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  1. #1
    Banned whodat?'s Avatar
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    Math help: what is a derivative of displacement?

    I know how to figure out derivatives now, but I dont know what people say when they say 'second derivative of displacement'?

    im trying to find velocity and acceleration...

    also I dont know wtf "ds/dt" and stuff like that means?
    Last edited by whodat?; 05-24-2007 at 03:45 AM.
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  2. #2
    Banned iBswole's Avatar
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    what ure looking for is :

    displacement is basicly distance it can be represented as a function of s -

    s = kt

    where k is a variabe in terms of time t - we'll consider k as position.

    then the rate of change of displacement = ds/dt will give us the rate of change of distance over time

    or v(t) = velocity

    if we look for the rate of change of velocity with repsect to time we get

    dv(t) / dt = accelleration

    so to recap

    d/dt of displacement (or the rate of change of disp(or distance in a specified direction)) = velocity
    d/dt of velocity (or the rate of change of velocity ) = acceleration
    Last edited by iBswole; 05-24-2007 at 03:59 AM.
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    Originally Posted by whodat? View Post
    I know how to figure out derivatives now, but I dont know what people say when they say 'second derivative of displacement'?

    im trying to find velocity and acceleration...

    also I dont know wtf "ds/dt" and stuff like that means?

    If you have a function let say y=x which is a func for displacement

    First Derivate will be y'=1 (is a function for VELCOITY)
    Second Derivative will be y''=0 (is a function for ACCELERATION)
    Third Derivative will be y'''= 0(for argument sake)(is a function for JERK)
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  4. #4
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    ahh i remember doing this cant remember what answer is though

    ds/dt maybe
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  5. #5
    Banned whodat?'s Avatar
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    Originally Posted by iBswole View Post
    displacement is basicly distance it can be represented as a function of s -

    s = kt

    where k is a variabe in terms of time t - we'll consider k as position.

    then the rate of change of displacement = ds/dt will give us the rate of change of distance over time

    or v(t) = velocity

    if we look for the rate of change of velocity with repsect to time we get

    dv(t) / dt = accelleration

    so to recap

    d/dt of displacement (or the rate of change of disp(or distance in a specified direction)) = velocity
    d/dt of velocity (or the rate of change of velocity ) = acceleration
    this makes no sense to me at all... i am math retarded or something

    i know how to find the derivative, but how do i find the 'second deriv of displacement'???

    example:

    s = -5t^2 + 120t + 500

    i know the deriv is -10x+120


    but i dont know what the second derivative of displacement is
    Last edited by whodat?; 05-24-2007 at 04:08 AM.
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  6. #6
    Registered LOLzers LOL MesoPeaks's Avatar
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    Is that physics?
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  7. #7
    Banned whodat?'s Avatar
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    Originally Posted by MesoPeaks View Post
    Is that physics?
    no, math class...practical application problem
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  8. #8
    Banned whodat?'s Avatar
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    whodat? is offline
    Originally Posted by MesoPeaks View Post
    Differential calculus?..if differential ..maybe i can you help you out..

    Can you post some of your lecture..?
    its a summer school class, and all of the people in it except me are re-taking it becaue they failed (im taking it to get ahead)

    so she teaches it like we already know how to do it. plus she is indian so its hard to understand here

    i know how to do pretty much everything except figure out acceleration, and acceleration is the second derivative of displacement and i dont know how to figure that out...

    the problem is:

    An object moves according to the distance (s), time (t) equation:

    s = -5t^2 + 120t + 500

    Find:

    1) the average velocity between t=0 and t=3 seconds
    answer: [delta]S / [delta]t which equals 105, got that problem right

    2) the velocity equation (the derivative)
    answer: -10x + 120

    3) the velocity at t=2 seconds
    answer: -10(2) + 120 = 100

    4) the accelration equation
    DONT KNOW

    5) acceleration at t=3 seconds
    DONT KNOW

    6) time to reach max height
    answer: havent figured out yet but I know how to

    7) max height above ground
    answer: havent figured out yet but i know how to
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  9. #9
    Banned whodat?'s Avatar
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    whodat? is offline
    I found a website that explained it.. correct me if I'm wrong...

    but all I do is take the origional derivative and break it down again?

    **** if its that ez then i was going ape over nothing
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  10. #10
    Registered LOLzers LOL MesoPeaks's Avatar
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    what website..?
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  11. #11
    Banned whodat?'s Avatar
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    Originally Posted by MesoPeaks View Post
    what website..?
    http://www.nipissingu.ca/calculus/tu...er_higher.html
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  12. #12
    Registered LOLzers LOL MesoPeaks's Avatar
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    MesoPeaks is offline
    y'- means first derivative
    y"-means second derivative
    y'"-third
    y""-72x their right, so 72<---------fourth derivative
    Last edited by MesoPeaks; 05-24-2007 at 05:04 AM.
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  13. #13
    Registered User bigbondo's Avatar
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    yes, for the second derivative, just take the derivative of the derivative

    so -10t+120 -> -10t
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  14. #14
    Banned whodat?'s Avatar
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    Originally Posted by bigbondo View Post
    yes, for the second derivative, just take the derivative of the derivative

    so -10t+120 -> -10t
    hehehe.... moron explanation is what makes the most sense to me

    I'm trying to follow the teacher and take notes and I'm just like 'W T F'

    Then I see that example and its like... just double down? I was gettin all worked up when I just do it twice?

    I learned more on friggin Yahoo! education in 10 minutes than I did in over 15 hours in that class..
    Last edited by whodat?; 05-24-2007 at 05:09 AM.
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    Thumbs up math help...

    go to www.mathhelpforum.com ... it really helped me through calc... people explain the answers and make everything very simple!!! You can thank me later
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  16. #16
    Banned whodat?'s Avatar
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    Originally Posted by xcelxp View Post
    go to www.mathhelpforum.com ... it really helped me through calc... people explain the answers and make everything very simple!!! You can thank me later
    thanks for the link, i'll keep that bookmarked
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