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Math Help - Motion

  1. #1
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    Motion

    A car starts from rest and maintains a uniform acceleration of 2.0 ms-2 along a straight road. As the car starts, another car moving in the same direction along a prallel road with a constant velocity of 80 kmh-1 passes it. When will the cars be level with each other again?

    I know that this is quite a simple question, but no matter what equations attempt to use I can't achieve the correct answer. I tried t = (v-u)/a, letting v = 22.22 ms-2, but got 11.11 s (which is incorrect). Does this mean that Car A was first passed by Car B after 11.11 seconds? Does this have any significance?

    Thank you for your patience.
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  2. #2
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    Re: Motion

    If the cars are parallel at t0, then the t you want is when their displacement is the same. What is d for a car with v0 and a=2.0ms^-2 ?
    Thanks from Fratricide
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  3. #3
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    Re: Motion

    Quote Originally Posted by mmonkman View Post
    If the cars are parallel at t0, then the t you want is when their displacement is the same. What is d for a car with v0 and a=2.0ms^-2 ?
    Do I need a value for t to solve this? If t = 0 and v = 0, then the equation for displacement simply equals 0 (i.e. they haven't begun moving yet).
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  4. #4
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    Re: Motion

    No you do not want to input t=0. We already know that the cars are parallel at that time. What other value for t gets car A and car B with the same displacement. i.e. d of car A = d of car B. So you would want to solve for t.
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  5. #5
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    Re: Motion

    What equation should I use?
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  6. #6
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    Re: Motion

    Quote Originally Posted by Fratricide View Post
    What equation should I use?
    Because the initial velocity is 0:

    d = (1/2)at^2 for the accelerated car
    d = vt for uniform motion of the second car

    Can you go from there?
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