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    • Thread Starter
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    I need help solving tis question, I know which formula to use but I just don't get the correct answer on the calculator.

    Here's the question:

    A particle is moving along a straight line OA with constant acceleration 2.5ms-1. At time t=0, the particle passes through 0 with speed 8ms-1 and is moving in the direction OA. The distance OA is 40m. Find the time taken for the particle to move from O to A.

    The formula I used is S=vt-1/2at^2 and I got the wrong answer.

    Any help?
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    been a while since i did any of this but might help if you post what answer you get so i can try to see where your going wrong
    • PS Reviewer
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    PS Reviewer
    Its fairly simple
    Use SUVAT since its constant acceleration
    s=40m
    U=8m/s
    V
    a=2.5m/s^2
    t=?
    Sub numbers into
    s=ut + 1/2at^2
    rearrange to get a quadratic and solve using quadratic equation or complete the square
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    The formula is s=ut+1/2at^2

    But, rather than using that formula and generating a very annoying quadriatic, it might be easier to use:

    Use v = sqrt (u^2 + 2as)
    v = sqrt (64 + 2*2.5*40)
    v = 16.248

    Then use t = (v-u)/a
    t = 16.248-8 / 2.5
    t = 3.3s


    If you wanted to use the first formula, do

    s = ut + 1/2 a t^2
    40 = 8t + 1.25t^2
    0 = 1.25t^2 + 8t - 40
    0 = 5t^2 + 32t - 40

    Use Quadriatic formula now (see why I suggested against this method?) to generate:

    t = -9.7s or 3.3s
    t = -9.7s or 3.3s (NEGATIVE VALUE IS INVALID)
    t = 3.3s
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    Have you tried: s = ut + {1/2}at² ?
    • Thread Starter
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    (Original post by innerhollow)
    The formula is s=ut+1/2at^2

    But, rather than using that formula and generating a pain of a quadriatic, it might be easier to use:

    Use v = sqrt (u^2 + 2as)
    v = sqrt (64 + 2*2.5*40)
    v = 16.248

    Then use t = (v-u)/a
    t = 16.248-8 / 2.5
    t = 3.3s
    Thanks, I see where I went wrong .. i'll rep you tomorrow
 
 
 
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