Suppose I was asked to find the Maximum acceleration of particle given the equation:

s(t) = -16t^2+7t+2

I know that

Velocity:

v(t) = -32t+7

and that

Acceleration

a(t) = -32

But how would I Maximize it? I know that when it comes to velocity I would just take the derivative of the given s(t) equation, then set that velocity v(t) equation equal to zero and solve for t. Then I would plug in that t value in the the original s(t) height equation; however I am unsure as to how I would go about doing this for acceleration. I know that F'' aka the acceleration double derivative deals with concavity. Would the best method be to set up a test line then test values below -32 and above -32 to see whether it is concave up or concave down. I guess that I would also have to test -32 to see whether or not it would be equal to. I am just wondering how I would go about "Maximizing Acceleration". Please let me know when you all get a chance. As always I appreciate all of your help and insight.

2. Originally Posted by qbkr21
Suppose I was asked to find the Maximum acceleration of particle given the equation:

s(t) = -16t^2+7t+2

I know that

Velocity:

v(t) = -32t+7

and that

Acceleration

a(t) = -32

But how would I Maximize it? I know that when it comes to velocity I would just take the derivative of the given s(t) equation, then set that velocity v(t) equation equal to zero and solve for t. Then I would plug in that t value in the the original s(t) height equation; however I am unsure as to how I would go about doing this for acceleration. I know that F'' aka the acceleration double derivative deals with concavity. Would the best method be to set up a test line then test values below -32 and above -32 to see whether it is concave up or concave down. I guess that I would also have to test -32 to see whether or not it would be equal to. I am just wondering how I would go about "Maximizing Acceleration". Please let me know when you all get a chance. As always I appreciate all of your help and insight.
that is the maximum acceleration. you can think of this as a falling body. it cannot accelerate faster than gravity causes it to

3. Re:

Thanks Jhevon. So the maximum acceleration for any particle on Earth is -32 feet per second, or -9.8 meters per second?

4. Originally Posted by qbkr21
Thanks Jhevon. So the maximum velocity for any particle on Earth is -32 feet per second, or 9.8 meters per second?
no, the maximum acceleration due to gravity is -32 feet per second, or -9.8 meters per second

the maximum velocity of this particle would be different

5. Re:

Sorry Jhevon that is what I meant. It has been a long day. I have had 4 tests already.

6. Re:

I will be back in about 2 hours I have History class. See you soon...

7. remember, acceleration is different from velocity. velocity is the rate of change of speed in a particular direction, acceleration is the rate of change of velocity.

to find the max of any function, you would do as you say, find its derivative, set it to zero, and examine the function at that point. if there is no variable in the derivative of a function, it means it is constant, so everything stays the same, and the max (and min for that matter) is just the value of the function

8. Originally Posted by qbkr21
Sorry Jhevon that is what I meant. It has been a long day. I have had 4 tests already.
yeech! Bless you son

9. Originally Posted by qbkr21
I will be back in about 2 hours I have History class. See you soon...
ok, i might not be here though, i have to leave soon my self, when i leave i won't be back for a few hours

10. Right... as Jhevon said, to maximize or minimize any function, take its derivative, find its critical values and examine the its increasing/decreasing behavior. You obviously know how to maximize velocity since you explained the process in the original post. To maximize acceleration, take the third derivative of position (or first derivative of F''), which I think is known as 'jerk' (but don't quote me on that), set it equal to zero, and use the same process. There is a slight difference though: F'' = 32 so F''' = 0 (i.e. the object is neither accelerating or decelerating). The max acceleration is therefore 32 m/s^2.

11. Originally Posted by Pulsar06
Right... as Jhevon said, to maximize or minimize any function, take its derivative, find its critical values and examine the its increasing/decreasing behavior. You obviously know how to maximize velocity since you explained the process in the original post. To maximize acceleration, take the third derivative of position (or first derivative of F''), which I think is known as 'jerk' (but don't quote me on that), set it equal to zero, and use the same process. There is a slight difference though: F'' = 32 so F''' = 0 (i.e. the object is neither accelerating or decelerating). The max acceleration is therefore 32 m/s^2.
I never heard of "jerk" before, is that an abreviation or mnemonic for something?

12. Originally Posted by qbkr21
Thanks Jhevon. So the maximum acceleration for any particle on Earth is -32 feet per second, or -9.8 meters per second?
Owie Mommy! That hurts!

Watch the negative signs. That assumes a specific coordinate system. It is MUCH better to be thinking of the magnitude of the acceleration due to gravity as 32 f/s^2 or 9.8 m/s^2. (Also, note the units!)

-Dan

13. Originally Posted by Pulsar06
...the third derivative of position (or first derivative of F''), which I think is known as 'jerk'
Originally Posted by Jhevon
I never heard of "jerk" before, is that an abreviation or mnemonic for something?
Given a displacement vector s, the velocity vector v, and acceleration vector a, we define a vector j:
j = da/dt = d^2v/dt^2 = d^3s/dt^3

as the "instantaneous jerk" on an object. Jerk doesn't stand for anything but how hard you might "yank" on a rope to make something move. (I love some Physics terminologies!)

Naturally then, the vector (Delta)a/(Delta t) is me, your "average jerk."

-Dan

14. Originally Posted by topsquark