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!

Kinematics of a point in rectilinear motion

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data
    A point Q in rectilinear motion passes through the origin at t=0, and from then until 5 seconds have passed, the acceleration of Q is 6 ft/s^2 to the right. Beginning at t = 5s, the acceleration of Q is 12t ft/s^2 to the left. If after 2 more seconds point Q is 13 feet to the right of the origin, what was the velocity of Q at t=0?


    2. Relevant equations
    x''=a
    x'=at+c1
    x=(a/2)t^2+(c1)(t)+c2


    3. The attempt at a solution

    At t=0:
    x''=6
    x'=6t+v0
    x=3t^2+(v0)(t)

    At t=5:
    x''=-12t
    x'=-6t^2+v1
    x=-2t^3+(v1)(t)+x1

    At t=7:
    x''=-12t
    x'=-6t^2+v1
    13=-2t^3+(v1)(t)+x1

    Now plug in t=5 s into the first set of equations and get:
    75+(5)(v0)=x1
    30+v0=v1

    Plug these two equations into 13=-2t^3+(v1)(t)+x1
    As for t in this equation, plug in 2 seconds because this new equation started at t=5 (and hence assume that to be the new starting point, thus 7-5=2).

    I get v0 = -15 m/s
    The book says the answer is +2. Obviously something went very wrong. Any help would be appreciated.
     
  2. jcsd
  3. Mar 9, 2012 #2
    How did you get the highlighted equation ?
     
  4. Mar 9, 2012 #3
    plug t=5 into x'=6t+v0
    Then knowing the fact that at t=5 the velocity is described by x'=-6t^2+v1, I set t=0 at this instant and thus 30+v0 = v1
     
  5. Mar 9, 2012 #4
    you are using same constants on integration in different equations. the motion from t=5 to t=7
    should have separate constants of integration from the motion t=7 onwards. thats why you are
    mixing equations and got the wrong answer
     
  6. Mar 13, 2012 #5
    But according to the problem, the motion at t=7s follows the same equation as the motion at t=5s (x''=-12), so shouldn't it be the same equations? Can you show me the correct equation? Because from the way I'm reading the problem my way to me is correct, when in fact it's not. I can't see it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinematics of a point in rectilinear motion
  1. Rectilinear Motion (Replies: 9)

  2. Rectilinear Motion (Replies: 2)

  3. Rectilinear motion (Replies: 4)

Loading...