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A rocket going up with acceleration =

  1. Nov 10, 2015 #1
    1. The problem statement, all variables and given/known data
    A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ay=(2.90m/s^3)t, where the +y-direction is upward.

    What is the height of the rocket above the surface of the earth at t = 10.0 s ?
    Express your answer with the appropriate units.

    2. Relevant equations
    v=dy/dt
    a=dv/dt
    y =v*t + (1/2) a*t^2
    3. The attempt at a solution

    I drew a diagram. A rocker going up, from 0 -10 a= 2.90 *t
    from 10 and higher, a = -9.802
    I integrated the acceleration from 0-10, and got the velocity to be 2.90/2 t^2 I used the kinematics formula for position, height in this case and got
    h= 2.90/2 t^3 + .5*-9.802*t^2 = 959.9
    its not correct, what am I doing wrong? What should I do instead and why?
     
  2. jcsd
  3. Nov 10, 2015 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Not sure why you care what happens to the rocket for t > 10 sec. The problem asks you to find the height of the rocket at t = 10 sec.

    What happens to the rocket after that is immaterial.

    You only need to concern yourself with what happens to the rocket over the interval 0 ≤ t ≤ 10.

    If you can integrate acceleration once to find velocity, why can't you integrate velocity to find height?
     
  4. Nov 10, 2015 #3
    I'm sorry, I didn't not read carefully enough, Im pretty sleepy doing tons of home work, I did a previous question that asked for something past the original time interval, and I was in the same mind set with this one. Thanks!
     
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