- Introduction to one-dimensional motion with calculus
- Interpreting direction of motion from position-time graph
- Interpreting direction of motion from velocity-time graph
- Interpreting change in speed from velocity-time graph
- Interpret motion graphs
- Worked example: Motion problems with derivatives
- Motion problems (differential calc)
- Analyzing straight-line motion graphically
- Total distance traveled with derivatives
Analyzing straight-line motion graphically
Learn how to analyze a particles motion given the graph of its position over time. Created by Sal Khan.
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- What does this have to do with calculus?(0 votes)
- Calculus is the mathematics of change. Velocity is a derivative of position, etc. It is a rate too. It is how much does the position change per unit of time.(28 votes)
- Can someone summarize the meaning of this graph for me? I know the connection between the functions:
a(t). However, I am confused with the positive and negative thing.
For example: When
t > 3, the position decreases but why does the velocity slightly increase? If the velocity keeps increasing (for example
v > 0), what will happen to the position?(4 votes)
- Well, let's make up a function and some characteristics in order to more easily understand this. It goes from 0 to -1 during time (0,1), and then -1 to -1.5 during time (1,2). So, during the first time interval, the velocity is -1. In the second time interval, the position is still decreasing but at a slower rate, so the velocity is now -0.5. It is harder to see in this video, but you can see that the position is decreasing at a slower and slower rate as time goes by. That is why the velocity then becomes less and less negative, and the acceleration actually becomes slightly positive.(5 votes)
- At3:26, why isn't 1 included but 0 is?(2 votes)
- We include 0 because the graph indicates positive velocity at that point, which means the particle is moving to the right. We exclude 1 because the graph indicates velocity is zero at that point, so the particle can't be moving to the right (it's not moving at all).(2 votes)
- I understand how calculus is used to analyze particle motion here, but, in the real world, how would you get the position function in the first place?(3 votes)
- In essence, start with F=ma and integrate twice.(6 votes)
- Why does the acceleration cross the x axis at 2 instead or 1? Surely the acceleration would be 0 at t=1?(2 votes)
- At t=1 velocity (red line) keep changing, So the acceleration cannot be 0 at this moment. It seems that velocity get to a constant level (no changes) closer to t=2.(1 vote)
- Hello, how can a velocity (also acceleration) be negative (at some points)? thanks Irena(1 vote)
- A negative velocity means that you are going backward. A negative acceleration means you are slowing down.(3 votes)
- For the last part of the question, finding total distance traveled, Sal used the position function. While we don't have the necessary equations, would he have gotten the same answer if found the absolute value of the area under the velocity curve over the same time period?(1 vote)
- Yes. Because the velocity is the derivative of the position function, the position function is the "antiderivative" of the velocity function, also known as the indefinite integral of the function. This indefinite integral can be used to find the area underneath the curve of the velocity curve and will give you the total distance traveled.(2 votes)
- In what kind of situation do these values actually come from. Literally particles such as electrons and stuff moving? Or is this all made up for the sake of the question?(0 votes)
- Well, these ideas can be translated to any time of movement or graph that represents movement. It can be represented about our movement in a car for instance. However, this isn't always 'fun' in mathematics so we use particles.
But it is very, very important in physics and science in general.(4 votes)
- Sal,in the 2nd question for the time interval, why is only the zero in square brackets and not one?(0 votes)
@ t = 1, v = 0
Which means the direction, in this case evaluating for right direction, is when
v > 0and
v ≠ 0; to translate these restrictions in the time domain, then we can have
t ≥ 0and
t < 1or
0 ≤ t < 1.
The interval notation
t ∈ [0, 1)is equivalent to
0 ≤ t < 1.(4 votes)
- Why the position equation is quadratic?(1 vote)
A particle moves along a number line not shown for t is greater than or equal to 0. Its position function, s of t, is shown in blue. So this is its position as a function of time. Its velocity function, v of t, is in red. That's velocity. And its acceleration function, a if t, is in green. All are graphed with respect to time t in seconds. With the graphs as an aid, answer the questions below. So that's what's going on here. So it's position as a function of time. Actually, let me just draw their number line that they did not depict just so we can really think about this. So let's say that's our number line. Let's say this right over here is 0. That's 1. That is 2. This is negative 1. So we're defining going to the right as the positive direction. So what's happening here? So at time equals 0 right over here, s of 0 is 0. And then as time increases, our position increases all the way until time equals 1. At time equals 1, our position is 2. So at time equals 1, our position is 2. And then our position, s of t, starts decreasing. So one way to think about it is, and you see we move up, we move to the right really fast. We get to 2. We stop at 2, and then we start moving to the left. So at time equals 0, the first second looks like this. We go zoom, oh, slow down and stop. And then we start moving the other way. And then we start drifting. Notice our position is decreasing. So our position is decreasing, but it's decreasing at ever slower, slower, and slower and slower rates. It's not clear if we'll ever get back to the origin. So that's what's going on here. And we see that no matter which graph we look at. Our position function is definitely telling that story. Our velocity function, which is the derivative of the position function, is telling that story. Out the gate, we have a high positive velocity, but we decelerate quickly. And at 1 second, our velocity is 0, and then we start having a negative velocity, which means we're moving to the left. So fast rightward velocity, but we decelerate quickly, stop at time equals 1 second, and then we start drifting to the left. And the acceleration also shows that same narrative. But anyway, let's actually answer the questions. The initial velocity of the particle is blank units per second. I encourage you to pause this video and answer that. Well, we just said the velocity, let's see, at time equals 0, we're at 8 units per second. So we'll just put 8 right over there. The particles moving to the right when t is in the interval, and since they're doing this as a member of, they really want this kind of in the set notation t is a member of the interval. Well, when are we moving to the right? We already went over that. We're moving it to the right-- there's a couple of ways to think about it. When our velocity is greater than-- so we're moving to the right when v of t is greater than 0. When v of t is less than 0, we're moving to the left. When v of t is equal to 0, we're stationary. So when is v of t greater than 0? Well, it's between t being 0, velocity is definitely positive, all the way to t is 1, but not including t is 1. So I'll put a parentheses there. So this is equivalent to saying-- so t is a member of that interval is equivalent to saying that 0 is less than or equal to t is less than 1. Once again, the first second, at time 0, we're going fast, slow down, and then stop for an infinitesimal moment, and then we start drifting back. That happens at time equals 1. We start drifting back. The total distance traveled by the particle for t in the interval between 0 and 3 is blank units. So once again, I encourage you to pause the video and try to answer that, the total distance. So this is interesting. Don't get distance confused with displacement. If I were to move three to the right and then I were to move back one to the left, the total distance I've traveled is four. The distance I traveled is four, while the displacement would be a positive. We could maybe put a minus 1 there. We moved one to the left. So three to the right, one to the left. Well, our displacement would be a net of positive 2. So they're asking, what's the total distance traveled? So between time 0 and time 1, we have moved two to the right. And then between time 1 and time 3, we move back, or to the left, we move half. So to the left, we move half. So what's our total distance? It's going to be two to the right plus half to the left, which is going to be 2.5 units. And we're done.