Kinematics: Position-Time and Velocity-Time Graphing Guide

Principles of Uniform Motion and Position vs. Time Graphs

  • Uniform motion refers to movement occurring at a constant speed along a path.
  • In a one-dimensional visualization of multi-object motion:
    • Objects starting at a position of 0 m0\,\text{m} and traveling to a finish line at 60 m60\,\text{m} can be represented on a single vertical position axis (xx).
    • Stacking objects vertically represents their motion along the vertical axis while shifting the graph horizontally simulates the progression of time (tt).
    • The resulting position versus time graph produces straight lines whose steepness directly corresponds to the relative speeds of the objects.
    • A steeper slope indicates a faster speed; for instance, a fast blue car yields the steepest line, while a slow purple car yields the least steep line.
  • Slope in graph analysis is defined as the rate of change of the vertical axis relative to the horizontal axis:
    • On a position versus time graph, the vertical axis represents position (xx) and the horizontal axis represents time (tt).
    • The rate of change of an object's position with respect to time equals its velocity (vv).
    • Therefore, the slope of a position versus time graph is equivalent to the velocity of the moving object.
  • Mathematical formula for calculating slope:
    • Slope=RiseRun\text{Slope} = \frac{\text{Rise}}{\text{Run}}
    • Rise\text{Rise} represents the vertical change (change in position, Δx\Delta x).
    • Run\text{Run} represents the horizontal change (change in time, Δt\Delta t).
    • Velocity=RiseRun\text{Velocity} = \frac{\text{Rise}}{\text{Run}}

Velocity Calculations from Position vs. Time Graphs

  • Step-by-step procedure for determining velocity from a position versus time graph:

    • Identify two distinct, known coordinate points (t1,x1)(t_1, x_1) and (t2,x2)(t_2, x_2) along the line on the graph.
    • Calculate the horizontal movement (Run\text{Run}) by moving from the initial time t1t_1 to the final time t2t_2: Run=t2−t1\text{Run} = t_2 - t_1.
    • Calculate the vertical movement (Rise\text{Rise}) by moving from the initial position x1x_1 to the final position x2x_2: Rise=x2−x1\text{Rise} = x_2 - x_1.
    • Divide the Rise\text{Rise} by the Run\text{Run} to determine velocity: v=RiseRunv = \frac{\text{Rise}}{\text{Run}}.

  • Calculation Example: Velocity of the Blue Object

    • Selected Point 1: (0 s,50 m)(0\,\text{s}, 50\,\text{m})
    • Selected Point 2: (1.8 s,10 m)(1.8\,\text{s}, 10\,\text{m})
    • Calculation of Rise\text{Rise}: 10 m−50 m=−40 m10\,\text{m} - 50\,\text{m} = -40\,\text{m}
    • Calculation of Run\text{Run}: 1.8 s−0 s=1.8 s1.8\,\text{s} - 0\,\text{s} = 1.8\,\text{s}
    • Velocity calculation: v=−40 m1.8 s=−22.22 m/sv = \frac{-40\,\text{m}}{1.8\,\text{s}} = -22.22\,\text{m/s}
    • Interpretation: The negative velocity value indicates that the blue object is moving downward or in the negative direction.

  • Calculation Example: Velocity of the Orange Object

    • Selected Point 1: (0 s,10 m)(0\,\text{s}, 10\,\text{m})
    • Selected Point 2: (1.8 s,40 m)(1.8\,\text{s}, 40\,\text{m})
    • Calculation of Rise\text{Rise}: 40 m−10 m=30 m40\,\text{m} - 10\,\text{m} = 30\,\text{m}
    • Calculation of Run\text{Run}: 1.8 s−0 s=1.8 s1.8\,\text{s} - 0\,\text{s} = 1.8\,\text{s}
    • Velocity calculation: v=30 m1.8 s=16.67 m/sv = \frac{30\,\text{m}}{1.8\,\text{s}} = 16.67\,\text{m/s}
    • Interpretation: The positive velocity value indicates motion in the positive direction.

Constructing Velocity vs. Time Graphs from Linear Position Graphs

  • Fundamental conversion rules:

    • The numerical value of the slope at any segment on a position versus time graph determines the vertical coordinate (velocity value) on a velocity versus time graph.
    • A straight line on a position versus time graph possesses a constant slope.
    • A constant slope on a position versus time graph translates to a constant velocity value, represented as a flat horizontal line on a velocity versus time graph.

  • Example 1: Blue Car Graph Conversion

    • Motion duration: t=0 st = 0\,\text{s} to t=1 st = 1\,\text{s} (cuts off at 1 s1\,\text{s}).
    • Graph points: (0 s,0 m)(0\,\text{s}, 0\,\text{m}) to (1 s,60 m)(1\,\text{s}, 60\,\text{m}).
    • Rise=60 m−0 m=60 m\text{Rise} = 60\,\text{m} - 0\,\text{m} = 60\,\text{m}
    • Run=1 s−0 s=1 s\text{Run} = 1\,\text{s} - 0\,\text{s} = 1\,\text{s}
    • Slope=60 m1 s=60 m/s\text{Slope} = \frac{60\,\text{m}}{1\,\text{s}} = 60\,\text{m/s}
    • Velocity-Time representation: A horizontal line plotted at v=60 m/sv = 60\,\text{m/s} extending from t=0 st = 0\,\text{s} to t=1 st = 1\,\text{s}.

  • Example 2: Purple Car Graph Conversion

    • Motion duration: t=0 st = 0\,\text{s} to t=2 st = 2\,\text{s} (spans full length of graph).
    • Graph points: (0 s,0 m)(0\,\text{s}, 0\,\text{m}) to (2 s,60 m)(2\,\text{s}, 60\,\text{m}).
    • Rise=60 m−0 m=60 m\text{Rise} = 60\,\text{m} - 0\,\text{m} = 60\,\text{m}
    • Run=2 s−0 s=2 s\text{Run} = 2\,\text{s} - 0\,\text{s} = 2\,\text{s}
    • Slope=60 m2 s=30 m/s\text{Slope} = \frac{60\,\text{m}}{2\,\text{s}} = 30\,\text{m/s}
    • Velocity-Time representation: A horizontal line plotted at v=30 m/sv = 30\,\text{m/s} extending from t=0 st = 0\,\text{s} to t=2 st = 2\,\text{s}.

  • Example 3: Yellow/Orange Car Graph Conversion

    • Motion duration: t=0 st = 0\,\text{s} to t=1.5 st = 1.5\,\text{s} (cuts off at 1.5 s1.5\,\text{s}).
    • Graph points: (0 s,0 m)(0\,\text{s}, 0\,\text{m}) to (1.5 s,60 m)(1.5\,\text{s}, 60\,\text{m}).
    • Rise=60 m−0 m=60 m\text{Rise} = 60\,\text{m} - 0\,\text{m} = 60\,\text{m}
    • Run=1.5 s−0 s=1.5 s\text{Run} = 1.5\,\text{s} - 0\,\text{s} = 1.5\,\text{s}
    • Slope=60 m1.5 s=40 m/s\text{Slope} = \frac{60\,\text{m}}{1.5\,\text{s}} = 40\,\text{m/s}
    • Velocity-Time representation: A horizontal line plotted at v=40 m/sv = 40\,\text{m/s} extending from t=0 st = 0\,\text{s} to t=1.5 st = 1.5\,\text{s}.

Constructing Velocity vs. Time Graphs from Piecewise Position Graphs

  • When an object's position versus time graph exhibits multiple slope changes (complex motion):

    • Partition the time axis into distinct time intervals wherever the slope (velocity) changes.
    • Calculate the velocity (Slope=RiseRun\text{Slope} = \frac{\text{Rise}}{\text{Run}}) independently for each time section.
    • Plot horizontal constant velocity segments across each designated time interval on the velocity versus time graph.

  • Case Study: Three-Segment Piecewise Motion Analysis

    • Section 1 (t=0 st = 0\,\text{s} to t=0.6 st = 0.6\,\text{s}):

    • Coordinate points: (0 s,0 m)(0\,\text{s}, 0\,\text{m}) to (0.6 s,30 m)(0.6\,\text{s}, 30\,\text{m}).

    • Rise=30 m−0 m=30 m\text{Rise} = 30\,\text{m} - 0\,\text{m} = 30\,\text{m}

    • Run=0.6 s−0 s=0.6 s\text{Run} = 0.6\,\text{s} - 0\,\text{s} = 0.6\,\text{s}

    • v1=30 m0.6 s=50 m/sv_1 = \frac{30\,\text{m}}{0.6\,\text{s}} = 50\,\text{m/s}

    • Plot: Horizontal line at v=50 m/sv = 50\,\text{m/s} from t=0 st = 0\,\text{s} to t=0.6 st = 0.6\,\text{s}.

    • Section 2 (t=0.6 st = 0.6\,\text{s} to t=1.4 st = 1.4\,\text{s}):

    • Coordinate points: (0.6 s,30 m)(0.6\,\text{s}, 30\,\text{m}) to (1.4 s,40 m)(1.4\,\text{s}, 40\,\text{m}).

    • Rise=40 m−30 m=10 m\text{Rise} = 40\,\text{m} - 30\,\text{m} = 10\,\text{m}

    • Run=1.4 s−0.6 s=0.8 s\text{Run} = 1.4\,\text{s} - 0.6\,\text{s} = 0.8\,\text{s}

    • v2=10 m0.8 s=12.5 m/sv_2 = \frac{10\,\text{m}}{0.8\,\text{s}} = 12.5\,\text{m/s}

    • Plot: Horizontal line at v=12.5 m/sv = 12.5\,\text{m/s} from t=0.6 st = 0.6\,\text{s} to t=1.4 st = 1.4\,\text{s}.

    • Section 3 (t=1.4 st = 1.4\,\text{s} to t=1.8 st = 1.8\,\text{s}):

    • Coordinate points: (1.4 s,40 m)(1.4\,\text{s}, 40\,\text{m}) to (1.8 s,30 m)(1.8\,\text{s}, 30\,\text{m}).

    • Rise=30 m−40 m=−10 m\text{Rise} = 30\,\text{m} - 40\,\text{m} = -10\,\text{m}

    • Run=1.8 s−1.4 s=0.4 s\text{Run} = 1.8\,\text{s} - 1.4\,\text{s} = 0.4\,\text{s}

    • v3=−10 m0.4 s=−25 m/sv_3 = \frac{-10\,\text{m}}{0.4\,\text{s}} = -25\,\text{m/s}

    • Plot: Horizontal line at v=−25 m/sv = -25\,\text{m/s} from t=1.4 st = 1.4\,\text{s} to t=1.8 st = 1.8\,\text{s}.

Determining Displacement from Velocity vs. Time Graphs

  • Key Concept: The displacement (Δx\Delta x) of an object equals the area enclosed between the graph line and the zero velocity line on a velocity versus time graph.

  • Identifying the Zero Velocity Line:

    • The horizontal line on the velocity axis where velocity equals 0 m/s0\,\text{m/s} (the time axis).

  • Area Calculation Method:

    • For rectangular regions formed by constant velocities:
    • Displacement=Area=Width×Height\text{Displacement} = \text{Area} = \text{Width} \times \text{Height}
    • Width\text{Width} corresponds to elapsed time (Δt\Delta t).
    • Height\text{Height} corresponds to velocity (vv).

  • Verification using the 60-Meter Race Objects:

    • Blue Car Displacement:
    • Width=1 s\text{Width} = 1\,\text{s}
    • Height=60 m/s\text{Height} = 60\,\text{m/s}
    • Displacement=1 s×60 m/s=60 m\text{Displacement} = 1\,\text{s} \times 60\,\text{m/s} = 60\,\text{m}
    • Orange/Yellow Car Displacement:
    • Width=1.5 s\text{Width} = 1.5\,\text{s}
    • Height=40 m/s\text{Height} = 40\,\text{m/s}
    • Displacement=1.5 s×40 m/s=60 m\text{Displacement} = 1.5\,\text{s} \times 40\,\text{m/s} = 60\,\text{m}
    • Purple Car Displacement:
    • Width=2 s\text{Width} = 2\,\text{s}
    • Height=30 m/s\text{Height} = 30\,\text{m/s}
    • Displacement=2 s×30 m/s=60 m\text{Displacement} = 2\,\text{s} \times 30\,\text{m/s} = 60\,\text{m}

  • Practice Problems for Area and Displacement:

    • Problem 1: Blue Object
    • Given: Graph line from t=0 st = 0\,\text{s} to t=1.6 st = 1.6\,\text{s} at v=15 m/sv = 15\,\text{m/s}.
    • Width=1.6 s−0 s=1.6 s\text{Width} = 1.6\,\text{s} - 0\,\text{s} = 1.6\,\text{s}
    • Height=15 m/s\text{Height} = 15\,\text{m/s}
    • Displacement=1.6 s×15 m/s=24 m\text{Displacement} = 1.6\,\text{s} \times 15\,\text{m/s} = 24\,\text{m}
    • Problem 2: Orange Object
    • Given: Graph line from t=0 st = 0\,\text{s} to t=0.8 st = 0.8\,\text{s} at v=−10 m/sv = -10\,\text{m/s}.
    • Width=0.8 s−0 s=0.8 s\text{Width} = 0.8\,\text{s} - 0\,\text{s} = 0.8\,\text{s}
    • Height=−10 m/s\text{Height} = -10\,\text{m/s}
    • Displacement=0.8 s×(−10 m/s)=−8 m\text{Displacement} = 0.8\,\text{s} \times (-10\,\text{m/s}) = -8\,\text{m}
    • Significance of negative sign: Indicates displacement in the left (negative horizontal) direction.

Constructing Position vs. Time Graphs from Velocity vs. Time Graphs

  • Process and Mechanics:

    • Divide the position versus time graph along the time axis using vertical boundary lines at every point where velocity changes on the velocity versus time graph.
    • Calculate the area (Displacement=Width×Height\text{Displacement} = \text{Width} \times \text{Height}) for each temporal section.
    • Identify the initial starting position (x0x_0):
    • A velocity versus time graph only provides displacement information and cannot inherently indicate starting position.
    • The starting position must be explicitly provided in the problem statement or chosen arbitrarily to fit the available axis bounds.
    • Plot the initial point, then add each section's displacement sequentially to find the subsequent position coordinate and draw connecting straight lines between interval endpoints.

  • Step-by-Step Reconstruction of Four-Section Complex Motion:

    • Initial Condition: Initial position x0=−6 mx_0 = -6\,\text{m} at t=0 st = 0\,\text{s}.

    • Section 1 (t=0 st = 0\,\text{s} to t=0.6 st = 0.6\,\text{s}):

    • Width=0.6 s\text{Width} = 0.6\,\text{s}

    • Height=15 m/s\text{Height} = 15\,\text{m/s}

    • Displacement=0.6 s×15 m/s=9 m\text{Displacement} = 0.6\,\text{s} \times 15\,\text{m/s} = 9\,\text{m}

    • End Position: x(0.6)=−6 m+9 m=3 mx(0.6) = -6\,\text{m} + 9\,\text{m} = 3\,\text{m}

    • Graph line: Connect (0 s,−6 m)(0\,\text{s}, -6\,\text{m}) to (0.6 s,3 m)(0.6\,\text{s}, 3\,\text{m}).

    • Section 2 (t=0.6 st = 0.6\,\text{s} to t=1.0 st = 1.0\,\text{s}):

    • Width=1.0 s−0.6 s=0.4 s\text{Width} = 1.0\,\text{s} - 0.6\,\text{s} = 0.4\,\text{s}

    • Height=0 m/s\text{Height} = 0\,\text{m/s}

    • Displacement=0.4 s×0 m/s=0 m\text{Displacement} = 0.4\,\text{s} \times 0\,\text{m/s} = 0\,\text{m}

    • End Position: x(1.0)=3 m+0 m=3 mx(1.0) = 3\,\text{m} + 0\,\text{m} = 3\,\text{m}

    • Graph line: Connect (0.6 s,3 m)(0.6\,\text{s}, 3\,\text{m}) to (1.0 s,3 m)(1.0\,\text{s}, 3\,\text{m}) (horizontal flat line indicating zero velocity).

    • Section 3 (t=1.0 st = 1.0\,\text{s} to t=1.4 st = 1.4\,\text{s}):

    • Width=1.4 s−1.0 s=0.4 s\text{Width} = 1.4\,\text{s} - 1.0\,\text{s} = 0.4\,\text{s}

    • Height=5 m/s\text{Height} = 5\,\text{m/s}

    • Displacement=0.4 s×5 m/s=2 m\text{Displacement} = 0.4\,\text{s} \times 5\,\text{m/s} = 2\,\text{m}

    • End Position: x(1.4)=3 m+2 m=5 mx(1.4) = 3\,\text{m} + 2\,\text{m} = 5\,\text{m}

    • Graph line: Connect (1.0 s,3 m)(1.0\,\text{s}, 3\,\text{m}) to (1.4 s,5 m)(1.4\,\text{s}, 5\,\text{m}).

    • Section 4 (t=1.4 st = 1.4\,\text{s} to t=2.0 st = 2.0\,\text{s}):

    • Width=2.0 s−1.4 s=0.6 s\text{Width} = 2.0\,\text{s} - 1.4\,\text{s} = 0.6\,\text{s}

    • Height=−20 m/s\text{Height} = -20\,\text{m/s}

    • Displacement=0.6 s×(−20 m/s)=−12 m\text{Displacement} = 0.6\,\text{s} \times (-20\,\text{m/s}) = -12\,\text{m}

    • End Position: x(2.0)=5 m+(−12 m)=−7 mx(2.0) = 5\,\text{m} + (-12\,\text{m}) = -7\,\text{m}

    • Graph line: Connect (1.4 s,5 m)(1.4\,\text{s}, 5\,\text{m}) to (2.0 s,−7 m)(2.0\,\text{s}, -7\,\text{m}).