I’m prepping for a test and my brain keeps doing cartwheels over velocity–time graphs. In theory I get that slope = acceleration and the area under the graph = displacement. But when I stare at an actual graph, I start second-guessing everything, especially when the line dips below the time axis.
For example: imagine a velocity–time graph that goes from 0 m/s at t=0 up to 10 m/s at t=4 s (straight line), then stays flat at 10 m/s until t=8 s, and then slopes down and crosses into negative velocity, ending at −5 m/s by t=12 s. If I want the total displacement and the total distance traveled over the whole 0 to 12 seconds, what exactly should I do with that part below the axis? Do I subtract that area for displacement but take the absolute value for distance? And do I have to split it into neat triangle/rectangle/trapezoid chunks, or is there a quicker way I’m missing?
Also, another thing that scrambles me: some graphs show a sudden vertical drop from, say, 10 m/s to 0 in no time at all. Is that even allowed? If a velocity–time graph has a vertical jump or a sharp corner, how am I supposed to talk about the acceleration at that point? Do I just say it’s undefined, or is there a standard test-friendly interpretation?
One more side quest: when the graph is curvy, I know the slope at a point is the instantaneous acceleration, but for average acceleration between two times, am I supposed to use the slope of the straight line between those points, or something with areas? I keep mixing up which quantities come from slopes and which come from areas.
And the big confusion: for average speed versus average velocity over the whole interval, do I use the signed area divided by total time for one, and the absolute area divided by total time for the other? Or am I sneaking in a physics rule that doesn’t actually belong here? Follow-up: if the axis label said “speed–time” instead of “velocity–time,” would that change how I treat the parts below the axis (like, would there even be any)?
Could someone explain, in a clean checklist kind of way, how to read these: (1) displacement vs distance from the graph (especially with negative sections), (2) what to do with vertical jumps and corners, and (3) the right way to get average speed and average velocity from the graph? I don’t need the numbers worked out-just how to think about it so I stop tripping over the same ideas right before this test.
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3 Responses
You’re already thinking about it the right way: for a velocity–time graph, areas give you displacement and slopes give you acceleration. Here’s the tidy way I keep it straight: (1) Displacement over an interval is the signed area under v(t): anything below the time axis counts negative, anything above positive. Distance is the total area with sign removed, i.e., integrate |v|, so practically you split the time interval wherever the graph crosses v = 0, find the geometric areas piece by piece, and add their absolute values. A quick trick for straight segments is to use the trapezoid formula: area = (v_start + v_end)/2 × Δt; for displacement keep the signs, for distance use absolute value and still split at zero crossings. In your 0–12 s example, you’d compute the trapezoids/rectangles on each straight bit, subtract the below-axis chunk for displacement, and add its magnitude for distance. (2) Sharp corners mean the instantaneous acceleration (the derivative) doesn’t exist at that point, but that’s fine for a graph; vertical jumps in velocity are “impulsive” changes-mathematically the acceleration is undefined (physically it would be infinite for an instant), and the vertical jump itself contributes no area, so displacement is still determined only by the horizontal pieces. On a test, it’s safe to say “acceleration is undefined at that instant,” but you can still compute average acceleration across any interval that includes it. (3) Instantaneous acceleration is the slope of the tangent (derivative dv/dt); average acceleration from t1 to t2 is the secant slope Δv/Δt-no areas needed. Average velocity over the whole interval is displacement divided by total time, i.e., the signed area/Δt; average speed is total distance divided by total time, i.e., the absolute area/Δt. And if the axis says speed–time instead of velocity–time, the graph won’t dip below the axis (speed is nonnegative), and the area directly gives distance; you lose any information about direction, so displacement can’t be recovered. I used to color these in: green above the axis and red below; during one quiz I confidently added green + red and got a “negative distance,” panicked, and then realized I needed to flip the red to positive for distance-splitting at the zero crossings saved me from that facepalm.
Here’s the clean checklist idea. From a velocity–time graph, displacement is the signed area under the curve (above the axis counts positive, below counts negative). Distance is the area under the speed curve, so take the absolute value of velocity: reflect any part below the axis and add that area. Average velocity over [t1, t2] = displacement/(t2 − t1); average speed over [t1, t2] = distance/(t2 − t1). Instantaneous acceleration is the slope of the v–t curve at a point; average acceleration between two times is the slope of the secant line, (v2 − v1)/(t2 − t1). In your example, yes: subtract the below-axis area for displacement, but flip-and-add it for distance. Practically, you split wherever the graph changes formula (or crosses the axis), then add areas of triangles/rectangles/trapezoids. If you have an explicit v(t), integrate v to get displacement and |v| to get distance. A concise refresher is here: The Physics Classroom on area under v–t graphs (https://www.physicsclassroom.com/class/1DKin/Lesson-4/Meaning-of-Area-on-a-v-t-Graph).
About corners and vertical jumps: a sharp corner means acceleration at that instant is undefined (no single slope), though it’s fine on either side. A true vertical jump (instantaneous change in velocity) isn’t physically realistic; mathematically it’s a jump discontinuity, so acceleration at that instant doesn’t exist (you can think of an “impulse”). For test purposes: say “acceleration is undefined at the jump/corner”; average acceleration over any interval is still (Δv)/(Δt). The vertical jump contributes zero to displacement because it has zero time-width; only the horizontal extent makes area. If the axis says speed–time, the graph never goes below zero; you can get distance and average speed directly from its area, but you cannot get displacement or average velocity without extra direction information.
Would it help if I mark up your specific timeline (0–4–8–12 s), show exactly where to split for the zero crossing, and write the area expressions you’d add/subtract?
Here’s the clean way I think about velocity–time graphs, especially the “below the axis” bits: displacement comes from signed area (regions above the axis count positive, below count negative), while distance comes from total area with everything made positive; practically, you split the time axis wherever v(t) changes sign, then add those areas with signs for displacement and with absolute values for distance. For your piecewise‑linear example, that means: triangle up, rectangle flat, then on the sloping-down part find the zero‑crossing time and treat the portion above the axis and the portion below as separate triangles; subtract the below‑axis triangle for displacement, add it for distance. Average velocity over the whole interval is total displacement divided by total time (i.e., signed area / Δt), while average speed is total distance divided by total time (absolute area / Δt); note that average speed is not the absolute value of average velocity. Instantaneous acceleration is the slope of the graph at a point; average acceleration between t1 and t2 is the secant slope (v2 − v1)/(t2 − t1)-no areas needed. Sharp corners mean the slope jumps, so the instantaneous acceleration at the corner is undefined (but averages over any nonzero interval are fine); a vertical jump in velocity would imply infinite/impulsive acceleration and, in a test setting, you can say “acceleration is not defined at that instant” (the displacement doesn’t change there because the time width is zero). As for a “quicker way”: if the graph is piecewise linear, adding triangles/rectangles/trapezoids is the quickest; if you have a formula for v(t), integrate v(t) for displacement and integrate |v(t)| for distance. And if the label said speed–time, the graph would never dip below the axis (speed is nonnegative); the area would give distance only, and you can’t recover displacement or average velocity without direction info. When I first learned this, I kept mixing up which thing was “slope” and which was “area” until I wrote a tiny checklist on my note card: slope → acceleration, signed area → displacement, absolute area → distance; I still mutter that to myself before I start adding triangles.