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Homework Help: Mechanics (Components of Acceleration)

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data

    The following is a solved problem:

    http://img600.imageshack.us/img600/6503/questiono.jpg [Broken]

    I can't understand how they worked out these two components.

    2. Relevant equations

    [itex]\vec{a}(t) = \frac{dv}{dt}\vec{T} + v \frac{d \vec{T}}{dt}[/itex]


    [itex]\vec{T}(t) = \frac{d \vec{r}(t)/dt}{||d \vec{r}(t)/dt||} = \frac{\vec{v}(t)}{v(t)}[/itex]

    3. The attempt at a solution

    In the equation I posted above, I believe the first term is the tangential component of acceleration and the second one is the normal component. I want first to find the tangential component:

    [itex]\vec{v}(t) = d(t^2 \vec{i} + 3t+2 \vec{j})/dt = 2t+3[/itex]

    [itex]|| \vec{v}(t) || = \sqrt{4t^2+9}[/itex]

    So substituting in

    [itex]\frac{dv}{dt}\vec{T} = \frac{d (\vec{r}(t)/dt)}{dt} . \frac{d \vec{r}(t)}{dt}.\frac{1}{||\vec{r}(t)||}[/itex]


    [itex]= \frac{4t+6}{\sqrt{4t^2+9}}[/itex]

    But this is wrong. The correct answer must be


    What did I do wrong? :confused: How can I get to the correct answer? Any guidance is greatly appreciated.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 21, 2012 #2
    You have speed: [itex]\vec v= \frac{dr}{dt}=(2t,3)[/itex]
    and acceleration: [itex]\vec a= \frac{dv}{dt}=(2,0)[/itex]

    then you've got to find the projection of [itex]\vec a[/itex] onto [itex]\vec v[/itex]:
    [tex]\vec a_t = \frac{(\vec a \cdot \vec v)}{||\vec v||^2}\vec v= \frac{1}{4t^2+9}(8t^2,12t)[/tex]
    The normal component of acceleration is then:
    [itex]\vec a_n= \vec a - \vec a_t[/itex]

    The given answers seem to be wrong.
    Last edited by a moderator: May 5, 2017
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