Chopsen, you are brilliant! 
Maths Help! • Page 3

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White_Shadow 2,538 posts
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Dangerous_Dan 2,378 posts
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Registered 4 years ago@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 24102012 
sensoji 6,137 posts
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Registered 7 years agoI 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 7,155 posts
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Registered 9 years agoNo, not 40% of the time, just 40% of the distance. PSN: DonOsito

Saucy 2,639 posts
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Registered 5 years agoIf 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 15,955 posts
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warlockuk 19,260 posts
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Registered 11 years agosensoji wrote:
Interesting.
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.
Spent ages trying to work this out, think I'm missing somethingI'm a grumpy bastard.

warlockuk 19,260 posts
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Registered 11 years agoOh 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 25102012I'm a grumpy bastard.

Alastair 16,630 posts
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Registered 13 years agoWhat level of maths are we talking? GCSE or Alevel?
Somehow I wonder if this is related to integrating/differentiating wrt time or summat. 
warlockuk 19,260 posts
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Registered 11 years agoChecks 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 25102012I'm a grumpy bastard.

warlockuk 19,260 posts
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Registered 11 years agoAlastair wrote:
I think it's technically Alevel but once you realise it's a straightline and geometry (mentally) fits in you can squeeze it into GCSE maths or Physics.
What level of maths are we talking? GCSE or Alevel?
Somehow I wonder if this is related to integrating/differentiating wrt time or summat.I'm a grumpy bastard.

mal 23,248 posts
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Registered 13 years agoSaucyGeoff wrote:
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).
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.
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 25102012Cubby didn't know how to turn off sigs!

speedofthepuma 13,329 posts
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Registered 9 years agoGod damn we are a sexy bunch! I lurk. If I've spoken to you, I'm either impassioned, or drunk.

warlockuk 19,260 posts
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Registered 11 years agoStop fapping over my profile picture. I'm a grumpy bastard.

Saucy 2,639 posts
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Registered 5 years ago@mal Ah, I see. Rookie mistake. I remember doing this kind of question during ALevels. Used to know all these formulae! 
Is the sprinter on a treadmill wearing a spacesuit? AVOID

Alastair 16,630 posts
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Registered 13 years agoneilka wrote:
Only when Lutz is answering...
Is the sprinter on a treadmill wearing a spacesuit? 
sensoji 6,137 posts
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Registered 7 years agoThanks guys  Warlockuk got the answer right
The question is from Alevel maths Mechanics 1 module. 
boo 12,053 posts
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Registered 11 years agoI'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 (3552x) women.
(x/9) received £45
((3552x) / 12) received £60
Total paid = (45 * (x/9)) + (60 * (3552x) / 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 
Dangerous_Dan 2,378 posts
Seen 6 months ago
Registered 4 years ago@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 08012013 
sport 12,892 posts
Seen 8 hours ago
Registered 9 years ago@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 
Thanks much! 
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