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  1. MHB Apprentice

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    #1
    A particle travels srating with a initial speed u, with uniform acceleration a. Show that the distance travelled during the nth second is u+an-.5a. (b) if the particle travels 17m in the 2nd second of motion and 47m in the 7th second of motion how far will it go in the (i) 10th second of motion (ii) nth second of motion

    i worked out the first bit by letting s=un+.5an^2. i then found s=u(n+1)=5a(n+1)^2. and i took these away from each other and i was left with the answer. i no not know what to do for part b though

  2. Pessimist Singularitarian
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    #2
    a) We begin with the kinematic equation:

    $ \displaystyle s(n)=\frac{a}{2}n^2+un$

    And so:

    $ \displaystyle \Delta s_n=s(n)-s(n-1)=\frac{a}{2}n^2+un-\frac{a}{2}(n-1)^2-u(n-1)=an+u-\frac{a}{2}$

    b) We are given:

    $ \displaystyle \Delta s_2=2a+u-\frac{a}{2}=\frac{3}{2}a+u=17$

    $ \displaystyle \Delta s_7=7a+u-\frac{a}{2}=\frac{13}{2}a+u=47$

    Multiplying both equations by 2, we obtain the 2X2 system:

    $ \displaystyle 3a+2u=34$

    $ \displaystyle 13a+2u=94$

    To proceed, subtract the former equation from the latter, eliminating $u$ and solve the result for $a$. Then use either equation to find $u$ using the value you find for $a$. Once you have $a$ and $u$, you will be able to express $\Delta s_n$ in terms of $n$ alone.

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