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Homework Help: Rectilinear motion

  1. Mar 19, 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=5 seconds, 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

    3. The attempt at a solution
    Assume right is positive and left is negative.

    t=0 to t=5

    From t=5 onwards
    x=-2t^3+(V1)t+[75+5(V0)t] (parts in brackets [] obtained by using the x=3t^2+(V0)t equation setting t=5)

    we are given 13=-2t^3+(V1)t+[75+5(V0)t] at t=7s (from beginning) or t=2s (considering the equation changes at t=5)

    thus plugging in t=2 (since x''=-12t begins at t=5) get:


    Now need another equation to solve variables. Equating the equations x'=6t+(V0) (t=0 to t=5) and the equation x'=-6t^2+(V1) (t=5 onwards)

    Thus plug in t=5 for x'=6t+(V0) and t=0 for x'=-6t^2+(V1) and then equate them to get


    Solve and get V0 = -8.83 ft/s

    Book says answer is 2ft/s to the right.

    Something went terribly wrong...
  2. jcsd
  3. Mar 20, 2012 #2


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    hi xzibition8612! :smile:

    hmm :rolleyes: … this is exactly why you should never take short-cuts :redface:
    that 5(V0)t should be 5V0, shouldn't it? :wink:

    in between those two paragraphs, you should have paused to state explicitly the final values of x and v, that you intend to use later​
  4. Mar 20, 2012 #3
    Yuo're right. But then I get 2(30+V0)+5(V0)=-46

    Hence V0 = -15 m/s

    which is still wrong. No idea where i went wrong.
  5. Mar 20, 2012 #4


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    hi xzibition8612! :smile:

    it's very difficult to check your work :redface:

    (i don't see a 73 anywhere)

    can you write it out in one continuous proof?​

    (and please try using the X2 button just above the Reply box, instead of ^ :wink:)
  6. Mar 21, 2012 #5
    tiny-tim did you get 2 ft/s as the correct answer? If you did, can you just show me how to do it? Because I'm afraid its very difficult to communicate online and if I see what you did then I can figure out where I went wrong thanks.
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