Maths Help! Page 3

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  • White_Shadow 24 Oct 2012 22:51:49 2,534 posts
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    Chopsen, you are brilliant!
  • Dangerous_Dan 24 Oct 2012 23:13:40 2,175 posts
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    @White_Shadow With your provided numbers I come up with about 1323.13 people who visited both displays. This assumes that there's nobody who visited neither of the displays. If you assume that some people stayed away from both displays then there is not enough information provided to calculate the exact number but an interval can be calculated.

    Damn Sir Chopsen was faster!

    Edited by Dangerous_Dan at 23:14:45 24-10-2012
  • senso-ji 25 Oct 2012 09:37:55 5,303 posts
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    I have a mechanics maths problem if anyone cares to help:

    Assume a sprinter runs 100m in 9.83 seconds. He accelerates constantly for the first 60m and then runs at his maximum speed for the remaining 40m.

    Calculate:

    A - The maximum speed
    B - The acceleration

    This is an actual exam question I've seen and I'm flumoxed as to how to work it out. I know all the motion in a straight line formulas but can't seem to work out how to use them to get both those answers.
  • dominalien 25 Oct 2012 09:55:18 6,741 posts
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    No, not 40% of the time, just 40% of the distance.

    PSN: DonOsito

  • SaucyGeoff 25 Oct 2012 09:57:12 2,479 posts
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    If his maximum speed is A m/s, then he covers 40% of the course at speed A, and 60% and 1/2 A (constant acceleration). So I reckon,

    Average Speed = 0.4*A + 0.6*1/2 A
    100/9.83= 0.7 A
    A = 100/(9.83*0.7)

    Can't remember straight line formulae I'm afraid, but think acceleration should be fairly simple from here.
  • X201 25 Oct 2012 10:05:19 14,696 posts
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    Post deleted
  • warlockuk 25 Oct 2012 11:48:41 19,101 posts
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    senso-ji wrote:
    I have a mechanics maths problem if anyone cares to help:

    Assume a sprinter runs 100m in 9.83 seconds. He accelerates constantly for the first 60m and then runs at his maximum speed for the remaining 40m.

    Calculate:

    A - The maximum speed
    B - The acceleration

    This is an actual exam question I've seen and I'm flumoxed as to how to work it out. I know all the motion in a straight line formulas but can't seem to work out how to use them to get both those answers.
    Interesting.

    Spent ages trying to work this out, think I'm missing something :p

    I'm a grumpy bastard.

  • warlockuk 25 Oct 2012 11:58:29 19,101 posts
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    Oh wait, it's easy.

    His constant acceleration as someone said means he averages half the final speed in the first 60metres

    So 60 metres at v/2 + 40m at v

    or if it was equal, 120m at v and 40m at v would mean at the actual speed he would've covered 160, not 100 metres.

    So, 160/9.83 = 16.27670397 for his final speed.
    At that speed, the 40 metres end piece would take 2.4575 seconds leaving 7.3725 seconds to accellerate to 16.27670397 m/s

    Acceleration is finalvel - start vel / time
    16.27670397 / 7.3725 = 2.2077591 metres per second per second.

    So (Fixed as I put A+B in the wrong order originally)
    A: 16.27670397 m/s for the final top speed
    B: 2.2077591 m/s/s for the initial acceleration

    Edited by warlockuk at 12:06:12 25-10-2012

    I'm a grumpy bastard.

  • Alastair 25 Oct 2012 12:01:05 14,953 posts
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    What level of maths are we talking? GCSE or A-level?
    Somehow I wonder if this is related to integrating/differentiating wrt time or summat.

    Not as nice as I used to be

  • warlockuk 25 Oct 2012 12:02:06 19,101 posts
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    Checks out, too.
    Acceleration at (B) for 7.3725 seconds = (B*T^2) / 2 = 60 metres
    Velocity = (B) * (T) = A :) \o/ Qapla' !

    *edit fixed for answering A+B the wrong way round before

    Edited by warlockuk at 12:04:02 25-10-2012

    I'm a grumpy bastard.

  • warlockuk 25 Oct 2012 12:02:55 19,101 posts
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    Alastair wrote:
    What level of maths are we talking? GCSE or A-level?
    Somehow I wonder if this is related to integrating/differentiating wrt time or summat.
    I think it's technically A-level but once you realise it's a straight-line and geometry (mentally) fits in you can squeeze it into GCSE maths or Physics.

    I'm a grumpy bastard.

  • mal 25 Oct 2012 12:03:30 21,944 posts
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    SaucyGeoff wrote:
    If his maximum speed is A m/s, then he covers 40% of the course at speed A, and 60% and 1/2 A (constant acceleration). So I reckon,

    Average Speed = 0.4*A + 0.6*1/2 A
    100/9.83= 0.7 A
    A = 100/(9.83*0.7)

    Can't remember straight line formulae I'm afraid, but think acceleration should be fairly simple from here.
    I think you've made the mistake dominalien pointed out - that you've averaged the speed assuming he spent 60% of his time accelerating, rather than 60% of the distance (or 60m in this example).

    Using the formula:
    speed = distance/time then
    distance = time * speed
    Also, distance = time * average speed for varying speeds

    So 40 = b * n and 60 = a * n/2 where a and b are the times taken and n the the maximum speed
    Rearrange those expressions to express a and b in terms of n, then replace them in the time equation (a+b=9.87). Therefore you have an expression with just n as a variable.

    Converting that to acceleration means you need to know how long he spent accelerating. That's a, which now you know n you can work out. Working out the actual acceleration from there is left as an exercise to the reader (mainly because I've forgotten all my acceleration formulas).

    Edit: Yeah, warlock got it :)

    Edited by mal at 12:05:06 25-10-2012

    Cubby didn't know how to turn off sigs!

  • speedofthepuma 25 Oct 2012 12:15:58 13,246 posts
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    God damn we are a sexy bunch!

    I've turned off all the avatars and crap, so don't expect me to be impressed by yours.

  • warlockuk 25 Oct 2012 12:19:04 19,101 posts
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    Stop fapping over my profile picture.

    I'm a grumpy bastard.

  • SaucyGeoff 25 Oct 2012 12:28:17 2,479 posts
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    @mal Ah, I see. Rookie mistake. I remember doing this kind of question during A-Levels. Used to know all these formulae!
  • neilka 25 Oct 2012 12:34:09 14,942 posts
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    Is the sprinter on a treadmill wearing a spacesuit?
  • Alastair 25 Oct 2012 13:06:12 14,953 posts
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    neilka wrote:
    Is the sprinter on a treadmill wearing a spacesuit?
    Only when Lutz is answering...

    Not as nice as I used to be

  • senso-ji 25 Oct 2012 13:14:38 5,303 posts
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    Thanks guys - Warlockuk got the answer right :)

    The question is from A-level maths Mechanics 1 module.
  • boo 8 Jan 2013 11:25:00 11,604 posts
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    I'm ashamed at how rusty I've become regarding maths.

    This was a puzzle I saw the other day, and while I have the answer, I can't get my head around the 'x cancels out' bit. Can anyone clarify for me.

    Original puzzle is:

    Rich person decides to give 45 to each man and 60 to each woman.
    Only 1/9th of men claim their prize.
    Only 1/12th of women claim their prize.
    There are 3552 people who can claim.
    How much is paid out?


    Solution.

    If there are x men, then there must be (3552-x) women.
    (x/9) received 45
    ((3552-x) / 12) received 60

    Total paid = (45 * (x/9)) + (60 * (3552-x) / 12)

    x cancels out, leaving

    5 * 3552 = 17,760



    How does x cancel out? If you add x to the right hand side of the equation, that would leave

    (45 * (x/9)) + x

    /is baffled

    Just Another Lego Blog

  • Dangerous_Dan 8 Jan 2013 11:41:15 2,175 posts
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    @boo The result is 17.760 if you assume that one half of the people are men and the other half are women.

    So 1776 men and 1776 women.

    Number of men who get their 45 pounds: 1776/9
    Money they receive all men together: (1776/9)*45 = 8880 pounds

    Same goes for the women which ends up all in all in (1776/12)*60 = 8880 pounds

    Which added up results in 17760 pounds.

    After getting this out of the way, look at that divided by 9 and multiplied by 45 in case of men and divided by 12 and multiplied by 60 again. they are the same factor, namely 45/9 or 60/12 which is the factor 5 in both cases.

    So no matter how many women or men there are among the people, the average man receives the same amount of money as does the average women which is 5 pounds on average. 3552*5=17760

    That being said, is that excercise out of some men/women equlity social engineering book? ;)

    Edited by Dangerous_Dan at 11:43:45 08-01-2013
  • sport 8 Jan 2013 11:42:15 12,045 posts
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    @boo - if you do the division on each side of the plus sign you're left with:

    5x + 17760 - 5x,
    therefore
    5x cancels -5x
    and you're left with
    17760
  • boo 8 Jan 2013 11:46:46 11,604 posts
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    Thanks much!

    Just Another Lego Blog

  • LeoliansBro 8 Jan 2013 11:49:27 41,863 posts
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    So, yeah, I can do this.

    x=men
    y=money

    (45x/9) + (60(3552-x)/12) = y

    (45x/9) = -(60(3552-x)/12) + y

    5x = -((213,120 - 60x)/12) + y

    5x = -(17,760 - 5x) + y

    5x = -17,760 + 5x + y

    5x + 17,760 - 5x = y

    17,760 = y

    LB, you really are a massive geek.

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