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Abstract applied maths

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  • 05-10-2015 6:14pm
    #1
    UhmMaths?!
    Registered Users Posts: 12 ✭


    We have this question to do and I don't have a clue how to do it....

    Q5 A car has to travel a distance s on a straight road. The car has maximum acceleration of a and maximum deceleration of b. It starts and ends at rest.

    I) Show that, if there is a speed limit of v m/s, the time taken to complete the journey is given by v/2a+v/2b+s/v (x/y this is a fraction format where x is above y)

    ii) Show that if there is no speed limit, the time is given by
    (this is very difficult to write but it's a square root and inside it is:

    2s (a+b)/ab

    (so in English it would be the square root of 2s and (a+b) over ab)

    If that makes sense to anybody I would need the solution and your workings please

    Thanks


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    #2
    murphm45
    Registered Users Posts: 505 ✭✭✭murphm45


    This is actually a really awkward one to try and answer using a standard keyboard. I'll do what I can to terry to highlight what to do but you'll need to work through the specifics.

    For the first part the first thing you should do is replace the deceleration rate 'b' with a negative acceleration, call it 'B' so that b = -B.

    To solve it you need to split out into three parts, the time while you are accelerating the time you're at the speed limit and the time you're decelerating for.

    To get the time accelerating/decelerating you need to remember that acceleration is the derivative of velocity (the taste of change of velocity over time). So velocity is the integral of the acceleration rate so you need to find a time so that the integral of the acceleration rate over a specific time equals the velocity (you have the velocity and acceleration what you need to find is time)

    Last thing you need to do is find the time you're travelling at v. To so this you need to find the distance covered in acceleration/deceleration and subtract this from s (is the formula below) and then divide by v.

    Distance = u*t + 0.5*a*t^2
    Where
    u is the starting velocity
    t is the time accelerating
    a is the constant rate of acceleration

    Does this help?

    The second bit is harder to explain but I want to make sure you follow the first half before I try!


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    #3
    MathsManiac
    Registered Users Posts: 1,595 ✭✭✭MathsManiac


    UhmMaths?! wrote: »
    We have this question to do and I don't have a clue how to do it....

    ...

    If that makes sense to anybody I would need the solution and your workings please

    Thanks

    OP: please read the forum charter. While it's ok to seek some help and guidance with homework questions or assignments, it's not ok to look for full solutions. We expect you to say what you've tried already and where you got stuck. If you have no clue at all as to where to start, then it would be more appropriate to ask for some help to get you going. If you then make an effort and come back when you get stuck, you'll find people more than willing to continue to help you along.

    Murphm45 has given you some good guidance on getting started, so I hope that gets you on the road to a solution.


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    #4
    Yakuza
    Registered Users Posts: 5,141 ✭✭✭Yakuza


    It's also a generalisation of the problem I outlined the solution to in this thread

    Edit : Latex is great for displaying mathematical answers like [Latex]\sqrt{\frac{2ab}{(a+b)}}[/Latex]

    The above was obtained by embedding 
    [Lat3x]\sqrt{\frac{2ab}{(a+b)}}[/Latex]

    in the post (change Lat3x to Latex above to get it to render, I had to deliberately change the e to a 3 or it would have just drawn the expression with the square root again!


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