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 all! We have been experiencing an issue on site where threads have been missing the latest postings. The platform host Vanilla are working on this issue. A workaround that has been used by some is to navigate back from 1 to 10+ pages to re-sync the thread and this will then show the latest posts. Thanks, Mike.
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 leaving cert 2013

  • 06-09-2013 7:08pm
    #1
    Registered Users Posts: 2


    hi guys,in the marking scheme of the applied maths lc 2013 question 1)the time in the first part is half(t-t1)can someone tell why its this?


Comments

  • Registered Users, Registered Users 2 Posts: 712 ✭✭✭MmmPancakes


    This is a Junior Cert Forum, don't expect answers here, mods will move the thread to the LC forum where you'll get some answers :)


  • Banned (with Prison Access) Posts: 209 ✭✭yoho139


    The acceleration f is equal to the deceleration. Since it accelerates to a certain speed, then stays at that speed for t1 seconds, the remaining time (for both acceleration and deceleration) is t-t1 (where t is the total time). Since the rate of acceleration and deceleration are equal, the remaining time is split equally between the two, giving 1/2(t-t1) for acceleration and 1/2(t-t1) for deceleration.

    E: Also, please post a link to the marking scheme with the page number next time, it'll make it easier for people trying to help you.


Advertisement