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## Straight-line motion

Current time:0:00Total duration:10:09

# Total distance traveled with derivatives

## 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.