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HL maths question help

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  • 03-01-2014 12:23pm
    #1
    Registered Users Posts: 306 ✭✭


    Hey guys, trying to do maths papers, stuck on a particular q I'm wondering can anyone help me. Its Sec set A 2013 pilot paper 1 , question 8 (d) , about a raindrop falling and its speed , question is The raindrops falls vertically from a height of 620 m , how long does it take to reach the ground, I'll attach the formulas given as I'm too lazy to type them :pac:
    Please if anyone knows how to do it I'd really appreciate help :)


Comments

  • Banned (with Prison Access) Posts: 209 ✭✭yoho139


    That image is more than slightly illegible. Should really just be subbing into the formulas though, by the looks of it.


  • Registered Users Posts: 896 ✭✭✭Ompala


    s(t) = 6t + 0.3t^2 – 0.01t^3 0 ≤ t ≤ 10
    and s(t) = k(t-10) t>10

    First thing to do is find k. Find this through continuity of first derivatives at t=10.
    Take the first derivatives to get
    s'(t) = 6 + 0.6t -0.03t^2 0 ≤ t ≤ 10
    s'(t) = k t>10

    At t=10 the derivatives will have the same value, since the derivative here will represent the terminal velocity of the raindrop and it reaches terminal velocity at t =10.
    So, 6 + 0.6(10) - 0.03(10)^2 = k which gives k = 9 (or k = 9m/s if you want to be really accurate)

    Now we can write the equations of motion as
    s(t) = 6t + 0.3t^2 – 0.01t^3 0 ≤ t ≤ 10
    and s(t) = 9(t-10) t>10

    Asked for time taken to fall 620m.
    Find the distance it falls in the first 10s from 1st equation.
    s(10) = 6(10) + 0.3(10)^2 - 0.01(10)^3 = 80m

    It falls 80m in first 10s so that means it fall 620-80 = 540m in the time after 10s, so we use the 2nd equation to solve for the time
    9(t-10) = 540
    t-10=60
    t=70s
    Answer is 70s


  • Banned (with Prison Access) Posts: 209 ✭✭yoho139


    Good eyes, and the reasoning is sound so far as I can tell.


  • Registered Users Posts: 306 ✭✭chatterboxxx95


    Sorry for the illegible photo, the original was okay but I clicked into it and see what you mean! :pac: Thanks a mill, I just couldn't get my head around how to find k but I understand it now :D


  • Registered Users Posts: 306 ✭✭chatterboxxx95


    Okay, bumping my own thread but how would you find the integral value for x if volume eq = 4x^3 -130x^2 +1000x amd you're told volume is 1512cm3 ? Its the educate ie papers, sample 2 paper 1question 7 part c


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  • Registered Users Posts: 896 ✭✭✭Ompala


    Okay, bumping my own thread but how would you find the integral value for x if volume eq = 4x^3 -130x^2 +1000x amd you're told volume is 1512cm3 ? Its the educate ie papers, sample 2 paper 1question 7 part c

    Eh from how you describe it would you not just solve

    4x^3 -130x^2 +1000x = 1512?

    I don't see how integrals come into this, did you mean integer?


  • Registered Users Posts: 306 ✭✭chatterboxxx95


    Ompala wrote: »
    Eh from how you describe it would you not just solve

    4x^3 -130x^2 +1000x = 1512?

    I don't see how integrals come into this, did you mean integer?

    Nope, papers definately say integral value! That's why I'm so confused by it! I tried just solving but not getting answer printed in the back :?


  • Registered Users Posts: 896 ✭✭✭Ompala


    Nope, papers definately say integral value! That's why I'm so confused by it! I tried just solving but not getting answer printed in the back :?

    Sounds like a mistake to me, cos there is an integer solution to

    4x^3 -130x^2 +1000x = 1512 @ x = 2

    The other solutions are x = 21.85 and x = 8.649

    I don't have the exam papers myself, what answer they giving you?


  • Registered Users Posts: 306 ✭✭chatterboxxx95


    Ompala wrote: »
    Sounds like a mistake to me, cos there is an integer solution to

    4x^3 -130x^2 +1000x = 1512 @ x = 2

    The other solutions are x = 21.85 and x = 8.649

    I don't have the exam papers myself, what answer they giving you?

    I've put it away now but I think it was some dogey fraction, so possibly one of the non-integer roots, I think you're right though that it must be a must be a mistake! Im just gonna leave it solved as x=2 thanks for your help :D


  • Registered Users Posts: 4,080 ✭✭✭EoghanIRL


    Differentiate s of t to get speed.
    After 10 seconds it reaches terminal velocity . Find speed when t is 10 .
    Time is distance divided by speed.
    Distance is 620 divided by s. Speed I got as 9ms ^-1
    620/9 = answer


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  • Registered Users Posts: 896 ✭✭✭Ompala


    EoghanIRL wrote: »
    Differentiate s of t to get speed.
    After 10 seconds it reaches terminal velocity . Find speed when t is 10 .
    Time is distance divided by speed.
    Distance is 620 divided by s. Speed I got as 9ms ^-1
    620/9 = answer

    No, that wouldn't work, because you are assuming the object has constant velocity for the entire motion.

    It doesn't, this is piece wise motion.

    The answers are very similar though.


  • Registered Users Posts: 4,080 ✭✭✭EoghanIRL


    Ompala wrote: »
    No, that wouldn't work, because you are assuming the object has constant velocity for the entire motion.

    It doesn't, this is piece wise motion.

    The answers are very similar though.

    Yes youare right . I didn't have exam paper in front of me when answering.


  • Registered Users Posts: 896 ✭✭✭Ompala


    EoghanIRL wrote: »
    Yes youare right . I didn't have exam paper in front of me when answering.

    Jaysus I sounded awful cynical earlier, sorry lad! At least you understand the logic behind the stuff anyway.


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