Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Applied maths question help!

  • 03-10-2012 4:54pm
    #1
    Registered Users, Registered Users 2 Posts: 62 ✭✭


    Hey everyone


    I'm stuck on 1a(ii) 2010

    How is a=-1.17 ??


    Thanks for your help


Comments

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


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


  • Registered Users, Registered Users 2 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


  • 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


  • Registered Users, Registered Users 2 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


  • Registered Users, Registered Users 2 Posts: 7,220 ✭✭✭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


  • Advertisement
  • Registered Users, Registered Users 2 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


Advertisement