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### Course: Calculus, all content (2017 edition)>Unit 3

Lesson 11: Rectilinear motion

# 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?
• 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
• 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?
• Well, not all of us know that method. This is the derivatives section not integrals.
• Can this topic "motion of a particle along axis" be related to quantum mechanics?
• 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...
• 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?
• No, minima and maxima are points where the particle turns left from going right or turns right from going left.
• what was the point of drawing the velocity graph here?
• 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.
• 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?
• 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?
• If u integrate the velocity and find the area under the curve. Would it be equal to the answer sal got?
• 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?
• 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)?
• 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?
• 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.
• 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"
• 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)