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# Position vs. time graphs

AP.PHYS:
CHA‑4.A (EU)
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CHA‑4.A.1 (EK)
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CHA‑4.A.1.1 (LO)
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CHA‑4.A.2 (EK)
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CHA‑4.A.2.1 (LO)
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CHA‑4.A.2.3 (LO)
How to read a position vs. time graph. Using the graph to determine displacement, distance, average velocity, average speed, instantaneous velocity, and instantaneous speed. Created by David SantoPietro.

## Want to join the conversation?

• Why don't physicists put position on the x axis and time on the y axis? •  Because it leaves room for the graph to fail the vertical line test. Then it would not longer be a function; rather an inverse function.
• Wouldn't it be quite confusing to call the VERTICAL axis X? •   Yeah, I was going to mention this in the video but it got a little long. It is really unfortunate that the horizontal (independent variable) is always called x in math class. In physics there is a convention that if time is involved, the horizontal axis is almost always selected to be time. This means that if you graph any variable vs. time, including horizontal position which we call x, it will have to go on the vertical axis. This is definitely confusing, but that is the convention physicists have selected.
• Thank you so much. you're amazing.
I just have one question. What if we wanted the instantaneous velocity at 4 seconds. which slope should we get ? • The function isn't differentiable at that point, because the functions is switching sides. This is because the left derivative and the right derivative don't match, leaving an awkward function. It is continuous but not differentiable (or slopable, if you wish to call it that). This kind of motions is NOT possible irl. No thing can switch directions (forward and backward) at that speed that quickly. irl, there would be some kind of a curve.
• Hey i actually have 3 questions to ask ...
Q1 At he said the turtle is 3m , but 3m from where ?
Q2 When he says that the turtle travels 3m , from where should the distance travelled be counted from, the head or the end..??
Q3 When he wrote down the distance traveled by the turtle in 2 seconds i.e 8 m ,isnt it wrong because when we use the distance formula between any two we points the answer as =√68..?? and the points are (2,3) and (4,-5)..??
pls help im confused.. • A1: 3m from an arbitrary point in space. for example it could be 3m from a starting line on a track.
A2. From center of mass (note the white line through the turtles shell) is typically used, but as long as consistency is maintained any singular part of the turtle would work (i.e. measure from when his nose crosses the 3m mark or his tail, but not both). The underlying principle is to use the simplest model (a point mass) unless further detail is required (For example: if the turtle were racing a hare and we needed to know who crossed the 5m mark first, we'd use the head as it's the most forward part of the turtle's body).
A3. The distance traveled is only the x direction (the y axis on the graph- admittedly confusing, but it is the independent variable) so from the 2sec to the 4 sec points on the graph the x variable changes 8 points total (from positive 3 to negative five). So it's like the turtle started his race ahead of the starting line, turned around went back behind the starting line and traveled a magnitude of 8m in the process.
• Is a position vs time graph the same a distance vs time • No. Distance-vs.-time graphs only account for the total movement over time. Position-vs.-time graphs note one's position relative to a reference point (which is where x=0 on the graph in the video).

Here's an example of the difference: A tennis player hits a ball to a wall 5 meters away, and the ball bounces back the same distance. If the reference point is where the tennis player is standing, the position-vs.-time graph for the ball would start at 0, move up to 5 meters when the ball hits the wall, then drop back down to 0 on the vertical axis when the player catches the ball again. It's right back where it started.

However, the distance traveled would be 10 meters total: 5 meters forward and 5 meters back. (Total distance does not take direction of motion into account.)
• If the Motion line is denoted in a parabola....How will we fine the instantaneous velocity? Or anything else? • At time equals 2 and 4 seconds, we can't find the instantaneous velocity/ speed because it's not differentiable at those two points. I understand that mathematically, but I'm sort of struggling to grasp that idea intuitively. So, the concept of an instantaneous velocity is basically the derivative of the curve at a certain point right? And to find the derivative, we need to be able to find the limit as a certain value approaches zero at that specific point. At both 2 seconds and 4 seconds, the limits of the graph exist, but not the limits of the slopes at those points (am I stating this part correctly?). So basically, all I understand is that we can't find the instantaneous velocity at 2 and 4 seconds simply because finding the derivatives at those points isn't possible, but I sort of want a better picture or idea in my head as to why we can't find the instantaneous velocity at points with sharp turns. • "At both 2 seconds and 4 seconds, the limits of the graph exist, but not the limits of the slopes at those points (am I stating this part correctly?)."

You have got it right. The function is continuous but the limits of the slopes at the points in question doesn't exist because the left-hand limit and the right-hand limit are not equal. If you draw the tangents on both sides of the point, you get two lines headed in different directions and therefore failing to be a tangent by definition.

What does this all mean in the real world? It means you can't be driving your car north at one instant and magically switch direction toward the west in the next instant. There will always be a smooth curve that would indicate that you safely turned the car left. What the graph is showing is not physically possible and the particle fails to have an instantaneous velocity at the corner where the displacement function is not a smooth curve but a sharp turn. Note that from time t = 0 to an instant before t = 2, the turtle was stationary but as soon as t = 2 seconds, the same stationary turtle now also seems to have a huge velocity along the negative x axis. This can't be possible in the real world. Any amount of change, no matter how large or small, takes over some finite amount of time, however small the time interval may be. You can't have two states of motion in the same instant well, at least not in Newtonian mechanics. Though Quantum mechanics is another game altogether, but I digress. :)

Hope it helps.
• hey what is ''x'' in this video? the distance??   