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How do position-time and velocity-time graphs encode motion, and what do their slopes and areas tell you?

Interpret and sketch position-time and velocity-time graphs, reading slope as velocity or acceleration and area under a velocity-time graph as displacement (MA STE Introductory Physics, Motion and Forces).

A standard-level answer on motion graphs for the Massachusetts High School Introductory Physics MCAS: how to read position-time and velocity-time graphs, what slope and area mean, and how to sketch the motion they describe.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. Position-time graphs
  3. Velocity-time graphs
  4. Sketching motion
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What this topic is asking

Graphs are how the Massachusetts Introductory Physics MCAS shows motion without words, and reading them is one of the most-tested skills. You must interpret and sketch position-time and velocity-time graphs: what the slope means on each, what the area under a velocity-time graph means, and how to translate a graph into a description of the motion. This is the practice of analyzing and interpreting data applied to kinematics.

Position-time graphs

A position-time graph plots where an object is (vertical axis) against time (horizontal axis). The key to reading it is the slope:

slope=change in positionchange in time=ΔdΔt=velocity\text{slope} = \frac{\text{change in position}}{\text{change in time}} = \frac{\Delta d}{\Delta t} = \text{velocity}

So the slope of a position-time graph is the velocity. This gives you a quick visual language:

  • A steep line means a large velocity (covering a lot of distance in little time).
  • A shallow line means a small velocity.
  • A horizontal line (zero slope) means the object is at rest: its position is not changing.
  • A line sloping downward means the object is moving in the negative direction (back toward the start).
  • A curved line means the velocity is changing, so the object is accelerating.

Velocity-time graphs

A velocity-time graph plots velocity (vertical axis) against time (horizontal axis). It carries two pieces of information, and the MCAS uses both.

The slope is the acceleration. Since acceleration is the change in velocity over time,

slope=ΔvΔt=acceleration.\text{slope} = \frac{\Delta v}{\Delta t} = \text{acceleration}.

A horizontal line means constant velocity (zero acceleration). A straight line sloping up means constant positive acceleration (speeding up); sloping down means slowing down.

The area under the line is the displacement. Velocity times time gives distance, and on the graph that product is the area between the line and the time axis. For a rectangle (constant velocity) the area is base times height; for a triangle (uniform acceleration from rest) it is 12\tfrac{1}{2} base times height. Splitting a graph into rectangles and triangles and adding their areas is the standard way to find total distance.

Sketching motion

You may be asked to sketch a graph from a description, or describe the motion from a graph. Translate each phase:

  • "moves at constant speed" then "speeds up" then "stops" becomes, on a velocity-time graph, a horizontal line, then a line sloping up, then a vertical drop to zero (or a line sloping back down to zero).
  • "stays still" becomes a horizontal line on a position-time graph and a line on the time axis (zero) on a velocity-time graph.

Try this

Q1. On a velocity-time graph, what does a horizontal line tell you? [1]

  • Cue. The velocity is constant, so the acceleration is zero.

Q2. A position-time graph is a straight line sloping downward. Describe the motion. [2]

  • Cue. The object moves at constant velocity in the negative direction (back toward the start), because the slope is constant and negative.

Exam-style practice questions

Practice questions written in the style of MA DESE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

MA Physics MCAS (style)3 marksA velocity-time graph shows an object moving at a constant 8.08.0 m/s for 5.05.0 s, then slowing steadily to rest over the next 4.04.0 s. (a) Calculate the acceleration during the last 4.04.0 s. (b) Calculate the total distance traveled. (c) Describe the motion in the first 5.05.0 s.
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A 3-point item on analyzing and interpreting data from a velocity-time graph.

(a) 1 point: acceleration is the slope =0−8.04.0=−2.0= \dfrac{0 - 8.0}{4.0} = -2.0 m/s squared (slowing down).
(b) 1 point: distance is the area under the graph. The first part is a rectangle 8.0×5.0=408.0 \times 5.0 = 40 m; the second is a triangle 12×4.0×8.0=16\tfrac{1}{2} \times 4.0 \times 8.0 = 16 m. Total =40+16=56= 40 + 16 = 56 m.
(c) 1 point: in the first 5.05.0 s the object moves at constant velocity (zero acceleration), shown by the horizontal line. Markers reward "constant velocity" or "no acceleration."

MA Physics MCAS (style)2 marksOn a position-time graph, line A is steep and straight, line B is shallow and straight, and line C is horizontal. (a) Which line shows the fastest motion? (b) What does the horizontal line C represent?
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A 2-point item on reading slope as velocity on a position-time graph.

(a) 1 point: line A shows the fastest motion, because the steeper the slope of a position-time graph, the greater the velocity.
(b) 1 point: line C is horizontal, so the position is not changing with time, which means the object is stationary (at rest). Markers reward "at rest" or "not moving."

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