Another math problem.

Post by **Deji** » Thu Dec 01, 2005 7:47 pm

From destination A a bicycle left to point B. At the same time a motorcycle left from point B to point A. When the motorcycle had traversed 1/3 of the way, the bicycle still had 26km to go. When the bicycle had traversed 1/3 of the way, the motorcycle had 5km to go.Ffind the distance between the 2 points.

I've been at it from all angles, mainly using speed and distance and then trying to cancel out the speeds, but it all leads to nonsense answers, like 12.095. Help?

Post by **Survivor** » Thu Dec 01, 2005 8:00 pm

It's always bigger than 26 km but something seems missing

Post by **phantasmagoria** » Thu Dec 01, 2005 8:12 pm

Don't you need the speed of at least one of the objects?

Post by **Deji** » Thu Dec 01, 2005 8:17 pm

phantasmagoria wrote:Don't you need the speed of at least one of the objects?

That's what I thought, but apparently not. It's one of the harder questions for a 12th grade math exam.

Post by **mad** » Thu Dec 01, 2005 8:37 pm

this looks like a pretty simple question but i have forgotten all my mechanic maths

Post by **rgoer** » Thu Dec 01, 2005 9:16 pm

Hellchick is pretty smart, and she says it's 30km.

Post by **MKJ** » Thu Dec 01, 2005 10:19 pm

i think that answer wont cut it though, you usually have to tell them why its x

"the answer is 30km. i asked on the internet and hellchick (who's pretty smart cause she used to run PQ) says so"

Post by **hax103** » Thu Dec 01, 2005 10:56 pm

MKJ wrote:i think that answer wont cut it though, you usually have to tell them why its x

"the answer is 30km. i asked on the internet and hellchick (who's pretty smart cause she used to run PQ) says so"

Ahh, good ole algebra...

its prolly something like

d = distance between A and B

v = velocity of motorcycle
t = time for motorcycle to traverse 1/3 distance

u = velocity of bicycle
s = time for bicycle to traverse 1/3 distance

(d/3) = v * t
d-26 = u * t

(d/3) = u * s
d-5 = v * s

I'd assume after solving for s and t, iit works out to a quadratic equation for d

Post by **Sanction** » Fri Dec 02, 2005 12:06 am

(D-26)/3 = (D-5)/3
D=21

I think.

Post by **Zyte** » Fri Dec 02, 2005 12:12 am

the bicycle still had 26km to go.

Post by **menkent** » Fri Dec 02, 2005 12:42 am

assuming constant velocities, the ratio of distance traveled will be constant regardless of time. x(t1)/y(t1) = x(t2)/y(t2)

Post by **hax103** » Fri Dec 02, 2005 1:07 am

cool. it was a quadratic for d and hellchicky was right!

now lets suppose that the bicyclist is accelerating uniformly from 0.95c to 0.99c

menkent wrote:

assuming constant velocities, the ratio of distance traveled will be constant regardless of time. x(t1)/y(t1) = x(t2)/y(t2)

Guest · Post by **Guest** » Fri Dec 02, 2005 1:23 am

I think I found 33KM

Post by **andyman** » Fri Dec 02, 2005 1:43 am

It's like 20 minutes away man

Post by **Dark Metal** » Fri Dec 02, 2005 1:44 am

pete wrote:I think I found 33KM

Har har.

Post by **menkent** » Fri Dec 02, 2005 4:42 am

that some sort of french joke?

Post by **[xeno]Julios** » Fri Dec 02, 2005 5:49 am

hax103 wrote:cool. it was a quadratic for d and hellchicky was right!

now lets suppose that the bicyclist is accelerating uniformly from 0.95c to 0.99c

must add that to my creative suicide methods. (attempting to solve the problem, not accelerating to 0.99 c

)

Post by **mrd** » Fri Dec 02, 2005 5:58 am

Fuckin homework3world :\

Post by **MKJ** » Fri Dec 02, 2005 10:05 am

mrd wrote:Fuckin homework3world :\

at least he tried it before asking us
unlike toxigfag who's just like "yea i gotta write a thesis for tomorrow about [subject]. tell me all you know"

Post by **zewulf** » Fri Dec 02, 2005 11:11 am

Another vote for 30km :icon30: