Results 1 to 12 of 12
Like Tree1Thanks
  • 1 Post By Wilmer

Math Help - Trouble with some substitution.. please help!

  1. #1
    Junior Member
    Joined
    Oct 2010
    Posts
    27

    Trouble with some substitution.. please help!

    given:

    1. s= (-1/2)gt^2 + 10t + 2

    and

    2. v = 10 - gt

    i want to eliminate the t term to give:

    v^2 = 4g + 100 - 2gs

    i'm really struggling with this one.. i rearrange eq.2 for t: giving:

    t = v + 10 / (-g)

    subbing this into eq. 1. i'm a little overwhelmed where to start.. appreciate any help.

    thanks in advance,

    Euph.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Trouble with some substitution.. please help!

    Quote Originally Posted by euphmorning View Post
    given:

    1. s= (-1/2)gt^2 + 10t + 2

    and

    2. v = 10 - gt

    i want to eliminate the t term to give:

    v^2 = 4g + 100 - 2gs

    i'm really struggling with this one.. i rearrange eq.2 for t: giving:

    t = v + 10 / (-g) <--- that should be: t = (10-v)/g

    subbing this into eq. 1. i'm a little overwhelmed where to start.. appreciate any help.

    thanks in advance,

    Euph.
    Substituting t by the term (10-v)/g yields:

    s = -\frac12 \cdot g \cdot \left(\frac{10-v}{g} \right)^2+10\left(\frac{10-v}{g} \right)+2

    Expand the brackets and collect like terms. Afterwards you can factor the numerator.

    You should come out with:

    s = \frac{(v+10)(v-10)}{2g}+2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2010
    Posts
    27

    Re: Trouble with some substitution.. please help!

    thankyou for your informed reply. unfortunately i'm still a little stuck. evidently i really need to go over my algebra.. is there any chance of a work through what is going on to get this factorised as given, then rearrangment to v^2?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68

    Re: Trouble with some substitution.. please help!

    Do you realise that (v + 10)(v - 10) = v^2 - 100 ?
    If not, then you are unaware of basics, so need classroom help...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2010
    Posts
    27

    Re: Trouble with some substitution.. please help!

    Yes, i perfectly reason that Wilmer. That is not the problem i'm having. as i've stated, it's arriving at that which i'm having difficultly with.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68

    Re: Trouble with some substitution.. please help!

    Well, to start, do you agree that your t = v + 10 / (-g) is incorrect and should be: t = (10-v) / g;
    PLUS that you needed brackets with yours: t = (v + 10) / (-g) ?

    I'm not sure about Earboth's s = (v + 10)(v - 10) / (2g) + 2:
    that simplifies to v^2 = -4g + 100 + 2gs, NOT to v^2 = 4g + 100 - 2gs as in your initial post...
    did you make a typo? Maybe we'll wait to see what Earboth has to say...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68

    Re: Trouble with some substitution.. please help!

    Quote Originally Posted by earboth View Post
    s = \frac{(v+10)(v-10)}{2g}+2
    Had another look...above is incorrect (sorry Earboth!); should be:
    s = 2 - (v+10)(v-10) / 2g
    This leads to your v^2 = 4g + 100 - 2gs : so you made no typo!

    You want help in getting to above, right? Well, let's see:

    s = -g/2[(10-v)/g]^2 + 10[(10-v)/g] + 2 : that's what we start with, ok?
    Multiply by -2:
    -2s = g[(10-v)/g]^2 - 20[(10-v)/g] - 4 : changes -g/2 to g (easier!)
    Expand:
    -2s = g[(100 - 20v + v^2) / g^2] - (200 + 20v) / g - 4
    Multiply by g:
    -2gs = 100 - 20v + v^2 - 200 + 20v - 4g
    Simplify:
    -2gs = v^2 - 100 - 4g
    In terms of v^2:
    v^2 = 4g + 100 - 2gs : SUCCESS!!

    By the way, that was my way of doing it; there are other ways...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2010
    Posts
    27

    Re: Trouble with some substitution.. please help!

    Wilmer, Brilliant - exactly what I needed help with. Thanks!

    kind regards,

    Euph.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Trouble with some substitution.. please help!

    Here is what I did ... (if I've made amistake I apologize for the confusion!)
    Quote Originally Posted by earboth View Post
    Substituting t by the term (10-v)/g yields:

    s = -\frac12 \cdot g \cdot \left(\frac{10-v}{g} \right)^2+10\left(\frac{10-v}{g} \right)+2

    ...
    Quote Originally Posted by Wilmer View Post
    Well, to start, do you agree that your t = v + 10 / (-g) is incorrect and should be: t = (10-v) / g;
    PLUS that you needed brackets with yours: t = (v + 10) / (-g) ?

    I'm not sure about Earboth's s = (v + 10)(v - 10) / (2g) + 2:
    that simplifies to v^2 = -4g + 100 + 2gs, NOT to v^2 = 4g + 100 - 2gs as in your initial post...
    did you make a typo? Maybe we'll wait to see what Earboth has to say...
    My re-arrangerments of the equation above:

    s = - \left(\frac{10-v}{2g} \right)^2+20\left(\frac{10-v}{2g} \right)+2

    Factoring out \frac{10-v}{2g} :

    s =  \left(\frac{10-v}{2g} \right)(-(10-v)+20)+2

    s =  \left(\frac{10-v}{2g} \right)(10+v)+2

    I can't detect a mistake ... so sorry.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Trouble with some substitution.. please help!

    Quote Originally Posted by Wilmer View Post
    Had another look...above is incorrect (sorry Earboth!); should be:
    s = 2 - (v+10)(v-10) / 2g
    This leads to your v^2 = 4g + 100 - 2gs : so you made no typo!

    You want help in getting to above, right? Well, let's see:

    s = -g/2[(10-v)/g]^2 + 10[(10-v)/g] + 2 : that's what we start with, ok?
    Multiply by -2:
    -2s = g[(10-v)/g]^2 - 20[(10-v)/g] - 4 : changes -g/2 to g (easier!)
    Expand:
    -2s = g[(100 - 20v + v^2) / g^2] - (200 + 20v) / g - 4 <--- are you sure?
    Multiply by g:
    -2gs = 100 - 20v + v^2 - 200 + 20v - 4g
    Simplify:
    -2gs = v^2 - 100 - 4g
    In terms of v^2:
    v^2 = 4g + 100 - 2gs : SUCCESS!!

    By the way, that was my way of doing it; there are other ways...
    ...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68

    Re: Trouble with some substitution.. please help!

    [Earboth]:
    -2s = g[(10-v)/g]^2 - 20[(10-v)/g] - 4 : changes -g/2 to g (easier!)
    Expand:
    -2s = g[(100 - 20v + v^2) / g^2] - (200 + 20v) / g - 4 <--- are you sure?
    .................................................. .................................................. ....................................

    Whoops....typo: should be (200 - 20v)
    -2s = g[(100 - 20v + v^2) / g^2] - (200 - 20v) / g - 4 <--- are you sure?
    You can tell by the line that follows above:
    2gs = 100 - 20v + v^2 - 200 + 20v - 4g

    OK?!
    Btw, yes, I am SURE of final result: tested it.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68

    Re: Trouble with some substitution.. please help!

    Quote Originally Posted by earboth View Post
    s = - \left(\frac{10-v}{2g} \right)^2+20\left(\frac{10-v}{2g} \right)+2 *******1

    Factoring out \frac{10-v}{2g} :

    s = \left(\frac{10-v}{2g} \right)(-(10-v)+20)+2

    s = \left(\frac{10-v}{2g} \right)(10+v)+2 *******2

    I can't detect a mistake ... so sorry.
    *******1, *******2 : those 2 do not equate; try v=2 and g=2 :
    *******1: s = 38
    *******2: s = 26

    I think your mistake is going from (-g/2)[(10-v)/g]^2 to -[(10-v)/(2g)]^2

    The initial equation given by the OP is : s= (-1/2)gt^2 + 10t + 2
    I think the best/easiest way to START is divide through by (-1/2)g = -g/2:
    -2s/g = t^2 - 20t/g - 4/g
    Then ambiguosities have vanished!
    Last edited by Wilmer; August 17th 2012 at 09:16 AM.
    Thanks from earboth
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trouble with integral substitution
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 2nd 2010, 09:10 PM
  2. TI-89 Trouble
    Posted in the Calculators Forum
    Replies: 6
    Last Post: March 8th 2010, 02:27 PM
  3. Having some trouble
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 4th 2008, 04:49 AM
  4. I'm having trouble with this one
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 10th 2008, 02:05 PM
  5. Replies: 5
    Last Post: April 3rd 2008, 01:19 PM

Search Tags


/mathhelpforum @mathhelpforum