04474\,\text{hours} \approx 2.68\,\text{minutes} \approx 161.05\,\text{seconds}.

Problem 16: A sports car accelerates from rest (0 km/h0\,\text{km/h}) to 95 km/h95\,\text{km/h} in 6.2 seconds6.2\,\text{seconds}. The average acceleration in km/h⋅s\text{km/h}\cdot\text{s} is:

  • Calculation: a=vf−vit=95 km/h−0 km/h6.2 s≈15.32 km/h⋅sa = \frac{v_f - v_i}{t} = \frac{95\,\text{km/h} - 0\,\text{km/h}}{6.2\,\text{s}} \approx 15.32\,\text{km/h}\cdot\text{s}.

Problem 17: A sprinter accelerates from rest to 10 m/s10\,\text{m/s} in 1.35 seconds1.35\,\text{seconds}.

  • Part A (Acceleration in m/s2\text{m/s}^2): a=10 m/s−0 m/s1.35 s≈7.41 m/s2a = \frac{10\,\text{m/s} - 0\,\text{m/s}}{1.35\,\text{s}} \approx 7.41\,\text{m/s}^2

  • Part B (Acceleration in km/h2\text{km/h}^2):

    • Convert Velocity: vf=10 m/s=36 km/hv_f = 10\,\text{m/s} = 36\,\text{km/h}.

    • Convert Time: t=1.35 seconds=1.353600 hours=0.000375 hourst = 1.35\,\text{seconds} = \frac{1.35}{3600}\,\text{hours} = 0.000375\,\text{hours}.

    • Calculation: a=36 km/h0.000375 h=96000 km/h2a = \frac{36\,\text{km/h}}{0.000375\,\text{h}} = 96000\,\text{km/h}^2.

Problem 18: A hot air balloon ascends at 1.7 m/s1.7\,\text{m/s}. The captain releases a gust of hot air, causing an acceleration of 4.7×10−3 m/s24.7 \times 10^{-3}\,\text{m/s}^2 for 5 minutes5\,\text{minutes}.

  • Time Conversion: t=5 minutes×60 s/min=300 secondst = 5\,\text{minutes} \times 60\,\text{s/min} = 300\,\text{seconds}.

  • Part A (Change of Velocity Δv\Delta v): Δv=a⋅t=(4.7×10−3 m/s2)×300 s=1.41 m/s\Delta v = a \cdot t = (4.7 \times 10^{-3}\,\text{m/s}^2) \times 300\,\text{s} = 1.41\,\text{m/s}.

  • Part B (Final Velocity vfv_f): vf=vi+Δv=1.7 m/s+1.41 m/s=3.11 m/sv_f = v_i + \Delta v = 1.7\,\text{m/s} + 1.41\,\text{m/s} = 3.11\,\text{m/s} .