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Applied Maths

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  • 20-06-2003 4:50pm
    #1
    Closed Accounts Posts: 17,163 ✭✭✭✭


    Not bad at all, the collisions question was a bitch so i skiped it.
    Alot of stuff that came up before was on it, question one, the how close would he have gotthen question kicking myself about it, because i remembered how to do it at the end. got the right time and the acceleration but forgot what to do with them, aw well.


Comments

  • Closed Accounts Posts: 14 dosser


    oKaY

    What was the answer to question one part (b)???????

    I know it must be so obvious and yet...... :confused:

    My way of looking at things will change forever when i see the answer for this, i almost had a seizure during the exam.

    I need to know now!!!!

    :cool:


  • Closed Accounts Posts: 14 dosser


    im reading it wrong


  • Closed Accounts Posts: 14 dosser


    okay i know what it is..... whatever, i did seven questions anyway


  • Registered Users Posts: 1,328 ✭✭✭Sev


    Thats another A1 in the bag. Collisions was same as always, standard stuff. Although I made a tiny slip in forgetting to eliminate one range of values for 'e'. Only mistake I figure they can dock a mark on tho, unless I made others I'm unaware of. I don't think so tho, was a pretty straightforward paper.

    I have to admit tho, I spent some time staring at Question 1 part (b) being totally sure for 2hrs that there was some information missing in the question. Anyhow, copped it with 15 min to go at the end that there was a minor subtlety in the question that said that the man just reached the bus, which made it all make sense.


  • Registered Users Posts: 1,328 ✭✭✭Sev


    lol, trust me to read the thread before posting on it.

    Anyway, part 1 b.

    Basically the key I found, was in getting the acceleration of the train.

    You get some kind of expression for the speed of the man being u = 2 + 10a. Where 'a' is the acceleration of the bus. Anyway, the fact that he 'just' catches the bus, means that when he gets to it, the velocity of the man is the same as the velocity of the bus. Which works out that the acceleration of the bus is (1/5)m/s^2. That allows you to find the constant velocity of the man which is 4m/s.

    Then, get an expression for distance the bus has gone, and the distance the man has gone. Subtract. Differentiate with respect to t. Find what t is for min distance, throw that back into your equation and you get 17.5.


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  • Closed Accounts Posts: 17,163 ✭✭✭✭Boston


    that was q1 a with the train, q1b was with the bus, it was simple yet i just fogot how to finish it off. Collisions was the standard stuff, but my head was up for messing around with e u and v


  • Registered Users Posts: 1,328 ✭✭✭Sev


    Yeah, I meant bus :/


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