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Thread: journey

  1. #1
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    journey

    A car starts from rest at P and moves with constant acceleration k m/s2.
    Three seconds later another car passes through P travelling in the same direction with constant speed u m/s, (i) , show that the second car will just catch up on the first if u = 6k and that it will not catch up on it if u < 6k.

    using s=ut+.5at^2 i have formed the 2 equations s=.5kt^2 and s=6kt-18k for the 1st and 2nd car respectively. im not sure what to do next .
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  2. #2
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    Re: journey

    sometimes it helps keep things straight by writing

    $s_1(t) = 0.5 k t^2$

    $s_2(t) = 6kt - 18k$

    i.e. these are both functions of $t$

    so at some time $t$ in the future car 2 will catch up (or not) to car 1, i.e. $s_1(t)=s_2(t)$

    I'm going to let you finish but what I suspect you'll find is that this equation is true for some real value of $t$ for $u \geq 6k$ and but only for a complex value of $t$ for $u < 6k$
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  3. #3
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    Re: journey

    Quote Originally Posted by markosheehan View Post
    A car starts from rest at P and moves with constant acceleration k m/s2.
    Three seconds later another car passes through P travelling in the same direction with constant speed u m/s, (i) , show that the second car will just catch up on the first if u = 6k and that it will not catch up on it if u < 6k.

    using s=ut+.5at^2 i have formed the 2 equations s=.5kt^2 and s=6kt-18k for the 1st and 2nd car respectively. im not sure what to do next .
    Well, another of your algebra drills disguised as a kinematics problem ...


    displacement of the accelerating car is $d = \dfrac{k}{2}t^2$ ... in three seconds, it has traveled $d = \dfrac{9k}{2}$ and reaches a speed of $v = 3k$

    let $t = 0$ be the time the second car passes point P ...

    distance traveled by constant velocity car = head start distance + distance traveled by accelerating car

    $ut = \dfrac{9k}{2} + 3kt + \dfrac{k}{2}t^2$

    equation is quadratic in $t$, so a re-arrangement ...

    $0 = \dfrac{k}{2}t^2 +(3k-u)t + \dfrac{9k}{2}$

    multiply every term by 2 to clear the fractions ...

    $0 = kt^2 + 2(3k-u)t + 9k$

    solve for $t$ using the quadratic formula ...

    $t = \dfrac{2(u-3k) \pm \sqrt{4(3k-u)^2 -4k(9k)}}{2k} = \dfrac{2(u-3k) \pm \sqrt{u^2-6ku}}{2k}$

    for the time calculation to make sense, the discriminant, $(u^2-6ku)$, has to be $\ge 0$

    $u^2 - 6ku \ge 0$

    $u(u-6k) \ge 0$

    already know that $u > 0$, that means $u - 6k \ge 0 \implies u \ge 6k$
    Thanks from markosheehan
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  4. #4
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    Re: journey

    i am trying to use romseks way. once i get my value for t which is 6 what i do next
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  5. #5
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    Re: journey

    You found a solution for time when u = 6k

    You have two more cases to deal with ... u > 6k and u < 6k

    Buena suerte.
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    Re: journey

    So for example will I use 5k and then show that there is no value for t which this works
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  7. #7
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    Re: journey

    You can try that, but how do you know that 5.5k won't work? Are you going to test every single value less than 6k?
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  8. #8
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    Re: journey

    you are right but i cant think of any other way to show it that i understand
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