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    Theres a question in the edexcel book :-

    A particle moves with constant acceleration along the straight line OLM and passes through the points O,L,M at times 0s,4s and 10s respectively. Given that OL = 14m and OM= 50m find the acceleration of the particle.

    The answer is 0.5ms^-1. They got this by solving simultanously to eliminate v from the equation S=vt-1/2at^2. They used the values from OM and LM and I understand how they got the answer. But what I don't understand is why OM and LM, why not OL and LM or OM and LM ?? I've tried these two aswell and get a different answer . Anyone who could clear this up for me ??.

    Thanks James.
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    k we dont know the initial velocity so keep it as u. we got displacement and time
    we need to find a.
    use s = ut + 0.5at^2 for both distance OL and OM with their respective times.
    you should come out with 2 simultaneous equations with unknowns u and a.
    cancel out the u to find a using subtraction method or substitution.


    remember acceleration is the same so we can use OM and OL and can do it in other ways as you mentioned above. its just easier to do it this way because we know "u" is the same for both equations as they both start at O with the same velocity.
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    they use OM and LM coz the distances are given for OM and LM
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    (Original post by HassanBHS)
    they use OM and LM coz the distances are given for OM and LM
    The distances are given for OL and OM not LM that has to be worked out. Which makes me wonder why they use LM ? rather than OM and OL
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    (Original post by jhwilliam)
    The distances are given for OL and OM not LM that has to be worked out. Which makes me wonder why they use LM ? rather than OM and OL
    I don't think it matters which distances you use, so long as you have enough information to form equations.
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    (Original post by steve2005)
    I don't think it matters which distances you use, so long as you have enough information to form equations.
    So did I, but I tried it by forming an equation using OL and got a wrong answer. I'm probably being retarded.
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    (Original post by jhwilliam)
    So did I, but I tried it by forming an equation using OL and got a wrong answer. I'm probably being retarded.
    You can't use s = vt - 1/2at^2 with the distances OL and OM as v is different a L and M, since the particle is accelerating.

    You can only use s = vt - 1/2at^2 if both the distances end at M, as the velocities will be the same, so it can be cancelled out.

    Similarly you can use s = ut + 1/2at^2 if you use two distances that start at O, as the velocities will be the same for both equations.
 
 
 
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