1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Postion of an object at a given time?

  1. May 21, 2005 #1
    im trying to do this problem involving the postion of a rocket going straight up at a given time.

    it says a rocket initially at rest on the ground, accelerates straight upward from rest with constant net acceleration a, until time t(subscript 1) when the fuel is exhausted.

    the equation for this is: position at that time=initial position+initial velocity times t(time)+0.5 times 3 times 9.81m/s^2 times T^2

    i know initial velocity and position are zero, so all i got left over is the rest of the equation.

    a=3g(g being accel. due to gravity) and T(subscript 1)=5.00s
    it says to give ur answer in meters. i ended up with 368 meters. but it said that i forgot a numerical factor?
     
  2. jcsd
  3. May 21, 2005 #2
    [tex]x = \frac{1}{2} at^2 = \frac{1}{2} (3*9.8)(5)^2 = 367.5m[/tex]
     
  4. May 21, 2005 #3
    thats what i basically entered but it keeps saying that im off by a numerical factor? why the hell does it keep saying that.

    the height im supose to give is in this form H (subscript) 3= the answer.

    it also says express the answer numerically in meters using g=9.81 m/s^2. just to be clear on that. i dunno whats the problem
     
  5. May 21, 2005 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    I don't think this has to do with sig figs.

    Compute the velocity at the time when all the fuel is burnt.From then on,the rocket goes straight up due to inertia.Gravity will be then the only force acting on it.

    Your number ~368 is too small.

    Daniel.
     
  6. May 21, 2005 #5
    Are you trying to find the height when the fuel is burnt, or the final height? If its the first, then 368 is the answer.
     
  7. May 21, 2005 #6
    it says to look for the maximum height
     
  8. May 21, 2005 #7

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Voilà,so i was right.:approve:

    Daniel.
     
  9. May 21, 2005 #8
    In that case find the velocity at t_1, then use the same equation you gave above with v_0 = v(t_1) and a = -9.8

    Solve v_0 - at = 0 for a = 9.8 and v_0 = v(t1) to find where the velocity is 0 (the top). Use that value for t.

    This is the distance travelled after the fuel is burned out.
     
    Last edited: May 21, 2005
  10. May 21, 2005 #9
    how would i go about computing the velocity at T_1.?
     
  11. May 22, 2005 #10
    You underwent constnat acceleration of 3*9.8 for 5 seconds.
     
  12. May 22, 2005 #11
    thanks for the help. guys.

    i have another quetion, but i dunno how to type these leters the right way

    The next figure View Figure shows the velocity vectors corresponding to the upward motion of the power ball. Indicate whether its velocity and acceleration, respectively, are positive (upward), negative, or zero.
    Use P, N, and Z for positive (upward), negative, and zero, respectively. Separate the letters for velocity and acceleration with a comma.

    i tried typing in v p, a n, and other combnations but said it had the wrong number of terms seperated by commas. how do i go about typing these terms in?


    heres a the figure there referring to.
     

    Attached Files:

  13. May 22, 2005 #12
    I think its looking for just p, n
    Those are correct answers.
     
  14. May 22, 2005 #13
    thanks alot man, that helped with the next problem. i really didnt know how to input these terms in.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?