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Variable acceleration, time, distance

  1. Feb 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle moves along a straight line. At t=0 the acceleration is k(t+2), initial velocity = 0 and x=5m.
    a) Write down the position of the particle as a function of time.
    b) At what point in time does the particle reach the velocity 16m/s, and where is the particle at this time?

    2. Relevant equations

    [tex]v=\int a\textit{dt}=(kt^2)/2+2kt+C_1[/tex]
    [tex]x=\int v\textit{dt}=(kt^3)/6+kt^2+C_2[/tex]

    3. The attempt at a solution

    I think I got part a) With the above equations that I calculated and with the initial values inserted I get: [tex]x=(kt^3)/6+kt^2+5[/tex]

    for part b I'm more unsure that I'm correct. Entering the velocity as 16m/s I get the equation: [tex]16=(kt^2)/2+2kt[/tex] which when calculated gets the time. [tex]t=-2+4\sqrt{1+{64/k}^2}[/tex] and then this time is inserted into [tex]x=(kt^3)/6+kt^2+5[/tex] to calculate the distance. This gives me [tex]1/6\,k \left( -2+4
    \,\sqrt {1+64\,{k}^{-2}} \right) ^{3}+k \left( -2+4\,\sqrt {1+64\,{k}^{-2}} \right) ^{2}+5 [/tex] Seems a bit to complicated to be right? Or is it right and I'm just confused :confused:

    Thanks for the help!
  2. jcsd
  3. Feb 4, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    Note that
    [tex]x=\int v\textit{dt}=(kt^3)/6+kt^2+C_1 t + C_2,[/tex]
    you forgot to integrate the C1 term.

    Finally, you can probably simplify your answer a bit but your way of calculating it looks correct. I am afraid that you are just confused and it is just an ugly answer :smile:
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