Results 1 to 14 of 14

Math Help - Question about Acceleration

  1. #1
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Red face Question about Acceleration

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qbkr21 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    Thanks Jhevon. So the maximum acceleration for any particle on Earth is -32 feet per second, or -9.8 meters per second?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qbkr21 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    Sorry Jhevon that is what I meant. It has been a long day. I have had 4 tests already.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    I will be back in about 2 hours I have History class. See you soon...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qbkr21 View Post
    Sorry Jhevon that is what I meant. It has been a long day. I have had 4 tests already.
    yeech! Bless you son
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qbkr21 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Apr 2007
    Posts
    16
    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Pulsar06 View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by qbkr21 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by Pulsar06 View Post
    ...the third derivative of position (or first derivative of F''), which I think is known as 'jerk'
    Quote Originally Posted by Jhevon View Post
    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
    Follow Math Help Forum on Facebook and Google+

  14. #14
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by topsquark View Post
    ...me, your "average jerk."
    lol
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help please, acceleration/velocity question.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 19th 2009, 06:16 PM
  2. Need help with an acceleration question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 23rd 2009, 02:09 PM
  3. Simple Acceleration Question!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 7th 2008, 06:30 PM
  4. Acceleration Question
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: October 30th 2007, 10:02 AM
  5. acceleration question
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: September 10th 2007, 07:40 PM

Search Tags


/mathhelpforum @mathhelpforum