I’m wrestling with velocity–time graphs and my brain keeps swapping what slope and area mean. I feel like I almost get it, and then I do a calculation and the units come out weird and I realize I’ve mashed two ideas together.
Here’s a specific example I sketched: from t = 0 to 4 s, the velocity line ramps up linearly from 0 to 8 m/s. Then from 4 to 6 s it’s flat at 8 m/s. From 6 to 8 s it slopes down to −4 m/s, and from 8 to 10 s it stays at −4 m/s. I thought I could find how far the object traveled by breaking it into chunks.
My completely wrong attempt (I know it’s wrong, but I want to show how I’m thinking):
– For 0–4 s: the slope is (8 − 0) / 4 = 2 m/s². So I multiplied slope by the width: 2 × 4 = 8. I called that “8 m of distance.”
– For 4–6 s: the slope is 0, so I said area is 0 and “no distance there.”
– For 6–8 s: slope is (−4 − 8) / 2 = −6 m/s². I did −6 × 2 = −12 and called that “−12 m.”
– For 8–10 s: slope is 0 again, so I wrote 0.
Then I added them and got something like 8 + 0 − 12 + 0 = −4 m for the total distance (?!). The units don’t even make sense because I’m mixing m/s² × s = m/s, which isn’t meters. I can see the red flags but I don’t know where to fix my brain.
Another thing I tried: I said, okay, maybe I should do base × height for each rectangle/triangle shape under the line. But then I freak out when the graph goes below the time axis (negative velocity). Do I subtract that area or flip it to positive for “distance”? I tried making it all positive and got 16 + 16 + 12 + 8 = 52 m (I used triangles and rectangles, probably inconsistently). That also feels off because it looks like I double-counted something. And when I tried a trapezoid formula on the slanted parts, I got yet another number (24 m for the first two chunks combined), and now I don’t trust anything.
I also keep mixing up which thing shows acceleration. I’m thinking: isn’t the slope the acceleration? But then the “distance” seems to be area? But then the negative bits… are they “backwards distance”? Is that displacement? I’m after distance traveled, not the “net” from start to end, but I’m not sure how to do that on this kind of graph without messing up signs.
Analogy that might be totally wrong: I’m imagining the graph like a treadmill readout. The height of the line is how fast the belt is moving, and the area under it is like the total number on the odometer. But if the line goes below zero, is that like the treadmill running in reverse and the odometer going backwards? Do I treat that as taking steps back, or do I just add them because my legs still did the work? My gut says add for distance, but subtract for displacement… but I keep tripping over which rectangles/triangles to draw and which numbers to flip.
Could someone help me sort this out with my example? Specifically:
– How do I systematically break it into pieces and compute the numbers without mixing slope and area?
– What exactly should I do with the negative velocity segments if I want distance vs if I want displacement?
– Is there a clean way to check units so I know I’m not doing something silly?
I’m sure this is one of those “once you see it, you can’t unsee it” things, but right now I feel like I’m counting shadows instead of shapes. Any nudges appreciated!
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3 Responses
Think of a velocity–time graph like this: the height is “how fast and which way,” the slope is “how the speed is changing” (acceleration), and the area under the curve is “how far you’ve gone” but with a sign (displacement). So yes: slope = acceleration (units m/s²), and area = displacement (units (m/s)·s = m). If the graph dips below the axis, that means you’re moving in the opposite direction; for displacement you keep that area negative, but for distance traveled you flip it positive (because your legs still did the work). Multiplying slope by a time width gives you a change in velocity (m/s² × s = m/s), not a distance-that’s the unit check that tells you you were mixing ideas.
Now your example, done systematically by areas. From 0–4 s: triangle area = ½·4·8 = 16 m. From 4–6 s: rectangle = 8·2 = 16 m. From 6–8 s the line goes from +8 to −4, crossing zero at t = 6 + 8/6 = 7⅓ s. So split it: 6 to 7⅓ is a positive triangle: ½·(4/3)·8 = 16/3 m; 7⅓ to 8 is a negative triangle: −½·(2/3)·4 = −4/3 m. Net over 6–8 is 16/3 − 4/3 = 4 m. From 8–10 s: rectangle = −4·2 = −8 m. Add for displacement: 16 + 16 + 4 − 8 = 28 m. For distance, take absolute values on the signed pieces: 16 + 16 + (16/3 + 4/3) + 8 = 16 + 16 + 20/3 + 8 = 140/3 ≈ 46.7 m.
Quick unit sanity-checks: slope has m/s per second = m/s² (acceleration); areas have (m/s)·s = m (distance). If an operation lands you in m/s, you’ve computed a velocity change, not a distance. And any time a segment crosses the axis, split it at the zero so you can keep the signs straight for displacement and then, if you want distance, flip those negative bits to positive before adding.
On a velocity–time graph, slope is acceleration (m/s²) and area is displacement (meters) because velocity × time = distance; for distance traveled add the absolute areas, while negative areas subtract only for displacement-like a treadmill odometer that can run backward (displacement) versus a step counter that only increases (distance). To compute systematically, split at segment boundaries and where v crosses 0; for your sketch, displacement = 0.5·4·8 + 8·2 + [(8 + (−4))/2]·2 + (−4)·2 = 28 m, and distance = 16 + 16 + (16/3 + 4/3) + 8 = 140/3 ≈ 46.7 m; unit check: areas are (m/s)·s = m, slopes are m/s² (see https://www.khanacademy.org/science/physics/one-dimensional-motion/acceleration-tutorial/v/calculating-displacement-from-a-velocity-time-graph).
You’ve got the right instincts: on a velocity–time graph, the slope tells you acceleration (units m/s²), and the area under the curve tells you displacement (units m, because m/s × s). A clean unit-check is: rectangle area = velocity × time → (m/s) × s = m; triangle area = 1/2 × base × height → still meters. For distance traveled, you add the magnitudes of those areas (treat below-axis parts as positive); for displacement, you keep the sign (below-axis subtracts). Khan Academy has a nice walkthrough of this idea: https://www.khanacademy.org/science/physics/one-dimensional-motion/average-velocity-1/v/displacement-from-velocity-time-graphs
For your example, break it into time-chunks and use areas. From 0–4 s, it’s a triangle up to 8 m/s, so displacement is 1/2 × 4 × 8 = 16 m. From 4–6 s, a rectangle at 8 m/s for 2 s gives 8 × 2 = 16 m. From 6–8 s, you can view it as a trapezoid: average velocity is (8 + (−4))/2 = 2 m/s, over 2 s gives 4 m (this is signed, so it’s +4 m for displacement). From 8–10 s, a rectangle at −4 m/s for 2 s gives −8 m. Adding signed areas gives displacement: 16 + 16 + 4 − 8 = 28 m. For distance, one quick (but a bit sloppy) way is to just take absolute values of those chunk-areas: 16 + 16 + |4| + 8 = 44 m. A more careful way is to split 6–8 s at the zero crossing (the line hits v = 0 around t ≈ 7.33 s): the part from 6 to 7.33 s contributes about 16/3 m and from 7.33 to 8 s contributes about 4/3 m, so that whole slanted chunk is about 20/3 ≈ 6.7 m. Then total distance ≈ 16 + 16 + 6.7 + 8 = 46.7 m. Either way, slope is only for acceleration; the areas (with sign for displacement, absolute for distance) are what give you the meters.
Hope this helps!