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    This is the question

    A car is travelling along a straight horizontal road with constant acceleration. The car passes over 3 consecutive point A, B and C. AB = 100m, BC= 300m. At B the speed of the car is 14m/s and at C it is 20m/s. Find the acceleration and the tome taken for the car to travel form A to C.

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    I got 0.34 for acc which is correct. Then I used s = vt -0.5at^2 and got two answers. What would they actually mean in practice? Why would it pass C twice?

    The book solution first found the value of u at A and the used that in 20 = u + 0.34t thus getting only one value for t which was the lower of the two solutions I got.

    Thanks
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    You will frequently get 2 answers when you use  s=ut+1/2at^2 but normally it's just a case of discrediting one of the answers (like if you get t<0).
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    (Original post by B_9710)
    You will frequently get 2 answers when you use  s=ut+1/2at^2 but normally it's just a case of discrediting one of the answers (like if you get t<0).
    Thanks - I get that but on what grounds could I discredit the other value as it wasn't -ve. I suppose the smaller value MUST be right, logically, but is that all I could do?
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    (Original post by maggiehodgson)
    This is the question

    A car is travelling along a straight horizontal road with constant acceleration. The car passes over 3 consecutive point A, B and C. AB = 100m, BC= 300m. At B the speed of the car is 14m/s and at C it is 20m/s. Find the acceleration and the tome taken for the car to travel form A to C.

    -----------------------------------------------------------------------------------------------
    I got 0.34 for acc which is correct. Then I used s = vt -0.5at^2 and got two answers. What would they actually mean in practice? Why would it pass C twice?

    The book solution first found the value of u at A and the used that in 20 = u + 0.34t thus getting only one value for t which was the lower of the two solutions I got.

    Thanks
    I presume one of your answers is negative.

    The equation s = vt -0.5at^2 is a quadratic in t, either "u" shaped or "n" shaped, and for a given displacment (s) you often have two values for t.

    If the acceleration had remained constant throughout, then at some stage in the past (before t=0), the velocity would have been negative and the car would have been travelling backwards, first through C, then B, then A, eventually coming to a stop and then moving forward, and it's this latter part that the question is asking about.
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    (Original post by maggiehodgson)
    Thanks - I get that but on what grounds could I discredit the other value as it wasn't -ve. I suppose the smaller value MUST be right, logically, but is that all I could do?
    In this case the higher value relates to the time when the car first passed A, going backwards! Note there is no "u" in your equation. For the higher time u would have been negative, which is contrary to the question.

    The difference between the two times is the time from when the car was going backwards through A (prior to t=0), until it came to a stop, and started going forwards again and got to A (at t=0).
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    (Original post by ghostwalker)
    In this case the higher value relates to the time when the car first passed A, going backwards! Note there is no "u" in your equation. For the higher time u would have been negative, which is contrary to the question.

    The difference between the two times is the time from when the car was going backwards through A (prior to t=0), until it came to a stop, and started going forwards again and got to A (at t=0).

    Wow! Thanks
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    (Original post by maggiehodgson)
    Wow! Thanks
    Just to clarify, I was taking a few liberties with the definition of t there.

    For the actual question t=0 refers to when the car was moving forwards at A, giving you the lower time.

    The higher answer refers to t=0 when the car was moving backwards at A, hence the greater time.
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    (Original post by ghostwalker)
    Just to clarify, I was taking a few liberties with the definition of t there.

    For the actual question t=0 refers to when the car was moving forwards at A, giving you the lower time.

    The higher answer refers to t=0 when the car was moving backwards at A, hence the greater time.

    As usual I'm struggling to get this.

    When it was moving backwards at A, where was it coming from? Was it moving in the direction B-A and if so where did it stop and turn round? And why did it stop because the acceleration was +ve.

    I don't suppose you could give me a drawing could you?
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    (Original post by maggiehodgson)
    As usual I'm struggling to get this.

    When it was moving backwards at A, where was it coming from? Was it moving in the direction B-A and if so where did it stop and turn round? And why did it stop because the acceleration was +ve.

    I don't suppose you could give me a drawing could you?
    Here's a time displacement graph, set up for positive velocity at A at t=0.

    Displacment is on the vertical axis, and time on the horizontal.

    Edit: Since acceleration is constant, displacment against time is a parabola (or part of one) - as is always the case with suvat. It will be inverted if the acceleration is negative.

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