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# Motion problems: when a particle is speeding up

Video transcript

Let's say that we have
some particle that's moving along the number line. So let me draw a number
line right over here. So that's our number
line right over there. And let's say it starts
right over here at 0. And then as time passes,
this little point is going to move around. Maybe it moves to the right,
slows down, speeds up. Maybe it moves to the left,
slows down, speeds up. It might do all sorts of things. And to describe this motion, its
position as a function of time, we have a function s of t. This particle's position as a
function of time we're given is t to the third power
minus 6t t squared plus 9t. And we're going to restrict
the domain to positive time. So we're going to
assume that time is greater than or equal to 0. Now the question that we want
to answer in this video is, when is this
particle speeding up? So when are we speeding up? And I think that bears
a little clarification. What does it mean to speed up? Well, there's two scenarios. If the particle
is already moving in the rightward
direction-- and the way we would know it's moving in
the rightward direction is if its velocity
is greater than 0. If it's moving in the
rightward direction and it's also accelerating
in the rightward direction-- so if its acceleration is
also greater than 0-- then this is a situation
where we are speeding up. Now another scenario where
we would be speeding up is if we're moving in
the leftward direction. In that case, our velocity
is going to be negative. So if our velocity
is negative and we want to go faster in
the negative direction, then our acceleration
should also be negative. That would make our velocity
getting more and more and more negative with time. So then our acceleration
needs to also be negative if we still want
to be speeding up. If you have any other
combination here, if your velocity is negative but
your acceleration is positive, that means that your velocity
is becoming less negative, or you would be slowing
down the leftward direction. And vice versa, if your
velocity is positive and your acceleration
is negative, that means you're
going to the right but you are slowing down
in the rightward direction. So let's think about
these two scenarios. And since velocity
matters here so much, we just have to remind ourselves
that the velocity-- remember, a derivative is just the
rate of change with respect to a variable. So if you have your position
function, the derivative of position with
respect to time, this is really just what is the
instantaneous rate of change of position with
respect to time? Well, what is the change of
position with respect to time? Well that is just going to be
equal to our velocity function. That's going to be equal to
our velocity function, v of t. Or we could write s prime
of t, which could be also written this way, as ds dt,
is equal to our velocity as a function of time. So let's take the
derivative of this. Our velocity as a
function of time is going to be equal to 3t
squared minus 12t plus 9. So let's see if we can
graph this velocity function to start
making sense of it. When is the velocity positive? When is it negative? And what's the acceleration
doing in those intervals? And so to help me graph it,
we could say the v-intercept, or the vertical
intercept, when v of 0 is going to be equal to 9. So that'll help us graph it. That's where we intersect
the vertical axis. But also, let's plot--
let's figure out where it intersects the t-axis. So let's set this equal to 0. So 3t squared minus 12t
plus 9 is equal to 0. Let's see. To simplify this, I can
divide both sides by 3. And I get t squared minus
4t plus 3 is equal to 0. Now this is very factorable. This is t. Let's see. What two numbers, when
you take a product, get 3, and when you add them,
you get negative 4? Well, that's going to be
t minus 3 times t minus 1 is equal to 0. How can this expression
be equal to 0? Well if either of these are
equal to 0, if either t minus 3 is 0 or t minus 1 is 0,
it's going to be equal to 0. So t could be equal to 3,
or t could be equal to 1. If t is 3 or t is 1, either
of these are equal to 0, or this entire expression up
here is going to be equal to 0. And since our coefficient on
the t squared term is positive, we know this is going to be
an upward opening parabola. So let's see if we can plot
velocity as a function of time. So that is my velocity axis. This right over here
is my time axis. And let's say this
is 1 times 1 second, or I'm assuming this is
in seconds-- 2, 3, 4. Actually, let me spread
them apart a little bit more just because 1 and 3 are
significant-- 1, 2, and 3. And they're not
going to be-- I'm going to squash to the
vertical scale a little bit. But this right over here, let's
say that is 9, a velocity of 9. And so when t equals
0, our velocity is 9. When t equals 1, then our
velocity is going to be 0. We get that right over here. 3 minus 12 plus 9, that's 0. And when t is equal to 3
our velocity is 0 again. Our vertex is going to be
right in between those, when t is equal to 2-- right
in between these two 0's. And we could figure out what
that velocity is if we like. It's going to be 3 times
4 minus 12 times 2 plus 9. So what is that? That's 12 minus 24 plus 9. So that is negative 12 plus 9. So that's going to be
equal to negative 3. Did I do that-- 12,
yep, negative 3. So you're going to be--
negative 3 might be-- that's 9, so that's positive. So it might be
something like this. So the graph of our velocity
as a function of time is going to look
something like this. And we only care
about positive time. It's going to look
something like this. So let's think. Remember, this is velocity. This is our velocity
as a function of time. Now let's think about when
is the velocity less than 0 and the acceleration
is less than 0? So let's think about this
case right over here? When is this the case? Both of them are going
to be less than 0. Well, velocity is
the less than 0 over this entire interval,
this entire magenta interval. But the acceleration isn't
less than 0 that entire time. Remember, the acceleration is
the rate of change of velocity. We can write here
that acceleration as a function of time, this
is equal to the rate which velocity changes
with respect to time. Or we could write,
acceleration is equal to v prime of t,
which is the same thing as the second derivative of
position with respect to time. And so the acceleration,
you could really think of the slope of the
tangent line of the velocity function. And so over here, the place
where this is downward sloping, where this
has a negative slope, and the curve itself
is below the t-axis, that's only over this
interval right over here. Between this 0 right
over here and the vertex, we get to this point
right over here. And then our slope flattens out. So this interval
right over here is t is going to be
greater than 1, and it is going to be less than 2. That meets these constraints. Now let's think about where
our velocity is greater than 0 and our acceleration
is greater than 0. Well our velocity is
greater than 0 over here. But notice, our acceleration,
the slope here is negative. We're downward sloping,
so that doesn't apply. Here our velocity
is greater than 0, and the slope of the velocity,
the rate of change of velocity, the acceleration, is
also greater than 0. So that's this interval
right over here, where we're speeding up in
the rightward direction. So that interval is
t is greater than 3. So when are we speeding up? We're speeding up between
the first and second seconds, and then we're speeding
up after the third second.