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Simple projectile motion question

  1. Nov 11, 2014 #1
    1. The problem statement, all variables and given/known data
    A missle is fired vertically up from a ground with an initial velocity of 20m/s. The acceleration at any time is given by dv/dt = -10 + t/2 m/s^2.
    Find the height of the missile after two seconds
    Find the height after four seconds
    the direction the missile is traveling when t = 4

    2. Relevant equations
    Displacement = x
    Velocity = dx/dt
    acceleration = dv/dt

    3. The attempt at a solution

    dv/dt = -10 + t/2 m/s^2
    dx/dt = -10t + t^2/4 + c

    found c to be 20

    x = -5t + x^3/12 + 20t + c

    when t = 0 displacement = 0
    c = 0

    -5t + x^3/12 + 20t

    then I don't know what to do
     
  2. jcsd
  3. Nov 11, 2014 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    You've managed to mangle your calculus here, which is why you get confused.

    According to the OP, dv/dt = -10 + t/2 m/s2

    If you separate the variables and integrate, you get

    ∫ dv = ∫ (t/2 - 10) dt, or

    v = t2/4 - 10t + C m/s

    According to the OP, v @ t = 0 is equal to 20 m/s, so C = 20

    But, v = dx/dt = t2/4 - 10t + 20, so separating variables and integrating again we get:

    ∫ dx = ∫ [ t2/4 - 10t + 20] dt which gives

    x = ∫ [ t2/4 - 10t + 20] dt

    which I'll leave you to work out. You should be able to answer the questions about the height of the rocket very easily.
     
  4. Nov 11, 2014 #3
    Sorry about the multiple post, thanks for the help but
     
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