Results 1 to 2 of 2

Math Help - Calculus Motion Problem

  1. #1
    Junior Member
    Joined
    Oct 2006
    Posts
    36

    Calculus Motion Problem

    If someone could please help me with this problem, I'd greatly appreciate it.

    A particle moves along the x-axis so that its velocity at any time t≥0 is given by v(t) = (3t^2) -2t - 1 . The position x(t) is 5 for t = 2.
    a) For what values of t, 0 ≤ t ≤ 3, is the particle's instantaneous velocity the same as its average velocity on the closed interval [0,3] ?
    b) Find the total distance traveled by the particle from time t = 0 until time
    t =3.

    Thanks for any help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645
    Hello, turtle!

    A particle moves along the x-axis so that its velocity at any time t ≥ 0
    is given by: .v(t) .= .3t -2t - 1. .The position x(t) is 5 for t = 2.

    a) For what values of t, 0 ≤ t ≤ 3, is the particle's instantaneous velocity
    the same as its average velocity on the closed interval [0,3]?

    b) Find the total distance traveled by the particle from t = 0 to t =3.

    (a) The average velocity is the definite integral of v(t) from t = 0 to 3
    . . .divided by the width of the interval (3).

    . .
    (3t - 2t - 1) dt .= .t - t - t (from 0 to 3) .= .(27 - 9 - 3) - (0 - 0 - 0) .= .15

    . . . . Hence, the average velocity is: .15 3 .= .5

    . . When is v(t) = 5? . 3t - 2t - 1 .= .5 . . 3t - 2t - 6 .= .0
    . . . . . . . . . . . . . . . . . . . . . . . . __________
    . . . . . . . . . . . . . . . . . . . . 2 √2 - 4(3)(-6)
    . . Quadratic Formula: .t .= .---------------------
    . . . . . . . . . . . . . . . . . . . . . . . . 2(6)
    . . . . . . . . . . . . . . . . . .__
    . . . . . . . . . . . . . .1 + √19
    . . Therefore: .t .= .--------
    . . . . . . . . . . . . . . . .3


    (b) x(t) .= .
    (3t - 2t - 1) dt .= .t - t - t + C

    We are told that x(2) = 5: .2 - 2 - 2 + C .= .5 . . C = 3

    . . Hence: .x(t) .= .t - t - t + 3


    The particle stops and changes direction when v(t) = 0.
    . . 3t - 2t - 1 .= .0 . . t = -1/3, 1

    x(0) .= .0 - 0 - 0 + 3 .= .3 . . . The particle starts at +3.

    x(1) .= .1 - 1 - 1 + 3 .= .2 . . . The particle moves 1 unit to the left.

    x(3) .= .3 - 3 - 3 + 2 .= .18 . . . The particle moves 16 units to the right.


    Therefore, total distance is: .1 + 16 .= .17 units

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: February 8th 2010, 09:57 AM
  2. stochastic calculus. Brownian motion. SDE
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: November 11th 2009, 12:48 PM
  3. particle motion with calculus
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 21st 2008, 04:10 AM
  4. Replies: 2
    Last Post: November 29th 2007, 03:12 AM
  5. Calculus, motion problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2007, 07:36 PM

Search Tags


/mathhelpforum @mathhelpforum