Main content

## Calculus 1

### Course: Calculus 1 > Unit 4

Lesson 2: Straight-line motion- 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

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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?(34 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.(28 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?(2 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.(7 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)
- at5:15, 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)

- How is that possible that at t=0 disance is zero but velocity is not zero?(3 votes)
- It is NOT!

At1:35The velocity function is derived (derivative of position function) and if you input t=0, you get v=10. Sal even graphs this at4:10

In any event, we are interested in total*distance*, so how fast or slow the particle was traveling is irrelevant, we just want the total distance traveled between time t=0 and t=6(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.