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Anyone doing LC Applied maths? Please help stuck on a question? Uniform Acceleration.

  • 22-07-2013 4:39pm
    #1
    allyb17
    Registered Users Posts: 59 ✭✭


    P and Q are points 162m apart.
    A body leaves P with initial speed 5m/s and travels towards Q with uniform acceleration 3m/s^2.
    At the same instant another body leaves Q and travels towards P with initial speed 7m/s and uniform acceleration 2m/s^2.
    After how many seconds do they meet and what then is the speed of each body?
    (p 22 Fundamental Applied Maths by Oliver Murphy, new edition)
    Please show your workings as I would like to know how to do this, I have only commenced applied maths this summer so I am eager to catch up as soon as possible, thus this may seem simple to those who do it!?
    Kind regards, and thanks. :):confused:
    Tagged:


Comments

  • #2
    Jamie Dean
    Registered Users Posts: 2 ✭Jamie Dean


    Let t be the time of the meeting.

    S1 = 5t + 1/2 [3] t squared = 5t + 3/2 t squared

    S1 + S2 = 162 so 12t + 5/2 t squared = 162. t=6.

    At t=6, V1 = u + at = 5 + [3] [6] = 23 m/s.

    V2 = 7 + [2] [6] = 19 m/s.

    I hope this helps you.


  • #3
    Peregrine
    Moderators, Education Moderators Posts: 26,403 Mod ✭✭✭✭Peregrine


    ^^ Yup, basically that

    I have most of that chapter done out and I have access to all the answers on folensonline so if you need help with any other questions just post in on this thread.


  • #4
    allyb17
    Registered Users Posts: 59 ✭✭allyb17


    Thanks a million that really helped! A lot of it just seems to be manipulation of formulae. This is another question which I got out, but I don't think I used the right method?

    r is a vector of magnitude 10 units in a direction 70 degrees North of West.
    s is a vector in a direction 10 degrees North of East.
    Find the magnitude of s, correct to one decimal place, if (r+s) must be in a NE direction.
    The answer is 15.8 but i'm a bit confused as to the method and why I got this answer.


  • #5
    yoho139
    Banned (with Prison Access) Posts: 209 ✭✭yoho139


    r = -10Sin(70)i + 10Cos(70)j (resolve to i and j)
    s = xSin(10)i + xCos(10)j (likewise)
    R+S = yi + yj (must be equal as the direction is 45 degrees from the axis, both sides are therefore equal)

    Therefore -10Sin(70) + xSin(10) = 10Cos(70) + xCos(10)
    -10Sin(70) - 10Cos(70) = xCos(10) - xSin(10)
    -12.8171 = 0.8112x
    -15.8 = x
    x = 15.8

    The answer being negative at first has something to do with the fact that they're going opposite directions. However, it's half past 2 in the morning and I haven't looked at my Applied Maths stuff since the end of May, so I'm gonna leave it at that. Tell me if you want an explanation of any individual steps.


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