Results 1 to 4 of 4

Math Help - 3 second burn rocket problem

  1. #1
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    3 second burn rocket problem

    Pretty sure I have every part of this problem solved but my final answer concerns me as it seems a little high.

    a(t)=(3-t)42-9.8m/s^2 for 0<=t<=3
    and a(t)=-9.8m/s^2 for t>3

    I should find the maximum altitude and time in which it gets there. My answer is 56s and 18565.18meters.

    **additional info**
    initial velocity ( v_i is 366.4m/s)
    initial position ( x_i) is 0
    \int (3-t)42-9.8 dx=v(t_i)=126t-42t-9.8t+v_i
    after taking integral of v(t) x(t_i)=-7t^3+63t^2+366.4t+x_i
    **
    x(t_i)= position fxn until 3 seconds.
    x(t_i)=-7t^3+63t^2+366.4t=1477.2m after 3 seconds of burn

    x(t_m)=position fxn after 3s initial burn.
    x(t_m)=\frac{9.88t^2}{2}+555.4t+1477.2
    x(t_m)=\frac{9.88t^2}{2}+555.4t+1477.2=0 at t=56.2145s

    x'(t_m)where t=56.2145s
    x'(56.2145)=17087.98m

    17087.98+1477.2=18565.18meters

    Is it me or is that pretty dang high?
    Last edited by Bowlbase; November 9th 2011 at 04:43 PM. Reason: information/clarification
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    Re: 3 second burn rocket problem

    I guess I don't need to add the two x positions since the second one already has it in there. So 17087.98meters.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    Re: 3 second burn rocket problem

    a(t)=(3-t)42-9.8m/s^2 for 0<=t<=3
    and a(t)=-9.8m/s^2 for t>3
    So,
    initial velocity ( v_i is 366.4m/s)
    initial position ( x_i) is 0
    \int (3-t)42-9.8 dx=v(t_i)=-21t^2=116.2t+v_i
    after taking integral of v(t) x(t_i)=-7t^3+58.1t^2+366.4t+x_i

    Then, at 3 seconds:
    x(3)=-7(3)^3+58.1(3)^2+366.4(3)=1433.1=x_m
    and
    v(3)=-21(3)^2=116.2(3)+366.4=526=v_m

    so after 3 seconds
    a(t)=-9.8m/s^2
    \int a(t)=-9.8dx
    =-9.8t+v_m=-9.8t+526
    =v(t)=-9.8t+526
    \int v(t)=-9.8t+526dx = x(t)=\frac{9.88t^2}{2}+526t+1433.1

    Finally:
    =v(t)=-9.8t+526=0? at t=53.23seconds
    then
    x(53.23)=\frac{9.88t^2}{2}+526t+1433.1=15479.02

    So, it reaches a height of 15479.02m after flying for 53.23s. Still seems high to me...
    Last edited by Bowlbase; November 9th 2011 at 06:18 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    Re: 3 second burn rocket problem

    For some reason I switched from 9.8 to 9.88 as grav. constant. I adjusted them on my paper but don't have the time to do it again here. The values are very nearly the same anyway.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rocket linear equation problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 17th 2010, 04:15 PM
  2. Replies: 5
    Last Post: May 28th 2010, 02:36 AM
  3. Rocket problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 23rd 2010, 10:05 PM
  4. Rocket in the Air
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: April 6th 2010, 03:55 AM
  5. A Rocket in the Air
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: December 8th 2009, 01:59 AM

Search Tags


/mathhelpforum @mathhelpforum