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    A car is moving along a straight horizontal road at a constant speed of 18ms^-1. At the instant when the car passes a lay by, a motorcyclist leaves the lay by, starting from rest and moves with a constant acceleration 2.5ms^-2 in pursuit of the car. Given that the motorcyclist overtakes the car T seconds after leaving the lay by, calculate the value of T, and the speed of the motorcyclist at the instant of passing the car.

    The solution says to use s = d/t because of the cars constant speed which I understand, but it says "as the car and the motorcycle were level at the lay by, when the motorcycle overtakes the car, they have travelled the same distance", so equate the 2 distances each have travelled.

    But I'm having trouble understanding why they move the same distance? I mean I can learn this to apply it, but I can't make sense of it. Can anyone explain it clearly? How will they have travelled the same distance and how am I supposed to know this in a question? Like what in the question indicated this? is the distance travelled by the car by s = d/t the distance travelled starting from when it passes the lay by? I'm quite confused
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    The point at which the motorbike overtakes the car, lets call it z.

    Call the lay by, o.

    OZ is the distance. And thats what they've both travelled.

    If it asked for the distance before the lay by as well, then they'd be different.
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    (Original post by L'Evil Fish)
    The point at which the motorbike overtakes the car, lets call it z.

    Call the lay by, o.

    OZ is the distance. And thats what they've both travelled.

    If it asked for the distance before the lay by as well, then they'd be different.
    but surely at the point the bike overtakes the car, the motorbike would be ahead, or does overtake just mean slightly in front of?
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    (Original post by lollage123)
    but surely at the point the bike overtakes the car, the motorbike would be ahead, or does overtake just mean slightly in front of?
    Yeah, but for a moment they're at the same spot
 
 
 
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