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

    19
    ReputationRep:
    Question:
    An athlete runs along a straight road. She starts from rest and moves with constant acceleration for 5 seconds, reaching a speed of 8m/s. This speed is then maintained for T seconds. She then decelerates at a constant rate until she stops. She has run a total of 500m in 75(s).

    b) Calculate the value of T

    My issue:
    So I drew the graph (I can't post it due to technical difficulties) and I have two unknowns, T and T2. I then presumed I had 2 simultaneous equations.

    First equation -- T + T2 = 70 (added up the seconds, minused 5 from 75...)
    Second equation -- 8T + 4T2 = 480 (area under the graph, minused 20 from 500...)
    Third equation -- 4T + 4T2 = 280 (I multiplied the first equation by 4 so I cancel out T2)

    I then did my working out and I got T = 50 seconds. Is this correct?
    • Community Assistant
    Offline

    13
    ReputationRep:
    Community Assistant
    (Original post by Ze Witcher)
    Question:
    An athlete runs along a straight road. She starts from rest and moves with constant acceleration for 5 seconds, reaching a speed of 8m/s. This speed is then maintained for T seconds. She then decelerates at a constant rate until she stops. She has run a total of 500m in 75(s).

    b) Calculate the value of T

    My issue:
    So I drew the graph (I can't post it due to technical difficulties) and I have two unknowns, T and T2. I then presumed I had 2 simultaneous equations.

    First equation -- T + T2 = 70 (added up the seconds, minused 5 from 75...)
    Second equation -- 8T + 4T2 = 480 (area under the graph, minused 20 from 500...)
    Third equation -- 4T + 4T2 = 280 (I multiplied the first equation by 4 so I cancel out T2)

    I then did my working out and I got T = 50 seconds. Is this correct?
    Yes, your value of T is correct. Alternatively, you can express your T2 in terms of T as T_2 = 75 - (5+T). Since the deceleration begins after 5+T seconds. Then you should only need the equation formed from considering the area under the graph.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by SherlockHolmes)
    Yes, your value of T is correct. Alternatively, you can express your T2 in terms of T as T_2 = 75 - (5+T). Since the deceleration begins after 5+T seconds. Then you should only need the equation formed from considering the area under the graph.
    Thanks for that -- I tried to do that single equation but I thought I was wrong -- I made a silly mistake prior to that calculation which meant it didn't work but I see it now
 
 
 
  • 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
    Has a teacher ever helped you cheat?
    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

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