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Thread: I'm not sure how I can solve these two functions

  1. #1
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    I'm not sure how I can solve these two functions

    G(t)=2t-0,25 0<=t<=t_1
    G(t)=1,6t+2,15 6<=t<=t_2

    So I just wanna know how I could calculate what t would be when y axe = 25

    Typing this into a tool works perfectly, and the result would be: t = 14,29 when y = 25, but I just don't know how I could calculate this on paper.
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  2. #2
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    Re: I'm not sure how I can solve these two functions

    Quote Originally Posted by emilo0212 View Post
    G(t)=2t-0,25 0<=t<=t_1
    G(t)=1,6t+2,15 6<=t<=t_2

    So I just wanna know how I could calculate what t would be when y axe = 25

    Typing this into a tool works perfectly, and the result would be: t = 14,29 when y = 25, but I just don't know how I could calculate this on paper.
    you give two function statements for what I assume are different intervals of $t$, making $G(t)$ a piece-wise defined function ...

    $G(t) = 2t-0.25$ for $0 \le t \le t_1$ and

    $G(t) = 1.6t+2.15$ for $6 \le t \le t_2$

    I'm sorry, but something seems to be missing in the translation. I really don't understand what you mean by ...

    how I could calculate what t would be when y axe = 25
    Can you clarify?

    Also, what do $t_1$ and $t_2$ represent?
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    Re: I'm not sure how I can solve these two functions

    So let me try and explain it like this:

    So there is this biker, who's trying to achieve 25 meters as fast as possible.
    He starts off by biking this fast: G(t)=2t−0.25 for 0≤t≤t1 and afterwards he slows down: G(t)=1.6t+2.15 for 6≤t≤t2

    I've accidentally forgotten to add one thing: There is a secondary biker, trying to achieve the same goal, and his timer is represented with : t1. t2 is this bikers time.

    and the secondary bikers function, looks like this, but it's pretty irrelevant, since it has nothing to do with this function: f(t)=1,75t-0,21

    I've set up these functions in an application called: "Geogebra" and this is what I get as a result:
    (Keep in mind, Y axe = Meters while X axe = seconds)
    The first function goes up to this point: 6, 11,75
    The secondary function goes from 6, 11,75 to 14,29, 25

    That means, that the biker reaches the 25 meters in 14,29 seconds, and I simply can't achieve the same result through calculation.
    Last edited by emilo0212; Oct 15th 2017 at 09:58 AM.
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  4. #4
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    Re: I'm not sure how I can solve these two functions

    He starts off by biking this fast: G(t)=2t−0.25 for 0≤t≤t1 and afterwards he slows down: G(t)=1.6t+2.15 for 6≤t≤t2 ...
    (Keep in mind, Y axe = Meters while X axe = seconds)
    Is $G(t)$ a position function or something else?
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  5. #5
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    Re: I'm not sure how I can solve these two functions

    I understand that the biker's speed is given by
    g(t)=\begin{cases}2t-0.25 & (0 \le t \le 6) \\ 1.6 t + 2.15 &(6 < t) \end{cases}

    It's odd, because he's travelling backwards at t=0.

    Since this is not in a calculus forum, I guess that you are supposed to calculate the area under the graph and that the aim is determine a value for t at which the distance travelled is 25 meters.
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  6. #6
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    Re: I'm not sure how I can solve these two functions

    There is no way to solve this without knowing how t1 and t2 are related to 6.

    Trying to find t such that the first formula gives 2t- 1/4= 25, 2t= 25+ 1/4= 101/4 so t= 101/8= 12.625. But that first formula only applies for t between 0 and t1. Whether we can use that or not depends on t1. If t1 is larger than 12.625, that is a correct solution to this problem.

    Trying to find t such that the second formula gives 1.6t+ 2.15= 25. 1.6t= 25- 2.15= 22.85 so t= 22.85/1.6= 14.28125. That is larger than 6 so whether or not that is the correct solution depends upon what t2 is. If t2 is larger than 14.2815, that is a correct solution to this problem.

    There are four possibilities:
    1) t1 is less than 12.625 and t2 is less than 14.2815. There are no solutions.

    2) t1 is larger than or equal to 12.625 but t2 is less than 14.2815. t= 12.625 is the only solution.

    3) t1 is less than 12.625 but t2 is larger than or equal to 14.2815. t= 14.2815 is the only solution.

    4) t1 is larger than or equal to 12.625 and t2 is larger than or equal to 14.2815. There are two solutions: t= 12.625 and t= 14.2815 are both solutions.
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    Re: I'm not sure how I can solve these two functions

    14,2815 is the correct answer, so I suppose: 1.6t+ 2.15= 25. 1.6t= 25- 2.15= 22.85 so t= 22.85/1.6= 14.28125, is the correct formula, meaning that t2 must be larger than 14,2815, cause the correct answer is 14,2815
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