Results 1 to 3 of 3

Math Help - [SOLVED] Differential Equation

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    52

    [SOLVED] Differential Equation

    A particle moves along the +ve x-axis. At time t seconds after leaving a fixed point O from rest, the displacement of the particle from O is x cm.

    The acceleration, a cm s^(-2) of the particle is defined by


    a = { 5 - 3t , 0 <= t <= 1
    -(4x+1), t>1


    Find
    (a) the maximum velocity of the particle and its distance from O when this happens

    (b) the speed of the particle when x = 2.5

    The acceleration of the particle is defined as a piece-wise function.

    Forgive me for the rather unclear mathematical expressions. I'm new here and still learning the ropes of LATEX.

    Btw, I calculated the maximum velocity to be 0.5 cm per second at distance 0.25 cm from O.
    But the answer given is 3.5 cm per second at distance 2 cm from O.

    Need help for both (a) and (b) !
    Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello cyt91
    Quote Originally Posted by cyt91 View Post
    A particle moves along the +ve x-axis. At time t seconds after leaving a fixed point O from rest, the displacement of the particle from O is x cm.

    The acceleration, a cm s^(-2) of the particle is defined by


    a = { 5 - 3t , 0 <= t <= 1
    -(4x+1), t>1


    Find
    (a) the maximum velocity of the particle and its distance from O when this happens

    (b) the speed of the particle when x = 2.5

    The acceleration of the particle is defined as a piece-wise function.

    Forgive me for the rather unclear mathematical expressions. I'm new here and still learning the ropes of LATEX.

    Btw, I calculated the maximum velocity to be 0.5 cm per second at distance 0.25 cm from O.
    But the answer given is 3.5 cm per second at distance 2 cm from O.

    Need help for both (a) and (b) !
    Thank you!
    (a) For 0\le t \le 1, we have:
    a = \frac{dv}{dt}=5-3t
    For 0\le t \le 1, a > 0. So the particle's speed is increasing during this phase.

    Then:
    \frac{dv}{dt}=5-3t

    \Rightarrow v = 5t-\tfrac32t^2 + c

    v(0) = 0 \Rightarrow c = 0

    \Rightarrow v = 5t-\tfrac32t^2

    \Rightarrow v(1) = 5 -\tfrac32 = \tfrac72
    Then, writing v as \frac{dx}{dt}:
    \frac{dx}{dt}=5t-\tfrac32t^2

    \Rightarrow x = \tfrac52t^2 -\tfrac12t^3 + c

    x(0) = 0 \Rightarrow c = 0

    \Rightarrow x = \tfrac52t^2 -\tfrac12t^3

    \Rightarrow x(1) = 2
    Then for t > 1 and x>0, a = -(4x+1) <0. Hence the maximum positive velocity occurs at when t = 1; i.e. the maximum velocity is 3.5 cm/sec when x = 2 cm.

    (b)
    a = v\frac{dv}{dx}= -(4x+1)

     \Rightarrow \tfrac12v^2 = -2x^2-x + c
    When x = 2,\; v = \tfrac72
    \Rightarrow \frac{49}{8}= -8-2+c

    \Rightarrow c = \frac{129}{8}

    \Rightarrow \tfrac12v^2 = -2x^2-x+\frac{129}{8}
    When x = 2.5:
    \tfrac12v^2 = -\frac{25}{2}-\frac{5}{2}+\frac{129}{8}
    =\frac98
    So when x = 2.5 the speed is \sqrt{\frac94}=1.5 cm/sec.

    Does that agree with the answer you've been given?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    52
    Yup. That's the answer. Thanks a lot!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Differential equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 5th 2009, 04:44 AM
  2. [SOLVED] Differential equation
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: October 28th 2009, 05:16 PM
  3. [SOLVED] Differential Equation
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: February 8th 2009, 04:37 PM
  4. [SOLVED] Differential Equation
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: January 25th 2009, 04:33 PM
  5. [SOLVED] differential equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 13th 2007, 06:33 PM

Search Tags


/mathhelpforum @mathhelpforum