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Thread: Kinematics problem (uniform acceleration)

  1. #1
    Member astartleddeer's Avatar
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    Kinematics problem (uniform acceleration)

    Hello,

    I've hit rough terrain again, unfortunately. Before I begin I've posted the question I'm stuck on. I've solved part i but part ii involves part i, respectively.

    Q: A body starts at $\displaystyle 2\ ms^{-1}$ and moves with constant acceleration for 4 metres. If it then has a velocity of $\displaystyle 6\ ms^{-1}$ find:

    i) how long it will take to move the next 4 m? [A = 0.56 s]

    ii) how far it will move in the next 3 seconds? [A = 36 m]

    For part i of the question:

    Let $\displaystyle u = 2\ ms^{-1} $,$\displaystyle v = 6\ ms^{-1} $ and $\displaystyle s = 4m $

    Then, $\displaystyle s = \left(\frac{u + v}{2}\right)\cdot t \Rightarrow t = \left(\frac{2s}{u + v}\right) = \left(\frac{(2)(4)}{6 + 2}\right) = 1\ second $

    Now, $\displaystyle v = u + at $ therefore, $\displaystyle a \Rightarrow \left(\frac{v - u}{t}\right) = \left(\frac{6 - 2}{1}\right) = 4\ ms^{-2} $

    $\displaystyle v = \sqrt{u^{2} + 2as} = \sqrt{2^{2} + [(2)(4)(8)]} = 2\sqrt{17}\ ms^{-1}\ after\ 8m $

    From $\displaystyle v = u + at $, $\displaystyle t \Rightarrow \left(\frac{v - u}{a}\right) = \left(\frac{2\sqrt{17} - 6}{4}\right) = \left(\frac{-3 + \sqrt{17}}{2}\right) \approx 0.56\ seconds $

    For part ii of the question:

    Let $\displaystyle s = ut + \frac{1}{2}at^{2} $, then $\displaystyle s = [(2\sqrt{17})(3)] + \left[\left(\frac{1}{2}\right)(4)(3^{2})\right] \approx 42.73\ m \neq 36\ m $

    I don't understand why this does not equate to 36 m? Is the initial velocity for the next 3 seconds $\displaystyle 2\sqrt{17} $?

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  2. #2
    Senior Member BAdhi's Avatar
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    Re: Kinematics problem (uniform acceleration)

    I thinks you have much easier way to solve the first question,

    to find the acceleration, you have the following equation which doesn't need time,
    $\displaystyle v^2=u^2+2as$

    from substitution,

    $\displaystyle \begin{align*} 6^2&=2^2+2a(4)\\ 36&=4+8a\\ a&=\frac{36-4}{8}\\ &=4ms^{-2}\\ \end{align*}$

    now to find the time that takes the object to move next 4m, use $\displaystyle s=ut+\frac{1}{2}at^2$

    $\displaystyle \begin{align*}4&=6t+\frac{1}{2}4t^2\\ 2&=3t+t^2\\ t^2+3t-2&=0\\ \end{align*}$

    thus gives,
    $\displaystyle \begin{align*} t&=\frac{-3 \pm \sqrt{3^2-4(1)(-2)}}{2}\\ &=\frac{-3 \pm \sqrt{17}}{2} \\ \end{align*}$

    since t is positive,

    $\displaystyle t=\frac{-3 +\sqrt{17}}{2}$


    Quote Originally Posted by astartleddeer View Post

    For part ii of the question:

    Let $\displaystyle s = ut + \frac{1}{2}at^{2} $, then $\displaystyle s = [(2\sqrt{17})(3)] + \left[\left(\frac{1}{2}\right)(4)(3^{2})\right] \approx 42.73\ m \neq 36\ m $

    I don't understand why this does not equate to 36 m? Is the initial velocity for the next 3 seconds $\displaystyle 2\sqrt{17} $?

    I think the part ii is not related with the part i, thus makes the initial velocity $\displaystyle u=6ms^{-1}$

    assuming that let's solve the question,

    $\displaystyle \begin{align*} S&=ut+\frac{1}{2}at^2\\ &=(6)(3)+\frac{1}{2}(4)(3)^2\\ &=18+18 \\ &=36m \\ \end{align*}$
    Thanks from astartleddeer
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    Member astartleddeer's Avatar
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    Re: Kinematics problem (uniform acceleration)

    Looking back now, I've just misinterupted the question.

    I thought part ii, how far it will move in the next 3 seconds, was the add-on on the (v,t) graph after the 8m had been covered. In other words, 8m + 36m = the total distance travelled. 1 s + 0.56 s + 3 s = the total time taken to cover this distance. So to compute the next 3 seconds I thought $\displaystyle 2\sqrt{17}\ ms^{-1} $ was the initial velocity for the extra 3 seconds because it was the final velocity after 8m.

    Now re-reading the question, if it then has a velocity of $\displaystyle 6\ ms^{-1}$ find: part i and part ii.

    Ok no problem, thank you for your efforts
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