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

Lesson 11: Solving optimization problems

# 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?

• At from 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!
• 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 :)
• Is there a tutorial video on the equation of motion, such as average velocity and instantaneous velocity?
• Can't you just complete the square to find the vertex for the maximum value?
• 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.
• 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.)
• Do you the same thing when doing integrals? Example: find the maximum displacement given acceleration.
• 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
• 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.