Welcome guest, is this your first visit? Create Account now to join.
  • Login:

Welcome to the Online Debate Network.

If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed.

Results 1 to 4 of 4
  1. #1
    ODN Community Regular

    Join Date
    Aug 2004
    Location
    Wheaton, IL
    Posts
    13,845
    Post Thanks / Like

    Distance / speed puzzle

    Suppose you're standing at the origin of the X-Y plane, (0,0). A particle travels (positively / rightward) along the line y = mx+b; how fast is the particle moving toward you when it is at the point (x0,mx0+b)? What is its acceleration at that point?
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
    **** you, I won't do what you tell me

    HOLY CRAP MY BLOG IS AWESOME

  2. #2
    ODN Community Regular

    Join Date
    Dec 2005
    Location
    USA
    Posts
    5,617
    Post Thanks / Like

    Re: Distance / speed puzzle

    Quote Originally Posted by CliveStaples View Post
    Suppose you're standing at the origin of the X-Y plane, (0,0). A particle travels (positively / rightward) along the line y = mx+b; how fast is the particle moving toward you when it is at the point (x0,mx0+b)? What is its acceleration at that point?
    I think you thought you put more information into the question than you did, but the correct answer is:


    Let the x-position as a function of time be denoted by x = f(t), s.t. f(0) = x0. Then the position 2-vector is given by:


    (x(t), y(t) ) == ( f(t) , m f(t) + b ) .


    In other words there's a whole twice-differentiable function's worth of information that's unspecified. Then the velocity vector is given by


    v(t) = (u,v) == ( dx/dt , dy/dt ) = ( d/dt [ f(t) ] , m d/dt[f(t)] ) .


    The direction that you're looking for is parallel to the line in question. The component along that direction is the total norm of the function, however. So, while enforcing positivity as requested (physically this means it is right-moving, i.e. df/dt > 0), the the norm at time, t=0, gives us

    |v(0)| == sqrt{ u2 + v2 }= sqrt{ 1+m2} df/dt(0),


    The acceleration may be computed similarly:

    a(t) = (ax, ay) == ( du/dt, dv/dt ) = d2f/dt2 (1,m)

    And the norm at time, t = 0, is given as:

    |a(0)| == sqrt{ax2, ay2} = sqrt{ 1+m2} | d2f/dt2| (0)


    (We must enforce positivity on the second derivative of f(t) because nothing prevents it from being negative but the norm takes the absolute value, obviously)




    To make Clive happy:

    1.) f is a twice-differentiable, real-valued function: f: R -> R.

    2.) p = (x,y) is a vector living in Euclidean 2-space, RxR, with Euclidean norm d(,y): RxR -> R*, and additive identity 0 for the vector space, such that the Euclidean norm is defined by

    d(p1,p2) = sqrt{ (x1 - x2)^2 + (y1 - y2) }

    with p1 = (x1, y1) and p2 = (x2, y2).

    3.) Finally, define the unary operator |x| = d(x,0).
    Last edited by GoldPhoenix; May 25th, 2014 at 02:40 PM.
    "Those who can make you believe absurdities, can make you commit atrocities." --Voltaire

  3. #3
    Registered User

    Join Date
    Jan 2010
    Location
    Nashville, TN
    Posts
    447
    Post Thanks / Like

    Re: Distance / speed puzzle

    Quote Originally Posted by GoldPhoenix View Post
    I think you thought you put more information into the question than you did, but the correct answer is:


    Let the x-position as a function of time be denoted by x = f(t), s.t. f(0) = x0. Then the position 2-vector is given by:


    (x(t), y(t) ) == ( f(t) , m f(t) + b ) .


    In other words there's a whole twice-differentiable function's worth of information that's unspecified. Then the velocity vector is given by


    v(t) = (u,v) == ( dx/dt , dy/dt ) = ( d/dt [ f(t) ] , m d/dt[f(t)] ) .


    The direction that you're looking for is parallel to the line in question. The component along that direction is the total norm of the function, however. So, while enforcing positivity as requested (physically this means it is right-moving, i.e. df/dt > 0), the the norm at time, t=0, gives us

    |v(0)| == sqrt{ u2 + v2 }= sqrt{ 1+m2} df/dt(0),


    The acceleration may be computed similarly:

    a(t) = (ax, ay) == ( du/dt, dv/dt ) = d2f/dt2 (1,m)

    And the norm at time, t = 0, is given as:

    |a(0)| == sqrt{ax2, ay2} = sqrt{ 1+m2} | d2f/dt2| (0)


    (We must enforce positivity on the second derivative of f(t) because nothing prevents it from being negative but the norm takes the absolute value, obviously)




    To make Clive happy:

    1.) f is a twice-differentiable, real-valued function: f: R -> R.

    2.) p = (x,y) is a vector living in Euclidean 2-space, RxR, with Euclidean norm d(,y): RxR -> R*, and additive identity 0 for the vector space, such that the Euclidean norm is defined by

    d(p1,p2) = sqrt{ (x1 - x2)^2 + (y1 - y2) }

    with p1 = (x1, y1) and p2 = (x2, y2).

    3.) Finally, define the unary operator |x| = d(x,0).
    Took the words right out of my mouth.

  4. #4
    ODN Community Regular

    Join Date
    Dec 2005
    Location
    USA
    Posts
    5,617
    Post Thanks / Like

    Re: Distance / speed puzzle

    Errata: d(p1,p2) = sqrt{ (x1 - x2)2 + (y1 - y2)2 }
    "Those who can make you believe absurdities, can make you commit atrocities." --Voltaire

 

 

Similar Threads

  1. Distance from a point to a line
    By CliveStaples in forum Logical Riddles & Puzzles
    Replies: 3
    Last Post: May 5th, 2014, 06:58 PM
  2. 55 M.P.H. National Speed Limit
    By Snoop in forum General Debate
    Replies: 14
    Last Post: July 10th, 2008, 08:32 PM
  3. IRS to refund EXCISE TAX on Long Distance
    By ladyphoenix in forum Shootin' the Breeze / Off-Topic
    Replies: 3
    Last Post: January 19th, 2007, 03:06 PM
  4. Automatic speed adjustment?
    By FruitandNut in forum General Debate
    Replies: 11
    Last Post: March 7th, 2005, 10:25 AM
  5. Speed of light
    By Anywien in forum Hypothetical Debates
    Replies: 33
    Last Post: July 25th, 2004, 03:55 PM

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •