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Rearranging Formulae

  1. Apr 14, 2007 #1
    Hey there,

    I was unsure as where to put this as it seems to fit into quite a few categories, however since it is primarily an algebra question I decided on the algebra forum.

    I recently spent loads of time working out a formula which (if the object is steadily accelerating through a vacuum, has no outside influences, and if you know that acceleration and also the start and end velocity of the object) gives the distance the object travelled. Here's that formula:

    [tex] d = \frac {v(\frac {v - u} {a} + 1)} {2} [/tex]

    d = distance travelled
    v = final velocity
    u = start velocity
    a = acceleration

    My first problem is that I'm wanting to rearrange the formula to make each the final velocity (v), start velocity (u) and acceleration (a) it's subject, but as of yet I haven't been able to do so.

    My second question is whether I've just wasted my time doing this and there's already a formula to do what my formula does. This was the part of the post that made me unsure as to which forum to post this in as it's sort of a physics question.

    Sorry if this all seems really trivial to anybody, but I'm really only formally educated up to a high school level (seeing as I'm in high school at the moment) and everything else I know is just bits and pieces I've picked up from magazines and the like. Oh and don't worry this isn't homework.

    Thanks in advance.

    Indy

    [edit] I've just realised that it sounds very much like I'm asking you to do this for me, but I assure you I've searched high and low for a formula similar to mine and couldn't find any. Also in regards to rearranging, it would be great if you could do it for me (save me a lot of work :P), but I'd really like to learn to do this on my own, I haven't been able to find any sort of advanced rearranging tutorials anywhere, so if you know of any and can post them I'd be more than grateful.[/edit]
     
    Last edited: Apr 14, 2007
  2. jcsd
  3. Apr 14, 2007 #2

    cristo

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    There is already a formula; it is [tex] v^2=u^2+2ad[/tex] or, with d as the subject, [tex]d=\frac{v^2-u^2}{2a}[/tex] which you can see isnt the same as yours.

    How did you derive your equation? There may be a flaw we could spot.

    Here's a link to a page with this, and the other kinematic equations of motion: http://www.glenbrook.k12.il.us/gbssci/Phys/Class/1DKin/U1L6c.html
     
    Last edited: Apr 14, 2007
  4. Apr 14, 2007 #3
    Ah :blushing: well I feel a fool.

    Okay, I'll try to remember how I got through this...

    Well first of all I created 2 "results" tables to check whether my final formula was right, and to help with the 'formulating' of said formula. These were with the accelerations of [tex] 10m/s^2 [/tex] and [tex] 7m/s^2 [/tex].

    The tables went by seconds (i.e. at 1 second the distance travelled was 10m and the velocity 10m/s, then at 2 seconds distance = 30m and velocity 20m/s).

    From there I tried dividing the distance by the velocity at any given second and got a sequence which increased by 0.5 every time. Which had the formula:

    [tex] \frac {1} {2}n +0.5 [/tex]

    I replaced the 'n' with 't' for time and so had the formula:

    [tex] \frac {d} {v} = \frac {1} {2}t + 0.5 [/tex]

    I then looked for formulae which included 't' and came up with:

    [tex] s = \frac {d} {t} [/tex]

    and

    [tex] a = \frac {v - u} {t} [/tex]

    I decided that since the formula for average speed dealt with averages, and I was wanting to be as accurate as possible, the formula for acceleration would be more suited, also seeing as it involved 'u', 'v' and 'a'.

    So from there I rearranged the acceleration formula to make 't' the subject:

    [tex] t = \frac {v - u} {a} [/tex]

    and replaced the 't' in my original equation with that formula, leaving me with:

    [tex] \frac {d} {v} = \frac {1} {2}(\frac {v - u} {a}) +0.5 [/tex]

    I moved the 'v' over to the other side to leave d by itself and then simplified the formula to get:

    [tex] d = \frac {v(\frac {v - u} {a} + 1)} {2} [/tex]

    There, that was pretty much everything I did.

    Hope this helps.

    Indy.
     
  5. Apr 15, 2007 #4
    Oh wait, I realise what I've done now. In simplifying the formula I got it wrong. Ack, oh well never mind. Thanks for you help.

    Indy

    [edit] actually wait...it works doesn't it? [/edit]
     
    Last edited: Apr 15, 2007
  6. Nov 16, 2010 #5
    The formula for acceleration is therefore:

    a= 1/(2d/v)-1 . (v-u)

    Lets use an example, just say d=10, then going by the formula, v=10, u=8 and a=2 because 10((10-8)/2)+1)=20 and 20/2 obviously = 10

    Therefore if we fill in the formula a= 1/(2d/v)-1 . (v-u), then we have 1/((20/10))-1 which = 1
    and then 1 . (10-8) = 2 which is A.

    I know this post is 3 years old, although i came across this question and couldn't help but try and solve it.
     
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