Velocity–Time Graphs and Motion Problems

Objective: Calculate acceleration from a velocity graph.

On a one-dimensional velocity–time graph, slope Δv/Δt is average acceleration over a straight segment. A horizontal segment has zero acceleration but may have nonzero velocity and ongoing motion. A line with negative slope represents acceleration in the negative direction, not automatically slowing down; the speed trend depends on velocity sign. For velocity increasing from −2 m/s to +6 m/s in 4 s, acceleration is +2 m/s², and the object reverses when velocity passes through zero. Units of graph slope are m/s² when axes are m/s and s.

Objective: Find displacement and distance from simple graph areas.

The signed area between a one-dimensional velocity–time curve and the time axis gives displacement over an interval. A rectangle at +3 m/s for 4 s contributes +12 m. An equal rectangle at −3 m/s for 4 s contributes −12 m, so the combined displacement is zero but distance is 24 m. Total distance in straight-line motion is the sum of the absolute areas over intervals of constant sign, not the absolute value of their sum. A triangular region under a linearly changing velocity can be calculated with one-half base times height. Area units are (m/s)×s = m.

Objective: Match descriptions to position and velocity plots and catch common errors.

For a single one-dimensional motion, position–time slope equals velocity, while velocity–time slope equals acceleration and signed velocity–time area equals displacement. Consistency across graphs can be tested: a constant positive velocity requires a rising straight position–time line, whereas constant positive acceleration produces a rising straight velocity–time line and a curved position–time graph. A negative position is not the same as negative velocity, and a negative velocity is not necessarily decreasing speed. When a velocity crosses zero, position may have a turning point. State reference frame, axis direction, units, and time interval before comparing graphs.