Results 1 to 3 of 3

Thread: uniform acceleration

  1. #1
    Senior Member
    Joined
    May 2016
    From
    england
    Posts
    265
    Thanks
    2

    uniform acceleration

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    5,565
    Thanks
    2348

    Re: uniform acceleration

    what you've done isn't quite right

    the first second is from 0 to 1, the 2nd from 1 to 2, etc.

    so the $nth$ second is $t\in [n-1,n)$

    putting this into the equation of motion we get

    $d(n) = \dfrac{a}{2}(n^2) + u n - \dfrac{a}{2}(n-1)^2 - u(n-1) = \dfrac{a}{2}(n^2) + u n - \dfrac{a}{2}(n^2 - 2n + 1)- u n + u $

    $d(n) = a\left(n-\dfrac 1 2\right)+u$

    Now for (b) we note

    $d(2) = 17$

    $d(7) = 47$

    $\dfrac{3a}{2} + u = 17$

    $\dfrac{13a}{2} + u = 47$

    Solving this we get

    $5a=30$

    $a=6,~u=8$

    $d(10)=6(9.5)+8 = 65$

    $d(n) = 6\left(n-\dfrac 1 2 \right) + 8$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    890
    Thanks
    199

    Re: uniform acceleration

    If you did what you said, you should have ended up with u + an + 0.5a rather than u + an - 0.5a.

    The first second of motion happens between n=0 and n=1.
    The second second of motion happens between n=1 and n=2, etc.

    So the nth second of motion happens between n-1 and n (not n and n+1 which is what you have done)

    So, first of all, you need to fix up part (a).

    Part (b) uses the fact that the distance travelled during the nth second is u + an - 0.5a.

    So, when n = 2, u+an-0.5a = u+2a-0.5a = u+1.5a = 17

    and when n=7 , u+an-0.5a = ..... etc

    This gives 2 simultaneous equations so you can work out u and a. Then sub into the distance formula u+an-0.5a to get a formula in terms of n only. Then you can do (i) and (ii).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. uniform acceleration
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: Sep 12th 2016, 10:56 AM
  2. Uniform Acceleration
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: Jun 7th 2011, 07:19 PM
  3. non uniform acceleration
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: Feb 1st 2009, 07:15 AM
  4. uniform acceleration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jul 28th 2008, 10:39 PM

/mathhelpforum @mathhelpforum