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Math Help - SUVAT(time)

  1. #1
    Junior Member Mist's Avatar
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    SUVAT(time)

    A particle is moving along a straight line OA with constant acceleration 2ms-2. At OA the particle is moving towards A with speed 5.5ms-1. The distance OA is 20m. Find the time the particle takes to move from O to A.

    S=20 V=5.5 A=2 T=?

    I used the formula  s=vt-1/2at^2 but I got the wrong answer.

    Any help?
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mist View Post
    A particle is moving along a straight line OA with constant acceleration 2ms-2. At OA the particle is moving towards A with speed 5.5ms-1. The distance OA is 20m. Find the time the particle takes to move from O to A.

    S=20 V=5.5 A=2 T=?

    I used the formula  s=vt-1/2at^2 but I got the wrong answer.

    Any help?
    I can't see any other method to solve this. In this case I am using A to denote acceleration as to not confuse with a in the quadratic formula

    s = vt + \frac{1}{2}At^2

    \frac{1}{2}At^2 + vt -s = 0

    t = \frac{-v\pm \sqrt{v^2+2As}}{A}

    Sub in what you know and I get t = 8
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  3. #3
    Junior Member Mist's Avatar
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    Quote Originally Posted by e^(i*pi)
    Sub in what you know and I get t = 8
    Thats the wrong answer, the answer is suppossed to be 2.5

    The formula has been rearranged wrong
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mist View Post
    Thats the wrong answer, the answer is suppossed to be 2.5

    The formula has been rearranged wrong
    From what I can gather you've took the initial speed, u as the final speed, v. If we define O as the start point and O to A is the positive direction then at point O the speed given is the initial speed.

    My equation is fine, I didn't evaluate both roots in my previous post - the second root is 2.5

    s = ut+\frac{1}{2}at^2 (this is one of the suvat equations)

    s = 20, u = 5.5, t = t, a = 2

    20 = 5.5t+\frac{1}{2}(2)t = 5.5t+t^2

    As this is a quadratic rearrange into the form ax^2+bx+c=0

    t^2+5.5t-20 = 0

    Solving that using the quadratic formula gives t = 8 and t= 2.5.
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