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!

Variable acceleration

  1. Aug 8, 2012 #1
    An object starts from position X = 0 and moves along a straight line with its position given by; X = 20t - 5t^3 where t is the time taken to reach position X.

    Find;
    i) The time at which the velocity is zero
    ii) The time at which the acceleration is zero
    iii) The average velocity over the first three seconds of motion

    No answers were given with the questions, so would appreciate some answers to check against mine. I have 1.15s, 0s and -50m/s
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 8, 2012 #2
    yes,yes, no(-25)
     
  4. Aug 8, 2012 #3
    Can you please explain how you come to get -25. Thanks for your time :-)
     
  5. Aug 8, 2012 #4
    I must be doing this wrong:

    v = 20 - 15t^2
    After 1s, v = 20 - 15(1) = 5m/s
    After 2s, v = 20 - 15(4) = -40m/s
    After 3s, v = 20 - 15(9) = -115m/s

    Average speed = 5 - 40 - 115 =
    = -150 ÷ 3 = -50m/s
     
  6. Aug 8, 2012 #5

    lewando

    User Avatar
    Gold Member

    You are looking for average velocity. This is defined as displacement / time.
     
  7. Aug 9, 2012 #6
    avg. velocity=(∫vdt)/(∫dt),but here ∫vdt is x itself(given) so for three seconds
    x=-75
    t=3
    <v>=-25
     
  8. Aug 9, 2012 #7
    thank you
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Variable acceleration
  1. Variable acceleration (Replies: 1)

  2. Variable Acceleration (Replies: 5)

Loading...