Main content

## AP®︎ Calculus AB (2017 edition)

# Motion problems: when a particle is speeding up

The position of a particle moving along the x-axis is given by s(t)=t³-6t²+9t. Sal analyzes it to find the times when the particle is "speeding up.". Created by Sal Khan.

## Want to join the conversation?

- If acceleration = s
`(t) is the second derivative of s(t), whats are we finding out in the third derivative or even fourth derivative of s(t) ?`

(3 votes)- The third derivative is the rate of change in acceleration. For example, if you keep the accelerator pedal in a car pressed to the floor, the car will eventually reach maximum speed and stop accelerating (or minimum speed, crumpled against a tree). Anyway, the third derivative is often (but not universally) called jerk, and the rate of change in jerk is (again, not universally) called jounce. Jerk and jounce can be important in some systems, such as controlling the flight of drones.(15 votes)

- Shouldn't we be speeding up between second and third second?(5 votes)
- Think of it this way: Between second and third the particle is still moving in the left direction (because velocity is negative) but you are accelerating in the right direction (because acceleration is positive, i.e. the slope of velocity is positive). So think of it as the particle is slowing down in the left direction and therefore you are not speeding up, you are actually slowing down. It is only when the particle switches direction to the right that it is speeding up (i.e. velocity turns positive at t > 3). Hope this helps!(8 votes)

- At6:25, why did he plug in 2 and 4 into the original equation to find the minimum?? Shouldn't he have just picked one number??(7 votes)
- The derivative is a second degree polynomial thus it is a parabola .......... when sal found its two roots at 1 and 3 ........ it is understood that the vertex of parabola will exactly be between them because the symmetry of parabola.... i.e. x=2 ....and for y value he plugged x=2 .....3(2)^2-12(2)+9.......3(4)-12(2)+9....(2 votes)

- Is the acceleration decreasing or increasing in the interval 1<t<2? is the acceleration=0 at t=1 and at t=2 because the slopes at these points are zero. if between t=1 n t=2 if it is becoming more and more negative how can it become zero at t=2(slope =0) because it is increasing in the negative direction?(4 votes)
- Your question "Is the acceleration decreasing or increasing" is asking about the change in acceleration. That concept is not needed to answer the question of when is the particle "speeding up", but it is a good question. On the interval 1 < t < 2 the acceleration is negative, but it is increasing. You can see this by imagining the tangent line to the velocity function in that interval. That tangent line is slanted down, so the slope of that line is negative, so the acceleration is negative. But as you move the tangent line to the right, its slope becomes less and less negative, the slope is increasing. Acceleration is the derivative of velocity. Sal didn't do this, but you can take the derivative of the velocity function and get the acceleration function:

v'(t) = a(t) = 6t - 12

From looking at the acceleration function you can also figure out the acceleration is negative but increasing from t = 0 to t = 2. From t = 0 to 2, the acceleration is going to be negative, at t = 2 the acceleration is zero, and at t > 2 the acceleration is positive. The function for acceleration is a linear function with a slope of positive 6, so the function is always increasing. Another way to see the acceleration is always increasing is to take the derivative of the acceleration function:

a'(t) = 6

So the acceleration is always increasing at a rate of 6/1.(7 votes)

- Since
`v(t) = ds / dt`

and`a(t) = dv / dt`

, could you write it as`a(t) = d(ds / dt) / dt`

?(4 votes)- Yes, but it is more common to write the last expression as
`d²s/dt²`

.(7 votes)

- How did Sal automatically know the vertex of the parabola was at (2, -1)?(1 vote)
- The vertex is halfway between the two x intercepts in the horizontal direction, always. Then he found out the vertical component of the vertex by plugging the horizontal component into the function.(5 votes)

- So the largest exponent on the original equation points out how many times is the particle speeding up? And the coefficient on the first term shows the magnitude of the speeding?(2 votes)
- Well, the short answer is "no and no".

The original equation describing movement on a line was

s(t) = t³ - 6t² +9t

In this case, there were two times when the particle was speeding up

That would be where the velocity and acceleration are both positive OR both negative in sign.

t(1, 2) Time between 1 and 2 seconds

t(3, ∞) Time greater than 3 seconds.

We do not count anything that happens before the clock starts ticking at time = 0

The coefficient on the first term doesn't show the magnitude of speeding. The coefficient on the first term is one. That doesn't match any way I can interpret your words, "magnitude of the speeding".

If you look at the graph of the acceleration (the rate of change for the velocity), it has a slope of 6. The only equation that has a coefficient of 6 on the first term is the second derivative of s(t), which is the equation for the acceleration, and of course its slope would equal that coefficient.

a(t) = s"(t) = 6t - 12

However, if you look at the slope of the velocity, at every point it is changing: after all, it is a parabolic function. Sal walked us through how the slope is negative and flattens out to zero and then becomes a little positive, and then very steeply positive. So if you mean from that graph that it is speeding up according to the coefficient of**that**equation, that also doesn't work. The equation for the velocity is v(t) = s"(t) = 3t²- 12t + 9

The main shortcut that I know is to quickly do the single and double derivatives and examine the behaviors of those curves. The powerful one for this purpose is the curve of the velocity or s'(t) as Sal showed us.(3 votes)

- At8:25, why is the particle speeding up between 1<t<2? Shouldn't it be slowing down since the particle's velocity reaches 0 at t=2?(1 vote)
- The particle velocity is -3 at time 2 (
`v(2)= -3`

) That is the minimum velocity of the particle, so the particle was accelerating (negatively) up to that point.(4 votes)

- How can we say that velocity is increasing only when it goes to right direction and decreases when it goes to left?(2 votes)
- We don't. We say that the velocity is increasing when there is positive acceleration and when there is positive acceleration the velocity will either climb to higher negative numbers ( which is to say from -7 to -2 for example) if negative and will climb to higher positive numbers if positive. If your question was why did Sal define velocity increasing in the rightward direction, it is because in this video we picked the convention that if velocity is positive and increasing then it is travelling rightward.(1 vote)

- In this video Sal says that for the particle to be "speeding up" (or in other words the magnitude of the velocity must be increasing). He states that the particle is "speeding up" when 1 < t < 2. However surely the object is speeding up at t = 1 as the velocity = 0 but the acceleration is negative and so its velocity will increase. Similar logic should also apply when t = 3 Therefore shouldn't the answer be 1 ≤ t < 2, t ≥ 3? Thanks for your replies in advance!(2 votes)
- I’m not sure I got it right but here is my reasoning:

At t = t₀ (in our problem, 1 or 3), acceleration is non-zero, but velocity is zero. It means that our point is not moving − yet. It is about to move! At t = t₀ + δt, velocity becomes non-zero and we can at last say that the point is speeding up.(1 vote)

## 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.