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# Total distance traveled with derivatives

Given a function representing the position of a particle over time, how can you find the total distance traveled? Created by Sal Khan.

## Want to join the conversation?

• Hi I have a question. There was no explanation in the video why he used differential before solving problem ? I mean, what relation have between calculating distance of volacity of the fuction in the given arrange of t and using differential?
(35 votes)
• Velocity is rate of change of position with respect to time. In other words, the derivative of position wrt time
Hence he used differentials.
Hope this helped
(23 votes)
• Am I crazy or would simply taking the integral of 0<t<1 (multiplied by two because its symmetrical from the interval 5<t<6) and adding it to the positive integral from 1<t<5 not give the total distance traveled over the interval?
(12 votes)
• Well, not all of us know that method. This is the derivatives section not integrals.
(30 votes)
• Can this topic "motion of a particle along axis" be related to quantum mechanics?
(4 votes)
• Yes. That's essentially what quantum mechanics is about, finding the equations of motion for particles. But this is extremely simplistic compared to real quantum mechanics. This is more suitable to basic Newtonian physics.
Remember, we call just about everything that can move a "particle" in physics: cars, bugs, electrons, people, flying jar of pickled pineapple...
(17 votes)
• A few questions to help clarify the concept. First, v(6) would give the net distance, right? Second, would finding the arc length of s(t) be one of way solving this? Third, why and how are the maxima and minima of s(t) related to solving this problem?

Edited to Add: Actually, I just realized something. Do minima and maxima represent points at which the function s(t) crosses 0 on a number line?
(5 votes)
• No, minima and maxima are points where the particle turns left from going right or turns right from going left.
(5 votes)
• what was the point of drawing the velocity graph here?
(3 votes)
• Since the problem said that the particle moved in both directions, sal had to find out on what intervals of time it was moving in what direction. The graph allows you to visualize when the velocity of the particle is positive or negative (the particle is moving right or left). You then take the x-intercepts and the endpoints and find the current displacements using the original equation. You use the x-intercepts because these are the values of time at which the particle is changing direction and this will tell you the extremes of the displacement graph.
(8 votes)
• The derivative of position graph is the velocity graph, and the derivative of the velocity graph is the acceleration graph, and the derivative of the acceleration graph is something called jerk? Is that how everything relates to each other?
(4 votes)
• Yes - that is how they relate to each other via the process of differentiation.
Can you figure out how they relate to each other via the process of integration?
(3 votes)
• If u integrate the velocity and find the area under the curve. Would it be equal to the answer sal got?
(2 votes)
• Not quite, in this case, only because the velocity curve is both positive and negative on the interval. If you integrate the absolute value of velocity (which is speed), then you get the total distance traveled. If you integrate just velocity, you get total displacement (how far apart the starting and ending positions are from each other) rather than the total distance the particle moves between the starting and ending times. Does that help?
(5 votes)
• Wouldn't it make much more sense to use an integral? Is this just to help practice derivatives, or is there ever going to be an instance where I have to use a derivative instead of an integral to find distance traveled (aka area under velocity curve)?
(4 votes)
• at , the function appears on the graph to have a position of 10 at t=0, but in his chart he makes it have a position of 0?
(3 votes)
• In case you still haven't found an explanation, the graph Sal drew (upward-facing parabola, where v(0)=10) is the graph of VELOCITY, not position. That's why, although the POSITION at t=0(sec) is equal to 0, the velocity at t=0(sec) is equal to 10.

In this problem, the position is calculated using the formula: s(t)=2/3t^3-6t^2+10t (which indeed gives you 0 for t=0), while the velocity is given by v(t)=2t^2-12t+10. You get the first formula from the task and the second by finding the derivative ds/dt of the first.

Motivation behind this: By definition, velocity is the rate (and direction, which means sign here: "+"=to the right; "–"=to the left) of change of position. Derivative is the tool used to figure it out.
(2 votes)
• idk this was kinda of a bad tutorial.... I'm still extremely confused on why we used the derivative? Specifically I am trying to apply this concept to the problem " position function: t^2 - 6t ; find the distance traveled from t=2 to t=4"
(3 votes)
• using the same method

1. f(x)=t^2-6t
-> f'(x)=2t-6
-> when t=3, f'(x)=0 meaning the velocity is 0 and the direction of motion shifts
2. f(2)=-8, f(3)=-9, f(4)=-8
3. |f(2)-f(3)| + |f(4)-f(3)| = 1 + 1 = 2

thus, the total absolute distance the particle moves is 2 units
(1 vote)

## Video transcript

The position of a particle moving along a number line is given by s of t is equal to 2/3 t to the third minus 6t squared plus 10t, for t is greater than or equal to 0, where t is time in seconds. The particle moves both left and right in the first 6 seconds. What is the total distance traveled by the particle for 0 is less than or equal to t is less than or equal to 6? So let's just remind ourselves what they mean by total distance. If I were to say start there, and if I were to move 3 units to the right and then I were to move 4 units to the left, and I'll say negative 4 to show that I'm moving to the left, then my total distance right over here is 7. 3 to the right and 4 to the left. Even though my position right over here is going to be negative 1. Or you could say my net distance, or you could say my displacement is negative 1. I'm 1 to the left of where I started. The total distance is 7. So now we've clarified that. I encourage you to now pause this video and try to answer the question. What is the total distance traveled by the particle in these first 6 seconds? So the easiest way I can think of addressing this is to think about, well, when is this thing moving to the right and when is it moving to the left? And it's going to be moving to the right when the velocity is positive, and it's going to be moving to the left when the velocity is negative. So this really boils down to thinking about when is the velocity positive or negative. And to think about that, let's actually graph the velocity function or make a rough sketch of it. So this is the position function. The velocity function is going to be the derivative of the position function with respect to time. So the derivative of 2/3 t to the third is going to be 2t squared. And then we have minus 12t plus 10. And so let's just try to graph this. This is going to be an upward opening parabola. This is clearly a quadratic. And the coefficient on the second degree term, on the t squared term, is a positive number, so it's going to be an upward opening parabola. It's going to look something like this. And we're assuming that it switches direction. So it's going to be positive some of the time and negative for some of the time. So it should intersect the t-axis where it's negative. The function is going to be negative in that interval, and it's going to be positive outside of that interval. So the easiest thing I could think of doing is let's try to find what the 0's are. Then we can draw this upward opening parabola. So to find its 0's, let's just set this thing equal to 0 so we get 2t squared minus 12t plus 10 is equal to 0. Divide both sides by 2 just to get rid of this 2, make this leading coefficient a 1. We get t squared minus 6t plus 5 is equal to 0. That made it a lot easier to factor. This can be factored into t minus 1 times t minus 5. Negative 1 times negative 5 is 5. Negative 1 plus negative 5 is negative 6. This is equal to 0. So this left hand side of the equation is going to be equal to 0 if either one of these things is equal to 0. Take the product of two things equaling 0, well, you get 0 if either one of them is 0. So either t is equal to 1 or t is equal to 5. So now let's graph it. So let's draw our axes. So I could say that's my velocity axis. And let me draw the-- we only care for positive values of time. So let's draw something like this. Positive time. And let's see. Let's take that 1, 2, 3, 4, 5. We could keep going. So this is t equals 1. This is t is equal to 5. This is our t-axis. And let's graph it. So it's going to be an upward opening parabola. It's going to intersect both of these points. And so its vertex is going to be when t is equal to 3 right in between those points. So it's going to look something like this. That's the only way to make an upward opening parabola that intersects the t-axis at both of these points. So it'll go like that, and it'll go like this. It'll intersect. When t equals 0, we actually can figure out. When t equals 0 our velocity is 10. So the v-intercept, we could say, is 10 right over here. So that's what it looks like. So we see that the velocity is positive for time between 0 and 1. And it's also positive for time is greater than 5 seconds. And we see that our velocity is negative, or that we're moving to the left, between 1 and 5 seconds. The velocity is below the t-axis right over here. It is negative. So let's just think about what our position is at each of these points, at time 0, at time 1, at time 5, and what we care about time 6. And then think about what the distance it would have had to travel between those times. So let's think about it. So let's make a little table here. Let's make a little table. So this is time, and this is our position at that time. So we care about time 0, time 1, time 5 seconds, and time 6 seconds. So at 0 seconds, we know that our position is 0. S of 0 is 0. At 1 second, this is going to be 2/3 minus 6 plus 10. So it's going to be 4 and 2/3. So I'll write down 4 and 2/3. At 5 seconds, let's see, it's 2/3 times-- I'm going to write this one down. So it's going to be 2/3 times 125. That's the same thing as 250 over 3, which is the same thing. Let's see, 250 over 3. That's the same. 83 times 3 is 249, so this is 83 and 1/3. That's this first term. Minus 6 times 25. So that's going to be minus 150 plus 10 times 5. So plus 50. So this is going to simplify. Minus 150 plus 50, that's going to be minus 100. 83 and 1/3 minus 100. That's going to be negative 16 and 2/3. So negative 16 and 2/3 is its position after 5 seconds. And then at the 6 seconds, it's going to be 2/3 times 6 to the third. I have to write this one down. 2/3 times 6 to the third minus 6 times 6 squared. Well, that's just going to be minus 6 to the third again. 6 times 6 squared plus 60. And let's see. How can we simplify this right over here? Well, this part right over here we can rewrite as-- we could factor out as 6 to the third. This is 6 to the third times 2/3 minus 1 plus 60. Scroll down a little bit and get some more space. So it's going to be 6 to the third times negative 1/3 plus 60. And let's see. Let's write it this way. This is going to be 6 squared times 6 times negative 1/3 plus 60. This right over here is negative 2. So it's negative 2 times 36. This is negative 72 plus 60. So this is going to be negative 12 right over there. So now we just have to think, how far did it travel? Well, it starts traveling to the right. It's going to travel to the right 4 and 2/3. So let's write this down. So we're going to have 4 and 2/3. And then it's going to travel to the left. Let's see, to go from 4 and 2/3 to negative 16 and 2/3, that means you traveled 4 and 2/3 again. You traveled 4 and 2/3 to the left, and then you traveled another 16 and 2/3 to the left. Just a reminder, we're 4 and 2/3 to the right now. We have to go 4 and 2/3 to the left back to the origin, and then we have to go 16 and 2/3 again to the left. So that's why this move from here to here is going to be 4 and 2/3 to the left followed by 16 and 2/3 to the left. Another way to think about it, the difference between these two points is what? It's going to be 4 and 2/3 plus 16 and 2/3. If you do 4 and 2/3 minus negative 16 and 2/3, you're going to have, that's the same thing as 4 and 2/3 plus 16 and 2/3. And then to go from negative 16 and 2/3 to negative 12, that means you went another 4 and 2/3 now to the right. So now this is 4 and 2/3. Now you're moving 4 and 2/3 to the right. And so we just have to add up all of these. We just have to add up all of these values. So what is this going to be? So this is going to be 2/3 times 4, so this part of it right over here, the fraction part of it. 2/3 times 4 is 8 over 3. And let's see, 4 plus 4 plus 16 plus 4 is 28. So 28 and 8/3, that's a very strange way to write it. Because 8/3 is the same thing as 2 and 2/3. So 28 plus 2 and 2/3 is 30 and 2/3. So the total distance traveled over those 6 seconds is 30 and 2/3 units.