Negative velocity on v–t graph: subtract or take absolute area?

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1. Great question! On a velocity–time graph, the signed area (counting area above the axis as positive and below as negative) gives displacement-the net change in position from your start. That’s why subtracting the below-axis part can leave you with almost zero if you turned around and came back near where you began. For total distance traveled, you want the area under the speed curve, which is the same as taking the absolute value of velocity: split the time axis at the points where velocity crosses zero, compute the area on each piece, and add those areas without signs. Also, a small but important distinction: “speed” never goes below zero; what went negative was your velocity, meaning you were moving in the direction you chose as negative (say, back down the hill). A helpful way to think about it: imagine walking down a hallway past a line on the floor-your steps counted (distance) always go up, but your position relative to the line (displacement) can go up and then back down, even returning to zero if you retrace your path. So, for displacement: subtract areas below the axis; for total distance: add their absolute sizes.

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2. You’ve got the right idea, just a tiny sign-issue gremlin snuck in. On a velocity–time graph, the signed area gives displacement (where negative area counts as motion in the opposite direction). So yes: when the graph dips below zero, that means you’re moving back along your chosen axis (say, back down the hill if “uphill” is your positive direction). Speed itself is never negative-only velocity can be-so that below-zero part is telling you direction. For net displacement, you add the signed areas (positive above the axis, negative below). For total distance traveled, you add the absolute areas (flip the below-axis chunks up before adding). Rule of thumb: displacement = signed area; distance = area of |v|.

   Quick example: suppose you ride at +6 m/s for 30 s, then turn around and ride at −4 m/s for 20 s. The first rectangle’s area is 6 × 30 = 180 m. The second is −4 × 20 = −80 m. Displacement = 180 − 80 = 100 m (you end up 100 m from where you started). Distance traveled = 180 + 80 = 260 m (everything you actually rode, no matter the direction). So if you subtracted and got “almost no distance,” that just means your displacement was small-you nearly came back to where you started-even though you covered plenty of ground. Flip the negative parts to positive when you want total distance, and you’re golden.

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3. Short answer:
   – Below zero on a v–t graph means your velocity is in the opposite direction (you’re going “backwards” relative to whatever you chose as positive).
   – The signed area gives displacement (net change in position). That’s where you subtract the bits below the axis.
   – Total distance is the integral of speed, i.e., add the absolute value of each area piece. No subtracting.

   Longer, clearer version (with some common-sense checks):

   1) Speed vs velocity
   – Speed is how fast you’re moving. It’s never negative.
   – Velocity is speed with a direction. It can be negative if you’re going the opposite way to your chosen “positive” direction.
   – If your graph dips below zero, it’s not a speed–time graph. It’s a velocity–time graph. Rename your axis accordingly.

   2) What the area means
   – Area under a velocity–time graph (counting sign) = displacement. That’s your start-to-finish “as the crow flies along a straight line” difference, with direction.
   – Area under a speed–time graph (always nonnegative) = total distance traveled. This is what your bike’s odometer would count.

   3) What to do with negative parts
   – For displacement: subtract the areas below the axis (because that motion is in the negative direction).
   – For total distance: take the absolute value of those areas and add everything. That means you split the time axis at every zero crossing of v(t), take the area of each chunk, and add their magnitudes.

   4) Why your result “felt wrong”
   You subtracted the downhill/return portion and got “almost no distance.” That is perfectly normal for displacement if you went out and back roughly the same amount. Your net change in position can be near zero while the total distance is large. Think: out 10 km, back 10 km → displacement 0, distance 20 km.

   5) Quick mental model
   – Odometer thinking (distance): every wheel turn counts, so ignore signs. Add all areas.
   – Map pin thinking (displacement): only care where you end up relative to where you started, so subtract the backward bits.

   6) How to compute from your sketch
   – Mark the times where v(t)=0 (where you turn around or stop).
   – Break the graph into intervals where v is all positive or all negative.
   – Approximate each area (triangles/trapezoids).
   – For displacement: add positives and subtract negatives.
   – For distance: take the absolute value of each and add.

   7) Tiny example to lock it in
   Say you ride 10 s at +5 m/s, then 4 s at −3 m/s.
   – Displacement = (5×10) + (−3×4) = 50 − 12 = 38 m forward.
   – Distance = 50 + 12 = 62 m.

   8) A small practical tip
   If your graph wiggles around zero due to noise when you’re basically stopped, don’t let tiny negative slivers mess up your totals. Treat a small band around zero as zero or smooth the data a bit. Otherwise you’ll “invent” back-and-forth distance while standing still.

   So, to answer your key question directly: yes, below zero means you’re moving the opposite way. Subtract those areas for displacement; take absolute values and add them for total distance. I can’t see your exact plot, but I’m pretty confident that’s why your “distance” came out tiny-you accidentally computed displacement.

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