I keep tripping over distance–time graphs and I think I’m mixing up two ideas. In a quiz last week, the graph had a line that sloped downward and I confidently said, “Aha, negative speed!” …and got it wrong. I know slope is supposed to be speed (steeper = faster, horizontal = stopped), but then what does a downward slope even mean on a distance–time graph?
Here’s where my brain splits:
– If the y-axis is “total distance traveled,” that should never decrease, right? So the graph should never go down.
– But if the y-axis is “distance from home,” that can go down when I turn around and come back.
Example I keep using: I start at 0 m at t = 0 s. I walk away from home at 1 m/s for 10 s (so I’m at 10 m), stop for 5 s, then walk back toward home at 2 m/s for 4 s (so I’m at 2 m from home at t = 19 s). When I sketch this, I get a line up from 0 to 10, a flat bit, then a line down toward 2. That screams “downward slope,” which my brain wants to call negative speed, but speed shouldn’t be negative.
My partially correct attempt: I tried to compute average speed as (final y-value)/(total time). For the example that would be 2 m / 19 s, which feels wrong because I did a lot more walking than 2 m! So I must be using the wrong interpretation of the y-axis.
So here’s my direct question: On a distance–time graph, is a downward slope ever allowed? If it is, what exactly is on the y-axis? And when the graph has a downward segment (like in my example), how am I supposed to read speeds and compute the average speed correctly? I feel like I’m close, but I keep mixing up “distance traveled” and “distance from start,” and it’s messing with my slopes!
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3 Responses
You’re exactly right that two different y-axes get used: if the y-axis is total distance traveled, the graph never goes down and its slope is speed (nonnegative), but if the y-axis is position/distance from home, it can go down and the slope is velocity (can be negative) while speed is the absolute value of the slope.
Example: out at 1 m/s for 10 s (slope +1), stop 5 s (slope 0), back at 2 m/s for 4 s (slope −2), so speeds are 1, 0, 2 m/s; average velocity = (2 m)/19 s ≈ 0.105 m/s, while average speed = (10 + 8)/19 s ≈ 0.947 m/s-the downward segment just means “moving back toward home,” not negative speed.
Short answer: a line sloping down is fine if the y-axis is position (distance from a chosen reference point with a sign), but not if the y-axis is total distance traveled. Slope is actually velocity, not speed. So on a position–time graph, a downward slope just means you’re heading back toward home (negative velocity); your speed is the absolute value of that slope. On a distance-traveled–time graph (think car odometer), the graph never goes down because total distance can’t un-happen, and the slope there is the speed itself. Quick rule of thumb: if you ever see the graph go down, you’re not looking at “total distance traveled,” you’re looking at “where you are” relative to home. In your example: position–time view gives +1 m/s for 0–10 s (up), 0 for 10–15 s (flat), then −2 m/s for 15–19 s (down). Average speed = total distance/total time = (10 + 8)/19 = 18/19 ≈ 0.95 m/s. The 2/19 you computed is average velocity (net displacement/time), not average speed-close, but not the same. Analogy: your GPS coordinate can go up and down as you drive north and south (position graph), but your odometer only ticks upward no matter what (distance-traveled graph). For a clean refresher, see Khan Academy on average speed vs average velocity: https://www.khanacademy.org/science/physics/one-dimensional-motion/one-dimensional-motion-tutorial/a/average-velocity-and-average-speed
You’re mixing two different y-axes. If y means total distance traveled (the running tally of how much ground you’ve covered), the graph can never go down; its slope is speed and is always nonnegative. If y means position or “distance from home along a line” (how far you are from the start at that moment), the graph can go down when you head back; its slope is velocity, which can be negative. Negative slope there means “moving toward home,” not “negative speed.” Speed is the magnitude of the slope.
Worked example with your numbers:
– Position–time: 0→10 m in 10 s has slope +1 m/s; then flat for 5 s; then 10→2 m in 4 s has slope −2 m/s. Instantaneous speed is the absolute value of the slope: 1, 0, and 2 m/s. Displacement over the whole trip is 2 m, so average velocity = 2 m / 19 s ≈ 0.105 m/s.
– Distance-traveled–time: 0→10 m in 10 s (slope +1), flat for 5 s, then 10→18 m in 4 s (slope +2). Total distance is 18 m, so average speed = 18 m / 19 s ≈ 0.947 m/s. Your 2/19 calculation was average velocity, not average speed.
Rule of thumb: decide which quantity is on the y-axis. If it’s total distance traveled, the graph never decreases and slope = speed. If it’s position/distance-from-home, downward segments are allowed; slope = velocity and speed is its absolute value. A short refresher is here: https://www.physicsclassroom.com/class/1DKin/Lesson-1/Distance-and-Displacement