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    A cyclist rides a long a straight road. She passes three points A,B and C, in that order, where AB = 130m and BC = 120,. She passes B 10 seconds after she passed A and she passed C a further 15 seconds later. Assuming that the cyclist is travelling with constant acceleration find

    a) the speed of the cyclist as she passes A and her acceleration
    b) The distance AD where D is the point where the cyclist comes to rest

    I tried to work out the first bit by writing out SUVAT for point A,B and C

    Point A
    S = 130
    U = ?
    V =?
    A =?
    t = t

    Point B
    S = 140
    U = ?
    V =?
    A =?
    t = t - 10

    Point C
    S = 250
    U = ?
    V =?
    A =?
    t = t - 25

    I tried using lots of different SUVAT equations but I always have more than one unknown and I can't seem to get rid of any. So I can't work out any of the missing values. If anyone could help, that would be much appreciated.
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    (Original post by Jasminea)
    A cyclist rides a long a straight road. She passes three points A,B and C, in that order, where AB = 130m and BC = 120,. She passes B 10 seconds after she passed A and she passed C a further 15 seconds later. Assuming that the cyclist is travelling with constant acceleration find

    a) the speed of the cyclist as she passes A and her acceleration
    b) The distance AD where D is the point where the cyclist comes to rest
    Try this question by considering the motion between A and B and then the motion between A and C.

    A -> B:
    s = 130
    u = u
    a = a
    t = 10

    A -> C:
    s = 250
    u = u
    a = a
    t = 25

    The 'u' and 'a' here are the same for A -> B and A -> C so you should end up with simultaneous equations in 'u' and 'a' which you can solve. Please post your working if you get stuck.
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    speed at A = e

    speed at B = f

    speed at C = g

    using the SUVAT s = ut + 0.5at2

    130 = e*10 + 0.5*a*102

    250 = e*25 + 0.5*a*252

    from there you find e & a using simultaneous equations then use another SUVAT to find f & g
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    (Original post by notnek)
    Try this question by considering the motion between A and B and then the motion between A and C.

    A -> B:
    s = 130
    u = u
    a = a
    t = 10

    A -> C:
    s = 250
    u = u
    a = a
    t = 25

    The 'u' and 'a' here are the same for A -> B and A -> C so you should end up with simultaneous equations in 'u' and 'a' which you can solve. Please post your working if you get stuck.
    But in your working aren't you suggesting that A is the starting point when it doesn't state that in the question.

    The answers are a =-0.4 and u =15
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    (Original post by Jasminea)
    But in your working aren't you suggesting that A is the starting point when it doesn't state that in the question.

    The answers are a =-0.4 and u =15
    No I haven't. I have only said that the displacement from A to B is 130 and the time is 10. These are true whether the cyclist started from A or not.

    And the same for A to C : the displacement is 250 and the time is 25.

    I have left 'u' and 'a' as variables since they are unknown. If the cyclist started at A then u = 0 but we don't know that.
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    (Original post by notnek)
    No I haven't. I have only said that the displacement from A to B is 130 and the time is 10. These are true whether the cyclist started from A or not.

    And the same for A to C : the displacement is 250 and the time is 25.

    I have left 'u' and 'a' as variables since they are unknown. If the cyclist started at A then u = 0 but we don't know that.
    I understand it now. I just made a mistake in my workings.
    Thanks
 
 
 
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