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Applied maths question help!

  • 03-10-2012 5:54pm
    #1
    ampmm
    Registered Users Posts: 62 ✭✭


    Hey everyone


    I'm stuck on 1a(ii) 2010

    How is a=-1.17 ??


    Thanks for your help


Comments

  • #2
    aarond280
    Registered Users, Registered Users 2 Posts: 275 ✭✭aarond280


    Are you asking about the - ? If so its because it's because it's deceleration.


  • #3
    ampmm
    Registered Users Posts: 62 ✭✭ampmm


    aarond280 wrote: »
    Are you asking about the - ? If so its because it's because it's deceleration.



    No I know - is deceleration... I don't know how to get the 1.17 bit :p


  • #4
    aarond280
    Registered Users, Registered Users 2 Posts: 275 ✭✭aarond280


    oh :P, -7/6 = -1.17 when rounded down to two decimal places. Does that help :D


  • #5
    dtfo
    Registered Users Posts: 125 ✭✭dtfo


    In the extra second he has travelled 14 metres and thus has only 84 metres left to decelerate

    so plug 84 into the formula instead of 98, hope that helps


  • #6
    bren2001
    Registered Users, Registered Users 2 Posts: 6,893 ✭✭✭bren2001


    he does not apply the brakes for 1 second so how far does he travel in that time?

    s = ut +0.5at^2
    s=14(1)+0
    s=14 m

    the distance left for him to travel is 98-14=84

    final velocity is 0 as he stops:

    v^2 = u^2 +2as
    0 = (14^2)+2a(14)
    a=-1.17 m/s


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  • #7
    ampmm
    Registered Users Posts: 62 ✭✭ampmm


    bren2001 wrote: »
    he does not apply the brakes for 1 second so how far does he travel in that time?

    s = ut +0.5at^2
    s=14(1)+0
    s=14 m

    the distance left for him to travel is 98-14=84

    final velocity is 0 as he stops:

    v^2 = u^2 +2as
    0 = (14^2)+2a(14)
    a=-1.17 m/s



    Brilliant, thank you so much


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