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    I need to work out the following simultaneous linear equation:

    s=ut+1/2at²

    a is constant acceleration, u is initial velocity, t is time

    after times of 3 seconds and 4 seconds the distances travelled were 25mtr and 35mtr

    determine the value of u and a

    Please help I am real lost on this one.
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    (Original post by leeroy221)
    I need to work out the following simultaneous linear equation:

    [img]data:image/png;base64,iVBORw0KGgoAAAANSUhEU gAAAFoAAAAkBAMAAAD7rjCAAAAAAXNSR 0IArs4c6QAAADBQTFRFAAAAAAAATU1Nf Hx8aGhojIyMmpqavb29p6ensrKy0NDQ2 dnZx8fH8PDw4eHh6enp39MGnAAAAAF0U k5TAEDm2GYAAAAJcEhZcwAADsQAAA7EA ZUrDhsAAAAZdEVYdFNvZnR3YXJlAE1pY 3Jvc29mdCBPZmZpY2V/7TVxAAABIklEQVQ4T2NgoCYoJ8UwVnFS VDeQpJphZKiWJCUEgwSt4Mq5HANI0MrC q0eCagYGAyyqucEWMj5AkQIJ8mzAopp1 AhbVIMFVQPFGEYQWVnmGTAZBwQZUs/kMwzcABbkW7GPgcWAD2ikIBAIMDHP9GL QZGD+gmf3xwEKQ4EVBOQYuFI/K8gswPAKZDNIOMYSBIYJhC1gQBA6B7IU CbgcuA4YyMAfJl3IMM6CCQHEfIIa6hLO BZQJDOJpqPgEGLaggw3KwaihgndBc9UD mKqrZ/HLX5SZABBmWgEMACrhEv0sxTPRGc0ljQ tAHiCB+wJdISqpiTDi4gJCJyPKMBSSpJ snsYlKM5tcnRTXTA1JUx5CimCuAjYQwY REUJEE1Ke6gr1oAuhU5E6kNMhMAAAAAS UVORK5CYII=[/img]
    a is constant acceleration, u is initial velocity, t is time

    after times of 3 seconds and 4 seconds the distances travelled were 25mtr and 35mtr

    determine the value of u and a

    Please help I am real lost on this one.
    your image didnt come out right

    Edit: its fine now
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    (Original post by leeroy221)
    I need to work out the following simultaneous linear equation:

    s=ut+1/2at²

    a is constant acceleration, u is initial velocity, t is time

    after times of 3 seconds and 4 seconds the distances travelled were 25mtr


    25 = 3u + 1/2 * a * 9

    and 35mtr

    determine the value of u and a

    Please help I am real lost on this one.
    35 = 4u + 1/2 *a a * 16 = 4u + 8a

    That's two equations in two variables. Surely you can do simultaneous equations?
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    (Original post by leeroy221)
    x.
    So you can form 2 questions from this info;

     \displaystyle (25) = u(3) + \frac{1}{2}a(9) , \quad \text{call this equation}\: (1)

    and

     \displaystyle (35) = u(4) + \frac{1}{2}a(16), \quad \text{call this equation}\: (2)\:

    .
    .
    .

     \displaystyle (1) \: \text{simplifies to;}\quad 25 = 3u + \frac{9}{2}a

     \displaystyle (2) \: \text{simplifies to;}\quad 35 = 4u + 8a

    have you got his far?
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    (Original post by Zacken)
    [/color][/font][/size]

    25 = 3u + 1/2 * a * 9

    [size=3][font=Times New Roman][color=#000000]

    35 = 4u + 1/2 *a a * 16 = 4u + 8a

    That's two equations in two variables. Surely you can do simultaneous equations?
    Yes, I need to solve a pair of simultaneous equations. Unfortunatly I missed this section due to time off with an illness. If you could help with workings out it would be appreciated so I can understand it and complete the remaining questions I have.
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    (Original post by leeroy221)
    Yes, I need to solve a pair of simultaneous equations. Unfortunatly I missed this section due to time off with an illness. If you could help with workings out it would be appreciated so I can understand it and complete the remaining questions I have.
    Dylan did it.
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    (Original post by DylanJ42)
    So you can form 2 questions from this info;

     \displaystyle (25) = u(3) + \frac{1}{2}a(9) , \quad \text{call this equation}\: (1)

    and

     \displaystyle (35) = u(4) + \frac{1}{2}a(16), \quad \text{call this equation}\: (2)\:

    .
    .
    .

     \displaystyle (1) \: \text{simplifies to;}\quad 25 = 3u + \frac{9}{2}a

     \displaystyle (2) \: \text{simplifies to;}\quad 35 = 4u + 8a

    have you got his far?
    Thanks Dylan, so is that the end /answer?
    How do I find the values of u and a?
    Sorry, this topic is totally new to me
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    (Original post by leeroy221)
    Thanks Dylan, so is that the end /answer?
    How do I find the values of u and a?
    Sorry, this topic is totally new to me
    Not quite. From the second equation, we re-arrange to get 4u = 35 - 8a \Rightarrow u = \frac{35}{4} - 2a then plug this into the first equation to get:

    25 = 3\left(\frac{35}{4} - 2a\right) + \frac{9}{2}a which I'm sure you can solve for a and then plug it back into u = \frac{35}{4} - 2a to find u.
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    (Original post by leeroy221)
    Thanks Dylan, so is that the end /answer?
    How do I find the values of u and a?
    Sorry, this topic is totally new to me
    Those last two are basic simultaneous equations, you should be able to solve that :yep:
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    (Original post by Asuna Yuuki)
    Those last two are basic simultaneous equations, you should be able to solve that :yep:
    It's clear that he/she can't.
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    (Original post by leeroy221)
    Thanks Dylan, so is that the end /answer?
    How do I find the values of u and a?
    Sorry, this topic is totally new to me
    Don't be sorry, its great that you're asking for help

    We have two equations in front of us which are;

     \displaystyle (1) ; \quad 25 = 3u + \frac{9}{2}a

     \displaystyle (2) ; \quad 35 = 4u + 8a


    I will multiply (1) by 2 as its never nice to work with fractions

     \displaystyle (1) ; \quad 50 = 6u + 9a

     \displaystyle (2) ; \quad 35 = 4u + 8a

    This next step is very important, we have to remove a variable somehow, we have the choice to remove u or a, in this case it looks easier to remove u so heres what were going to do;


    multiply (1) by 2 and multiply (2) by 3 (you'll see why soon). (1) and (2) then become;

     \displaystyle (1) ; \quad 100 = 12u + 18a

     \displaystyle (2) ; \quad 105 = 12u + 24a


    Notice we've got the coefficient of u the same. So lets do  \displaystyle (2) - (1) (you can do (1)-(2) if you wish but it makes everything minus which is boke)

     \displaystyle (2)-(1); \quad 105-100 = 12u -12u +24a -18a

     \displaystyle (2)-(1); \quad 5 = 6a \implies \boxed{a = \frac{5}{6}}

    We now have a value of a, so now all thats left to do is find the value of u, so well go back to (1) and (2) and choose one to sub our value of a into (again choose whichever looks easiest.)

    Let's choose  \displaystyle (1) ; \quad 50 = 6u + 9a

    so subbing in our value for a we get;

     \displaystyle (1) ; \quad 50 = 6u + 9\left(\frac{5}{6}\right) \implies \boxed{u = \frac{85}{12}}

    This part is optional; sub your values for u and a into (2) to check if it works

     \displaystyle (2) ; \quad 35 = 4\left(\frac{85}{12}\right) + 8\left(\frac{5}{6}\right) which is correct, so were done

    Edit: holy mother of god i spent about 30mins on this... :cry: my latex is so slow

    Edit II: sorry for full solution btw but i felt OP would benefit more from seeing a worked example rather than small hints, but maybe that's not my decision to make

    Edit III; T.I.L. \boxed
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    (Original post by DylanJ42)
    Don't be sorry, its great that you're asking for help

    We have two equations in front of us which are;

     \displaystyle (1) ; \quad 25 = 3u + \frac{9}{2}a

     \displaystyle (2) ; \quad 35 = 4u + 8a


    I will multiply (1) by 2 as its never nice to work with fractions

     \displaystyle (1) ; \quad 50 = 6u + 9a

     \displaystyle (2) ; \quad 35 = 4u + 8a

    This next step is very important, we have to remove a variable somehow, we have the choice to remove u or a, in this case it looks easier to remove u so heres what were going to do;


    multiply (1) by 2 and multiply (2) by 3 (you'll see why soon). (1) and (2) then become;

     \displaystyle (1) ; \quad 100 = 12u + 18a

     \displaystyle (2) ; \quad 105 = 12u + 24a


    Notice we've got the coefficient of u the same. So lets do  \displaystyle (2) - (1) (you can do (1)-(2) if you wish but it makes everything minus which is boke)

     \displaystyle (2)-(1); \quad 105-100 = 12u -12u +24a -18a

     \displaystyle (2)-(1); \quad 5 = 6a \implies \boxed{a = \frac{5}{6}}

    We now have a value of a, so now all thats left to do is find the value of u, so well go back to (1) and (2) and choose one to sub our value of a into (again choose whichever looks easiest.)

    Let's choose  \displaystyle (1) ; \quad 50 = 6u + 9a

    so subbing in our value for a we get;

     \displaystyle (1) ; \quad 50 = 6u + 9\left(\frac{5}{6}\right) \implies \boxed{u = \frac{85}{12}}

    This part is optional; sub your values for u and a into (2) to check if it works

     \displaystyle (2) ; \quad 35 = 4\left(\frac{85}{12}\right) + 8\left(\frac{5}{6}\right) which is correct, so were done

    Edit: holy mother of god i spent about 30mins on this... :cry: my latex is so slow

    Edit II: sorry for full solution btw but i felt OP would benefit more from seeing a worked example rather than small hints, but maybe that's not my decision to make

    Edit III; T.I.L. \boxed
    Dylan, that it brilliant thank you so much.
    I understand it until where you get the value of U? How do you get that fraction?
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    (Original post by leeroy221)
    Dylan, that it brilliant thank you so much.
    I understand it until where you get the value of U? How do you get that fraction?
    Glad to have helped

    Oh sorry maybe I done that part too quickly,

    so once we get a value of a were going to want to sub it into either (1) or (2) to find our value of u. As I said before (1) looks easier so we'll sub  \displaystyle a= \frac{5}{6} into

    so subbing in our value for  \displaystyle a we get;

     \displaystyle (1) ; \quad 50 = 6u + 9\left(\frac{5}{6}\right) (notice how ive just replaced the letter "a" with 5/6)

     \displaystyle 50 = 6u + 7.5


     \displaystyle 6u = 42.5

     \displaystyle u = \frac{85}{12}

    anddd youre done
 
 
 
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