Read and sketch position, velocity and acceleration graphs, using slopes, areas and the sign of acceleration

Learning goal: Interpret position vs. time, velocity vs. time and acceleration vs. time graphs for motion that happens in stages and can have negative values; find instantaneous velocity from a tangent, acceleration and displacement from a velocity graph, and change in velocity from an acceleration graph; sketch the other two graphs when one is given; explain the sign of acceleration from the definition rather than from a memorized rule; and draw motion diagrams that show acceleration as well as velocity.

Three graphs of one motion. Any straight-line motion can be shown three ways: position vs. time, velocity vs. time and acceleration vs. time. The three graphs are not independent pictures; each is built from the one before it. Velocity is the rate of change of position, so the slope of the position graph at any instant is the velocity at that instant. Acceleration is the rate of change of velocity, so the slope of the velocity graph is the acceleration. Going the other way, the area between the velocity graph and the time axis over an interval is the displacement during that interval, and the area under the acceleration graph is the change in velocity. These are not rules to memorize. They follow from the definitions v = Δx/Δt and a = Δv/Δt, and every claim below can be rebuilt from those two definitions.

Position vs. time. A straight segment on a position graph means constant velocity; the steeper the segment, the faster the object. A horizontal segment (zero slope) means the object is at rest. A curve means the velocity is changing. An object speeding up produces a position graph that gets steeper as time goes on; an object slowing down produces one that gets less steep. For a curved graph, the velocity at one instant is the slope of the tangent line, the straight line that just touches the curve at that instant. Average velocity over an interval is the slope of the chord joining the two end points. For constant velocity the tangent and the chord are the same line, so instantaneous and average velocity agree; for an accelerating object they rarely do.

A worked model with data. A cart is released on a ramp and its position is recorded once per second: 0 m, 0.2 m, 0.8 m, 1.8 m and 3.2 m at t = 0, 1, 2, 3 and 4 s. Plot the points and draw a smooth curve; it bends upward, so the cart is speeding up. To find the velocity at t = 1 s, draw the tangent at (1, 0.2). A carefully drawn tangent passes through about (0, −0.2) and (2, 0.6), so its slope is (0.6 − (−0.2)) ÷ (2 − 0) = 0.4 m/s. At t = 3 s the tangent through (3, 1.8) passes through about (2, 0.6) and (4, 3.0), slope (3.0 − 0.6) ÷ 2 = 1.2 m/s. The average acceleration between those two instants is the change in velocity divided by the time: (1.2 − 0.4) m/s ÷ 2 s = 0.4 m/s². Notice the difference between the chord and the tangent: the average velocity from 0 to 2 s is 0.8 m ÷ 2 s = 0.4 m/s, which happens to equal the instantaneous velocity at the midpoint, 1 s; that is a special property of constant acceleration, not a general fact.

Velocity vs. time. On a velocity graph the slope is the acceleration, so constant acceleration gives a straight line, not necessarily horizontal. A horizontal line above the axis is constant positive velocity; a horizontal line on the axis is rest. An object whose speed is increasing has a velocity graph that moves away from the time axis; one whose speed is decreasing moves toward the axis. The area between the line and the axis is the displacement. Area above the axis counts as positive displacement and area below counts as negative. A velocity graph tells you displacement, never absolute position: two objects starting 100 m apart can have identical velocity graphs.

Take a car that starts from rest, speeds up steadily to 12 m/s in 4 s, holds 12 m/s for 6 s, then slows steadily to rest in 4 s. Stage 1 has slope 12 ÷ 4 = 3 m/s² and area ½ × 4 × 12 = 24 m. Stage 2 has zero slope, zero acceleration, and area 6 × 12 = 72 m. Stage 3 has slope −3 m/s² and area 24 m. The total displacement is 120 m. The position graph is a curve bending upward, then a straight segment, then a curve flattening out; the acceleration graph is three horizontal segments at +3, 0 and −3 m/s².

The sign of acceleration. Because a = Δv/Δt, acceleration is negative whenever velocity is decreasing in the algebraic sense, which can mean slowing down while moving in the positive direction or speeding up while moving in the negative direction. Consider a velocity graph that runs from +6 m/s at t = 0 to −6 m/s at t = 6 s as one straight line, crossing the axis at 3 s. The slope is −12 ÷ 6 = −2 m/s² throughout. From 0 to 3 s (quadrant I) the object moves in the positive direction and slows down; from 3 to 6 s (quadrant IV) it moves in the negative direction and speeds up. Same acceleration, opposite descriptions. Negative acceleration does not mean slowing down; it means velocity is becoming more negative. For this reason the word "deceleration" is avoided in this course. The displacement is +9 m for the first half and −9 m for the second, net zero, though the distance traveled is 18 m. That graph fits a ball thrown straight up at 6 m/s on a planet where g = 2 m/s², or a cart rolling up a ramp and back down.

Acceleration vs. time. For uniform acceleration this graph is a horizontal line. It sits on the axis for an object at rest or moving with constant velocity; a horizontal line along the time axis (zero acceleration) cannot tell you which. The area under the acceleration graph over an interval is the change in velocity, and nothing more: absolute velocity, displacement and position cannot be recovered from an acceleration graph alone.

Motion diagrams with acceleration. In physical science a motion diagram was a row of dots at equal time intervals with a velocity arrow at each dot. In this course an acceleration arrow is added. Compare the velocity arrows at two neighboring dots: the change from the earlier arrow to the later one, Δv, points the same way as the acceleration, so draw an arrow labeled a in that direction (once beside the diagram when the acceleration is constant, or between each pair of dots when it changes). For the car above, stage 1 has dots that spread out, velocity arrows that grow, and a forward acceleration arrow; stage 2 has evenly spaced dots, equal arrows and a = 0, written beside the diagram; stage 3 has dots that bunch up, arrows that shrink, and a backward acceleration arrow. For the ball thrown up at 6 m/s where g = 2 m/s², the acceleration arrow points down at every dot, including the top, where the velocity arrow shrinks to nothing for an instant. When the velocity and acceleration arrows point the same way the object is speeding up; when they point opposite ways it is slowing down, whichever direction is called positive.

Sketching the other two graphs. Given any one of the three, sketch the others by asking two questions in order: what is the slope, and what is the area? From a velocity graph, the slopes give the acceleration graph and the accumulated areas give the shape of the position graph. From a position graph, tangent slopes give velocity and the change in those slopes gives acceleration. From an acceleration graph, areas give changes in velocity, but you need a starting velocity and starting position to fix the other two graphs.

Two constant-velocity buggies. Two battery buggies travel toward each other on a track. Buggy A starts at x = 0 moving at +0.30 m/s; buggy B starts at x = 2.4 m moving at −0.50 m/s. Their position graphs are straight lines: x_A = 0.30t and x_B = 2.4 − 0.50t. They collide where the lines cross: 0.30t = 2.4 − 0.50t, so 0.80t = 2.4 and t = 3.0 s, at x = 0.90 m from A's start. Prediction before the run, measurement after: compare the measured meeting point with 0.90 m, and look for the cause of any difference in the buggies' speeds not being perfectly constant and in the uncertainty of the two measured speeds.

Investigation: measure changing velocity over 5 m of fall and 5 m of rise. Materials: a soft foam ball, a smartphone that records video at a known frame rate (30 or 60 frames per second), a tape measure, chalk or tape to mark every 0.50 m on a wall or post, and a stairwell landing or balcony at least 5.0 m above a clear, cordoned-off floor. Steps: mark the wall; have one student drop the ball from 5.0 m while another films from far enough back that the whole fall is in frame; then, standing on the ground, throw the ball straight up so that it rises at least 5.0 m past the marks (a launch speed near 10 m/s is needed, which a strong throw can give) and film again. Record: for each video, the frame number when the ball passes each mark, converted to time using the frame rate. Analyze: plot position vs. time, draw tangents at two instants, read the velocities and compute the average acceleration. A result that supports the claim: velocities that grow by roughly 9.8 m/s for every second of fall and shrink by the same amount during the rise, so both graphs give an acceleration near −9.8 m/s² in the upward-positive sense, whether the ball is rising or falling. Safety: nobody stands below the drop zone, the ball is soft, and the person filming stays clear of the throw. What the model cannot show: the frame rate limits time resolution to about 0.03 s, so tangents are approximate; and near 10 m/s air resistance on a light ball is not negligible: while the ball rises, drag points down along with gravity, so the measured acceleration is slightly larger than 9.8 m/s² in size; while it falls, drag points up, so the measured acceleration is slightly smaller.

Worked example 1 — write a driving scenario from a velocity graph. A velocity graph starts at 0, rises in a straight line to 15 m/s at t = 5 s, stays at 15 m/s until t = 20 s, drops to 0 at t = 25 s, stays at 0 until t = 30 s, and then goes to −5 m/s by t = 32 s and stays there until t = 40 s. A scenario that fits: a car pulls away from a stop sign and speeds up steadily for 5 s, cruises at 15 m/s for 15 s, brakes steadily to a stop over 5 s, waits 5 s, then reverses at a steady 5 m/s for 8 s, having taken 2 s to reach reversing speed. The displacement is ½ × 5 × 15 + 15 × 15 + ½ × 5 × 15 + 0 + (−½ × 2 × 5) + (−8 × 5) = 37.5 + 225 + 37.5 − 5 − 40 = 255 m. The area below the axis is subtracted, because the car is moving backward.

Worked example 2 — from a position graph to the other two. A position graph is a straight line rising from 0 to 20 m over 0 to 4 s, then horizontal at 20 m from 4 to 6 s, then a straight line falling from 20 m to 0 over 6 to 10 s. Velocity is the slope: +5 m/s, then 0, then −5 m/s. So the velocity graph is three horizontal segments at 5, 0 and −5 m/s. The acceleration graph is zero throughout, except at the instants 4 s and 6 s where the velocity changes abruptly; on an idealized graph these are spikes, and a real object would show a brief, large acceleration there. The object ends where it started, displacement zero, distance 40 m.

Worked example 3 — sign of acceleration in quadrants I and IV. A velocity graph is a straight line from −8 m/s at t = 0 to +8 m/s at t = 4 s. The slope is +16 ÷ 4 = +4 m/s², positive the whole time. From 0 to 2 s the object moves in the negative direction with decreasing speed; from 2 to 4 s it moves in the positive direction with increasing speed. It is at rest for an instant at t = 2 s but the acceleration there is still +4 m/s², because the velocity is changing through zero, not staying at zero. Displacement: −8 m for the first half and +8 m for the second, net zero.

Worked example 4 — average acceleration from position data. The cart data in the lesson gave tangent velocities of 0.4 m/s at t = 1 s and 1.2 m/s at t = 3 s. Average acceleration is Δv/Δt = (1.2 − 0.4) ÷ (3 − 1) = 0.4 m/s². As a check, the data fit x = 0.2t², whose velocity is 0.4t, giving exactly 0.4 and 1.2 m/s at those instants, and a constant acceleration of 0.4 m/s². Tangents drawn by hand will not land exactly on these values; work out the percentage difference for your own, and treat agreement within about 20 percent as a suggested classroom target, not a measured standard.

Representation task: three stacked graphs. Using the car in the lesson (speeds up to 12 m/s in 4 s, cruises 6 s, slows to rest in 4 s), draw the velocity graph on one set of axes, then directly below it, with the same time axis, draw the acceleration graph, and below that the position graph. Label the slope of each straight segment of the velocity graph, shade and label the three areas (24 m, 72 m, 24 m), and mark the positions 24 m, 96 m and 120 m on the position graph at 4 s, 10 s and 14 s. Write two sentences explaining why the position graph curves during the first and last stages and is straight in the middle. Finally, draw a motion diagram for the whole trip with dots every 2 s, a velocity arrow at each dot and an acceleration arrow for each stage.

Independent performance: unlabeled graphs to scenarios. You are given six unlabeled graph shapes: a constant positive slope, an increasing positive slope, a zero slope, a decreasing positive slope, a straight line that crosses the time axis from positive to negative, and a horizontal line below the axis. For each shape, state what it shows if the vertical axis is position and what it shows if the vertical axis is velocity, and give one real object that would produce each graph. Then take one velocity graph of your own design that has both positive and negative values and write a matching driving scenario. Success criteria: each description uses slope or area correctly to justify the claim; the sign of acceleration is explained from Δv/Δt and not from whether the object is slowing; at least one case shows negative acceleration with increasing speed; and the scenario's displacement is computed from areas with the sign of each area stated.