I get the basic idea that on a velocity–time graph, slope is acceleration and area is displacement. But the second the line dips below zero, I start second-guessing everything. Do I subtract that area or take absolute value? And when exactly is the object ‘turning around’ versus just slowing down?
Here’s the specific setup I’m working with (all in m/s and seconds):
– 0 to 4 s: velocity ramps linearly from 0 to +12
– 4 to 8 s: constant at +12
– 8 to 12 s: ramps linearly down to −6 (so it crosses zero somewhere in there)
– 12 to 14 s: constant at −6
– 14 to 16 s: ramps linearly back up to 0
I want three things: (a) total displacement at 16 s, (b) total distance traveled, and (c) plain-English intervals where it’s speeding up vs slowing down. I can chop the areas into triangles/rectangles, no problem. My confusion is the sign handling and the whole ‘decelerating’ wording when velocity is negative.
My attempt:
– For 0–4: triangle area = 1/2 * 4 * 12 = 24 (above zero).
– For 4–8: rectangle = 4 * 12 = 48 (above zero).
– For 8–12: I tried the trapezoid trick: average velocity (12 and −6) is 3, times 4 s gives +12 displacement. Then I got paranoid and split at the zero crossing: looks like about 2.67 s above (area ~16) and 1.33 s below (area ~4), so net +12. That seems consistent, but I’m not 100%.
– For 12–14: rectangle at −6 for 2 s → −12 displacement.
– For 14–16: triangle from −6 up to 0 over 2 s → −6 displacement.
If I add the signed areas I get a displacement of 66 m. If I take absolute areas to get distance, I get 110 m. Are those numbers actually right, or did I count a triangle twice or miss a sign flip somewhere?
Also, the wording messes with me: when velocity is negative and the slope is negative, the speed is increasing (just in the negative direction), which means it’s actually speeding up, not ‘decelerating’, right? Is there a quick rule-of-thumb like: ‘speeding up if velocity and acceleration have the same sign; slowing down if they have opposite signs’? And is there a fast way to eyeball distance vs displacement on these without splitting every segment by hand?
Any help appreciated!
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3 Responses
Pretty sure your totals are right: displacement 66 m (signed areas, so anything below zero subtracts) and distance 110 m (take absolute areas), and it “turns around” where v crosses 0 at t ≈ 10.67 s. Rule of thumb: speeding up when velocity and acceleration have the same sign, slowing when opposite-so speeding on 0–4 s and 10.67–12 s (I think also 14–16 s), slowing on 8–10.67 s, with constant speed on 4–8 and 12–14.
You’ve got the right ideas: on a velocity–time graph, signed area gives displacement (areas below zero subtract), and absolute area gives distance (treat below-zero as positive). For your data, the signed areas add to 24 + 48 + 12 − 12 − 6 = 66 m displacement; for distance, reflect the below-axis pieces: 24 + 48 + (16 + 4) + 12 + 6 = 110 m, so your 66 m and 110 m look consistent. The “turnaround” happens exactly when v crosses 0, at t = 32/3 ≈ 10.67 s during the 8–12 s ramp; at 16 s it’s just come to rest (I’m pausing to check myself: some people would also say it “turns around” there since the next instant it would go positive if the ramp continued, but within 0–16 s it hasn’t reversed again). For speeding up vs slowing down, use the quick rule you proposed: speed increases when velocity and acceleration have the same sign, and decreases when they have opposite signs. So: 0–4 s (v > 0, a > 0) speeding up; 4–8 s constant speed; 8–10.67 s (v > 0, a < 0) slowing down to a stop; 10.67–12 s (v < 0, a < 0) speeding up in the negative direction; 12–14 s constant speed; 14–16 s (v < 0, a > 0) slowing down to rest. Example: if v = −5 m/s for 3 s, displacement is −15 m while distance is 15 m; the minus just tells you the direction. For displacement on any straight (linear) segment, the trapezoid average works even if it crosses zero; for distance, you typically do need to split at the zero crossing (or, as a quick eyeball I sometimes use, take the average of endpoint speeds times the duration-though that can overcount when there’s a crossing). A concise refresher on reading velocity–time graphs and areas: https://www.khanacademy.org/science/physics/one-dimensional-motion/acceleration-tutorial/a/what-are-velocity-time-graphs
You’ve got the right instincts, and your numbers are spot on: total displacement is 66 m and total distance is 110 m. The rule is: for displacement, add signed areas (below the axis counts negative); for distance, take the absolute value of velocity, which means you must split anywhere the graph crosses zero and reflect that piece upward. In your 8–12 s segment, the zero-crossing is at t = 8 + 12/4.5 ≈ 10.67 s, so that chunk contributes +16 m above and 4 m below, giving +12 m displacement but 20 m distance-exactly what you found. “Turning around” happens when velocity changes sign, so here that’s only at t ≈ 10.67 s (not at 14–16 s; that’s just coming to rest). For “speeding up vs slowing down,” your thumb rule is perfect: speed increases when velocity and acceleration have the same sign, and decreases when they’re opposite. So: speeding up on 0–4 s (v > 0, a > 0) and 10.67–12 s (v < 0, a < 0); slowing down on 8–10.67 s (v > 0, a < 0) and 14–16 s (v < 0, a > 0); constant speed on 4–8 s and 12–14 s. Tiny language note: “decelerating” really means “speed going down,” not “acceleration negative,” which is why a negative velocity with negative slope is actually speeding up. Personal tangent: I used to shade my v–t graphs-green above, red below-and then “fold” the red up in my head to get distance; once I started doing that, the sign chaos finally stopped yelling at me.