Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    14. A particle moving in a straight line with constant acceleration takes 3s and 5sto cover two successive distances of 1m. Find the acceleration


    I got these 2 formulas


    1=3u+4.5a
    2=8u+32a

    Why cannot i use this method to find a?


    1=3u+4.5a *8
    2=8u+32a *3

    I've tried doing it this way but get the wrong answer and instead have to work it out by substitution and am wondering why this method doesnt work and when to know when to use which method
    Offline

    22
    ReputationRep:
    (Original post by ErniePicks)
    14. A particle moving in a straight line with constant acceleration takes 3s and 5sto cover two successive distances of 1m. Find the acceleration


    I got these 2 formulas


    1=3u+4.5a
    2=8u+32a

    Why cannot i use this method to find a?


    1=3u+4.5a *8
    2=8u+32a *3

    I've tried doing it this way but get the wrong answer and instead have to work it out by substitution and am wondering why this method doesnt work and when to know when to use which method
    You need to multiply both sides by 8 and 3.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    You need to multiply both sides by 8 and 3.
    sorry i have another suvat question

    for this question

    A train stops at two stations P and Q which are 2km apart. It acceleratesuniformly from P at 1ms-2 for 15s and maintains a constant speed for a timebefore deceleration uniformly to rest at Q. If the deceleration is 0.5ms-2 findthe time for which the train is travelling at a constant speed



    I calculated that during the first acceleration stage the V=15 m/s so for the constant speed phase i got

    U=15
    V=15
    A=0
    T=?

    Why cannot i use v=u+at to find t? is it because 0 doesnt count as constant acceleration? (i mean it's constantly zero..)
    Offline

    22
    ReputationRep:
    (Original post by ErniePicks)

    U=15
    V=15
    A=0
    T=?

    Why cannot i use v=u+at to find t? is it because 0 doesnt count as constant acceleration? (i mean it's constantly zero..)

    You can use it, 0 is constant acceleration. It's just that your equation is useless.

    It's like asking "why can't I use 1 + 1 = 2 to find t?"; 1 + 1 = 2 is a true equation but it's useless to the specific scenario at hand.
    Offline

    2
    ReputationRep:
    (Original post by ErniePicks)

    Why cannot i use v=u+at to find t? is it because 0 doesnt count as constant acceleration? (i mean it's constantly zero..)
    if a=o then v=u+at is just v=u since it is the same as v=u+0t and anything times zero is still zero
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brussels sprouts
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.