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### Course: AP®︎/College Calculus AB > Unit 5

Lesson 11: Solving optimization problems- Optimization: sum of squares
- Optimization: box volume (Part 1)
- Optimization: box volume (Part 2)
- Optimization: profit
- Optimization: cost of materials
- Optimization: area of triangle & square (Part 1)
- Optimization: area of triangle & square (Part 2)
- Optimization
- Motion problems: finding the maximum acceleration

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# Motion problems: finding the maximum acceleration

The speed of a particle moving along the x-axis is given by v(t)=-t³+6t²+2t. Sal analyzes it to find the time when the particle's acceleration attains its maximum value.

## Want to join the conversation?

- At3:35from what I understand Sal is trying to use the 2nd derivative to demonstrate that t=2 is the time of maximum acceleration, but it's not clear to me what he's using to arrive at that decision.

Can anyone point me in the right direction?

Thank you!(5 votes)- Whenever we think of the first derivative, we think of critical points and whenever we think of the second derivative, we think of inflection points. Since our problem is about acceleration, the thing we must realize is that Sal treats acceleration as if it was the function we started with. Thus, don't always think of acceleration as a derivative. In this problem, we think of acceleration as the starting equation. Now, in his past vids, we know that a function reaches a maximum or minimum value at a critical point (if you are new to this information, I suggest looking at his past vids on critical and inflection points). The critical point is just where the first derivative of a function equals 0. Thus, Sal finds where the derivative of acceleration equals 0 because all maximum/minimum values are critical points. Since he got only one answer which happened to be 2, he wanted to make sure it was a maximum point. He checked his answer by using the second derivative test which states only if the first derivative of a function equals 0 and its second derivative is greater than 0, then that value is a local minimum, but if the first derivative is equal to 0 and its second derivative is less than 0, then that value is a local maximum. The second derivative of acceleration would have been -6 which is less than 0, so according to the second derivative test, it proves that 2 was the maximum value of acceleration. Thus, it is important to not always think of acceleration as a derivative, but also as a starting equation for the remainder of the problem. Sorry I made it so long :)(18 votes)

- Is there a tutorial video on the equation of motion, such as average velocity and instantaneous velocity?(5 votes)
- There are some videos about that in the physics playlist.

I think this is the exact video you're looking for: https://www.khanacademy.org/science/physics/one-dimensional-motion/displacement-velocity-time/v/instantaneous-speed-and-velocity(4 votes)

- Can't you just complete the square to find the vertex for the maximum value?(3 votes)
- For many quadratics you can, however, like in real world some numbers and Quadratics aren't nice. Sometimes it is nice to have another tool in your toolbox.(7 votes)

- wouldnt the second derivative of acceleration be the jerk? If so what does it tell us about the particles movement(1 vote)
- The first derivative of acceleration is jerk, the second derivative is called jounce, or snap. What is tells us is how fast the jerk is changing (the more derivatives we take, the more abstractly we have to think to make sense of what they mean, so snap doesn't tell us very much, intuitively.)(4 votes)

- Do you the same thing when doing integrals? Example: find the maximum displacement given acceleration.(2 votes)
- A body starting from rest has an acceleration of 5metre per second square. How to calculate distance travelled by it in 4th second?? Please explain in detail.(1 vote)
- simply calculate the definite integral between t=0 and t=4....which will be distance traveled(3 votes)

- So, the maximum acceleration is at t=2. Is it always that the derivative of v(t) gives the maximum acceleration? What if when finding the minimum acceleration?(1 vote)
- The derivative of v(t) gives acceleration as a function of time. Once we have that function, we can use the calculus tools we're developing to find its maxima and minima.(2 votes)

- Is this formula specific for this situation or is it general and can fit any case?

I'm looking for a way to calculate the time required for an object to reach maximum velocity without using the kinematic formulas(1 vote)- No, this question just used a random function which we had to solve, it has nothing to do with physics.(1 vote)

- When you get a(t)=-3t^2+12t+2, can't you just find the maximum value with -b/2a for ax^2+bx+c(1 vote)
- So if the change in position per time is velocity, and the change in velocity per time is acceleration, and I know that the change in acceleration per time is the impulse, what is the change in impulse per time?(1 vote)
- Change in acceleration per unit time is "jerk", not impulse. Change in jerk per unit time is "snap". And it continues on from there:

http://iopscience.iop.org/article/10.1088/0143-0807/37/6/065008(1 vote)

## Video transcript

- [Voiceover] A particle
moves along the x-axis so that at any time T
greater than or equal to zero its velocity is given
by V of T is equal to negative T to the third power
plus six T squared plus two T. At what value of T does the particle obtain its maximum acceleration? So we want to figure
out when does it obtain its maximum acceleration. So let's just review what they gave us. They gave us velocity
as a function of time. So let's just remind ourselves. If we have let's say our
position is a function of time, so let's say X of T is
position as a function of time, then we if were to take
the derivative of that, so X prime of T, well that's going to be the
rate of change of position with respect to time or the
velocity as a function of time and if we were to take the
derivative of our velocity, then that's going to be the
rate of change of velocity with respect to time. Well that's going to be
acceleration as a function of time. So they gave us velocity. So from velocity, we can
figure out acceleration. So let me just rewrite that. So we know that V of T is equal to negative T to the third power
plus six T squared plus two T and so from that we can figure out the acceleration as a function of time, which is just going to be the derivative with respect to T of the velocity. So just use the power rule a bunch. So that's going to be, this
is a third power right there. So negative three T squared plus two times six is twelve T to the first plus two. So that's our acceleration
as a function of time and we want to figure out when we obtain our maximum acceleration and just inspecting this
acceleration function here, we see it's a quadratic, it has a second degree polynomial and we have a negative coefficient out in front of the highest degree term, in front of the second degree term. So it is going to be a
downward opening parabola. So it is going to be a downward opening, let me draw it in the same color, so it is going to have that general shape and so it will indeed take on, it will indeed take on a maximum value. But how do we figure
out that maximum value? Well that maximum value's going to happen when the acceleration value, when the slope of its
tangent line is equal to, when the slope of its
tangent line is equal to zero and we could also verify
that it is concave downwards at that point using the
second derivative test by showing that the second
derivative is negative there. So let's do that, let's look at the first and second derivatives of
our acceleration function. So and I'll switch colors. That one's actually a
little bit hard to see. So the first derivative, the
rate of change of acceleration is going to be equal to, so this is negative six T plus 12. Now let's think about when
does this thing equal zero? Well if we subtract 12 from both sides, we get negative six T
is equal to negative 12, divide both sides by negative six, you get T is equal to two. So a couple of things. You could just say alright look, I know that this is a downward opening parabola right over here. I have a negative coefficient
on my second degree term. I know that the slope
of the tangent line here is zero at T equals two. So that's gonna be my maximum point. Or you could go a little bit further. You could take the second derivative. Let's do that just for kicks. So we could take the second derivative of our acceleration function. So this is going to be equal
to negative six, right. The derivative of
-6t is -6 the derivative constant is just zero. So this thing, the second
derivative is always negative. So we are always, always concave downward and so by the second derivative
test at T equals two, well at T equals two our second derivative of our acceleration function's
going to be negative and so we know that this
is our maximum value. Our max at T is equal to two. So at what value of T does the particle obtain its maximum acceleration? At T is equal to two.