Interpret and construct position-time and velocity-time graphs to find velocity, acceleration and displacement for segments of uniform motion

Learning goal: Read velocity from the slope of a position-time graph and acceleration from the slope of a velocity-time graph; recognize rest, constant velocity and constant acceleration on each kind of graph; build the velocity-time graph that corresponds to a given position-time graph and the reverse; use the area under a velocity-time graph to find displacement; and measure the acceleration of a freely falling object.

Graphs are records of motion, not rules to memorize. A motion graph puts time on the horizontal axis and either position or velocity on the vertical axis. Everything below can be worked out from the definitions in the previous lesson, and you should be able to rebuild each statement rather than recite it. In this course graphs use positive time values and show only uniform segments: rest, constant velocity or constant acceleration, in a straight line.

Position-time graphs. The slope of a position-time graph between two points is rise over run: change in position divided by change in time, which is exactly the definition of average velocity. So slope means velocity. An object at rest has a horizontal line, because its position does not change. An object with constant velocity has a straight line that is not horizontal: the same displacement in every equal interval gives the same slope everywhere. A positive slope means motion in the positive direction; a negative slope means motion in the negative direction; a steeper line means a larger speed. An accelerating object produces a curved line, because its displacement per second keeps changing, so the slope keeps changing. Where a graph curves upward more and more steeply, the object is speeding up in the positive direction.

A worked model: reading velocity from slope. Suppose a walker's position-time graph goes in a straight line from (0 s, 0 m) to (4 s, 8 m), then stays level at 8 m until 6 s, then goes straight back down to (8 s, 0 m). First segment: slope = (8 − 0) m ÷ (4 − 0) s = +2 m/s. Second segment: slope = 0, at rest 8 m from the origin. Third segment: slope = (0 − 8) m ÷ (8 − 6) s = −8 m ÷ 2 s = −4 m/s, moving back toward the origin twice as fast as the walk out. The matching velocity-time graph is three horizontal steps: +2 m/s from 0 to 4 s, 0 from 4 to 6 s, −4 m/s from 6 to 8 s. The walker's distance is 16 m; the displacement is 0 m.

Velocity-time graphs. The slope of a velocity-time graph is change in velocity divided by change in time, which is the definition of average acceleration. So on a velocity-time graph, slope means acceleration. An object with no acceleration, whether at rest or moving with constant velocity, has a horizontal line; at rest the line sits on the time axis at v = 0, while constant velocity is a horizontal line above or below it. Constant acceleration is a straight, sloping line. Read the graph carefully: a horizontal line on a velocity-time graph is constant velocity, but a horizontal line on a position-time graph is rest. Mixing these two is the most common graph error, and it comes from memorizing shapes instead of asking what the vertical axis measures.

Displacement from a velocity-time graph. For an object moving at a constant +6 m/s for 2 s, the displacement is 6 m/s × 2 s = 12 m, which is the area of the rectangle under the line. That idea extends to a sloping line: the area under the velocity-time graph over an interval is the displacement during that interval. For constant acceleration from rest to 6 m/s over 3 s, the region under the line is a triangle of area ½ × 3 s × 6 m/s = 9 m. The same result comes from the average velocity: starting at 0 and ending at 6 m/s with a steady change, the average is 3 m/s, and 3 m/s × 3 s = 9 m. Areas below the time axis (negative velocity) count as negative displacement.

Corresponding graphs. Given any one of the graphs for uniform motion you can build the others. From a position-time graph, measure the slope of each segment to get velocities, then plot them as horizontal steps against time (this works for straight segments; a curved segment gives a sloping velocity line). From a velocity-time graph, the slope of each segment gives acceleration, and the accumulated area gives displacement, which becomes position once the starting position is known. Consider the velocity-time graph in this lesson's diagram: a straight line from (0 s, 0 m/s) to (3 s, 6 m/s), a horizontal line at 6 m/s until 5 s, then a straight line down to (8 s, 0 m/s). Accelerations: (6 − 0) ÷ 3 = +2 m/s², then 0, then (0 − 6) ÷ 3 = −2 m/s². Displacements: 9 m, 12 m and 9 m, total 30 m. The position-time graph therefore curves upward from 0 to 9 m during the first 3 s (positions 0, 1, 4, 9 m at 0, 1, 2, 3 s, since x = t² for this acceleration), runs straight from 9 m to 21 m between 3 and 5 s, and curves to level off at 30 m by 8 s. The acceleration-time graph is three horizontal steps: +2, 0 and −2 m/s².

Writing the story of a graph. A velocity-time graph tells a story if you read it segment by segment: what was the velocity at the start, was it changing, in which direction, and for how long. The diagram's graph could be a delivery cart that starts from rest, speeds up steadily for 3 s, cruises for 2 s, then brakes steadily to a stop in 3 s, ending 30 m from where it began. Telling that story, and drawing the graph from a story, are the two directions of the same skill.

Two objects on one position-time graph. When two objects move along the same line, plot both on one position-time graph using the same origin and clock. Where the two lines cross, the objects are at the same position at the same time: that is where and when they meet. Suppose battery car A starts at the origin and moves at +2 m/s, while car B starts 30 m away and moves toward A at −3 m/s, both released at t = 0. Car A's position is 2t and car B's is 30 − 3t (meters, with t in seconds). Setting them equal, 2t = 30 − 3t, so 5t = 30 and t = 6 s; the meeting point is 2 × 6 = 12 m from the origin. Check on the graph: A's line rises from (0 s, 0 m) to (6 s, 12 m) and B's falls from (0 s, 30 m) to (6 s, 12 m). If one object is released later, its line simply starts later on the time axis; the crossing point still marks the meeting. Traffic engineers and air traffic controllers reason about crossing points in exactly this way.

Accelerating down a ramp from rest. Motion in this course can also be up or down a straight incline. A cart released from rest at the top of a 1.60 m ramp passes marks at 0, 0.10, 0.40 and 0.90 m at 0, 0.5, 1.0 and 1.5 s. The displacements in successive half-seconds are 0.10, 0.30 and 0.50 m: they grow by the same 0.20 m each interval, the signature of constant acceleration, and the position-time graph curves upward. From rest with constant acceleration, v_f = 2d ÷ t: at 1.5 s, v_f = 2 × 0.90 m ÷ 1.5 s = 1.2 m/s, so a = 1.2 m/s ÷ 1.5 s = 0.8 m/s². Use that acceleration to predict the speed at the bottom: the cart reaches 1.60 m at 2.0 s (the pattern continues 0.70 m in the next half-second), so v = a × t = 0.8 m/s² × 2.0 s = 1.6 m/s, which a motion sensor at the bottom of the ramp can check. These are invented values chosen to show the method.

Investigation: the acceleration of a freely falling object. Materials: a small dense ball, a tape measure, a phone or camera recording video at a known frame rate (30 frames per second is common), a plain wall as background, a partner. Steps: mark a release height of 2.00 m above a padded landing spot; record the ball being released from rest; step through the video and count frames from the frame in which the ball is released to the frame in which it lands; repeat five drops; convert the frame count to time by dividing by the frame rate. For constant acceleration from rest, the average velocity is half the final velocity, so v_f = 2d ÷ t and a = v_f ÷ t. Record height, frame counts, times and computed accelerations. With invented values, 19 frames at 30 frames per second gives t = 19 ÷ 30 ≈ 0.633 s, v_f = 2 × 2.00 m ÷ 0.633 s ≈ 6.32 m/s and a = 6.32 m/s ÷ 0.633 s ≈ 10.0 m/s². A fall of 2.00 m at exactly 9.8 m/s² would take about 0.639 s, or 19.2 frames, so a count of 19 is as close as whole frames allow. A result clustered near 9.8 m/s² across the five drops supports the claim that a freely falling object near Earth's surface accelerates at about 9.8 m/s². What the method cannot show: it assumes air resistance is negligible, which is fair for a dense ball over 2 m but not for a feather or a sheet of paper; one frame of uncertainty in the count changes the time by about 0.03 s, so report the spread across drops, not a single value. A stopwatch is not adequate here because reaction time is a large fraction of 0.64 s.

Worked example 1 — from position-time to velocity-time. A position-time graph has three straight segments: from (0 s, 0 m) to (4 s, 8 m); level at 8 m from 4 s to 6 s; from (6 s, 8 m) to (8 s, 0 m). Slopes: segment 1, (8 − 0) ÷ (4 − 0) = +2 m/s; segment 2, 0 m/s; segment 3, (0 − 8) ÷ (8 − 6) = −4 m/s. The velocity-time graph is therefore a horizontal line at +2 m/s from 0 to 4 s, a horizontal line at 0 from 4 to 6 s and a horizontal line at −4 m/s from 6 to 8 s. Acceleration is zero within every segment. The object travels 8 m out and 8 m back: distance 16 m, displacement 0 m, average speed 16 m ÷ 8 s = 2 m/s, average velocity 0 m/s.

Worked example 2 — slope and area on a velocity-time graph. A velocity-time graph rises in a straight line from (0 s, 0 m/s) to (3 s, 6 m/s), stays at 6 m/s until 5 s, then falls in a straight line to (8 s, 0 m/s). Accelerations from slopes: (6 − 0) m/s ÷ 3 s = +2 m/s²; 0; (0 − 6) m/s ÷ 3 s = −2 m/s². Displacements from areas: triangle ½ × 3 s × 6 m/s = 9 m; rectangle 6 m/s × 2 s = 12 m; triangle ½ × 3 s × 6 m/s = 9 m; total 30 m. The object ends 30 m from the start in the positive direction. Check the first segment with average velocity: (0 + 6) ÷ 2 = 3 m/s, and 3 m/s × 3 s = 9 m, agreeing with the area.

Worked example 3 — recognizing rest, constant velocity and acceleration. Three velocity-time graphs are shown to a class. Graph A is a horizontal line at v = 0. Graph B is a horizontal line at v = +3 m/s. Graph C is a straight line from (0 s, 2 m/s) to (4 s, 10 m/s). A shows no motion: velocity is zero throughout. B shows constant velocity: the object moves steadily at 3 m/s in the positive direction with zero acceleration, and its position-time graph is a straight line of slope +3 m/s. C shows constant acceleration of (10 − 2) ÷ 4 = +2 m/s²; its position-time graph is a curve that gets steeper. Note that B is not "at rest" even though its line is flat: flat on a velocity graph means unchanging velocity, not unchanging position.

Worked example 4 — free-fall data. Five invented drops from 2.00 m give frame counts of 19, 19, 20, 19 and 18 at 30 frames per second, or times of 0.633, 0.633, 0.667, 0.633 and 0.600 s. Using a = 2d ÷ t², which is the same as a = (2d ÷ t) ÷ t: for t = 0.633 s, a = 4.00 m ÷ 0.401 s² ≈ 10.0 m/s²; for 0.667 s, a ≈ 9.0 m/s²; for 0.600 s, a ≈ 11.1 m/s². The mean of the five is about 10.0 m/s² with a spread of roughly ±1 m/s², consistent with 9.8 m/s² given that one frame changes the time by 0.03 s. The conclusion states the value with its spread and its condition: for a dense ball over a short drop where air resistance is negligible.

Worked example 5 — when and where do two cars meet? Car A leaves x = 0 at t = 0 moving at +0.5 m/s. Car B is released 2.0 s later from x = 10 m, moving at −0.5 m/s. At t = 2.0 s car A is at 0.5 × 2.0 = 1.0 m, so the gap is 10 − 1.0 = 9.0 m. The gap closes at 0.5 + 0.5 = 1.0 m/s, so it takes 9.0 s more: they meet at t = 11 s. Car A's position then is 0.5 × 11 = 5.5 m; check with B: 10 − 0.5 × (11 − 2.0) = 5.5 m. On a position-time graph, A's line starts at the origin, B's line starts at (2 s, 10 m), and they cross at (11 s, 5.5 m).

Representation task: three matching graphs. Using the velocity-time data in this lesson's diagram (0 to 6 m/s over 0–3 s, 6 m/s from 3–5 s, 6 to 0 m/s over 5–8 s), draw on graph paper the velocity-time graph, the acceleration-time graph (steps of +2, 0 and −2 m/s²) and the position-time graph (points at 0, 1, 4, 9, 15, 21 m for 0–5 s, then curving to 30 m at 8 s). Label every axis with quantity and unit. Beside each segment of the velocity-time graph write its slope with units and its area with units. Write a four-sentence story of an object whose motion matches the graphs, naming when it speeds up, cruises and slows.

Independent performance: measure free-fall acceleration and report. Carry out the free-fall investigation, or analyze a teacher-provided video, and write a laboratory report. Success criteria: the release height is measured and recorded with units; at least five drops are timed by frame counting with the frame rate stated; each drop's time, final velocity (v_f = 2d ÷ t) and acceleration (a = v_f ÷ t) are computed with units; the result is reported as a mean with its spread, not as a single number; the conclusion compares the result with 9.8 m/s² and states the condition under which the comparison is fair (negligible air resistance); the report explains why a stopwatch would have been inadequate; and it sketches the velocity-time graph of one drop, labeling its slope as the measured acceleration and its area as the drop height.