oh I see what your saying.. yeah that's true, the horseman does 225m total he's traveled 150m forward with the brigade when he hit the front.. but on the way back.. he's running the 100m and the brigade has moved 25m in thew opposite direction.. so he runs 75m on the way back
That would be true if the question said he was travelling at double the speed. But it says he's going 3x the speed! Triforce!!!111 edit: you ninja'ed me. The officer only covers 200m total, due to the cancelling out of extra distances, so 225m is incorrect.
well easier if make a diagram.. I dunno why but it bugs me you don't see it I put the speedometer on the brigade.. think of it in 2d might be easier- the horse will be running 30 mps in this instance but against the brigade, 20 (30-10)
Yes if the troupe travels 75m, then the officer travels 225m... my point was that 225m is proof that 75m is the incorrect answer to the question! I'm on the phone atm, so it would be a pita to draw and post a diagram.... I wonder if any conclusions from the rest of bit tech's mathematicians have now changed?
yeah sorry drew that up quick.. was watching survivor xD now look at it, it's not so clear in that diagram.. the speedometer and the horse are connected.. it's the speed of the horse against the brigade.. maybe that might help you see it better the 225 for the horse travel is correct.. 150 forward, 75 back
you can plug in any numbers.. let's say the speed of the brigade is 22mps and the horse runs 4x the speed of the brigade so the horse now runs 88mps now against the brigade he's running 66mps (because the brigade is moving forward at 22mps.. 88-22) now let's get the percentage 22/66=33.3333333% and that works with any speed if you do the math above.. on the way back the horse is running 4 times 22mps or any speed really, but the brigade is still moving forward at 22 in this example.. so 22 multiplied by 4 we get 88 again.. now add 22mps to 88 because of the brigades forward speed.. and we get 110mps for the horse so against the brigade, the horse is now running 110mps towards the back.. now let's get the percentage to figure out the back run.. 22/110=20% so now let's add up both percentages 33.3333% and 20% = 53.3333% multiply that by the length of the brigade for the answer on how far the brigade has traveled
So instead of a "suggest some physics problems I can use in class" thread, this is now a "here's some physics problems for you to solve/debate" thread .
trying to help mv.. he was going to lose his 460 over some physics problem any elementary student in gate could solve I used to be stubborn like that too.. that's what cool about math- actually like journeys answer it's a lot more simple- but trying to explain it in a deeper way is always hard to do.. why I'm not a teacher (my girl is though)
Pff, I worked out the answer properly within about 3 mins of getting it. I however used maths, as opposed to heresay, conjecture and opinion The guy who got it in this thread did it a much simpler way than me too, which im a bit miffed about. The graph is a thing of beauty!
It is quite lovely. I just scribbled a bit of crap out on paper between jobs at work and got it epically wrong.
lmao that's funny Lets up the stakes, incorrect answer buys the other player a complete high end pc, including ssd's, big ips monitors, gtx 590, i7, whatever you like Wanna play?
aye I already explained it best I could.. make a bet with bungle- I'm sure he'd be happy to take a rig off your hands don't do it mv! =-] I could try to explain it another way but I'm busy tonight.. my moms coming over from alaska to visit maybe one of these other guys would explain it a bit better too
Too late now, I've invested more than a moment's thought to the problem, and made a quick scribble on a bit of paper, and I now have irrefutable proof that the troupe cover 75m by the time that fcuking annoying officer completes his Mel Gibson impersonation This could have been bit-tech's super month of give aways Pfft @ppl that can't give away crt's, I can't even give away new hardware!
I was hoping the thread starter had the correct answer in the first place My first approach was drawing it too (though I did in in excel) The second approach was using relativistic speed Both give out 75m of troup-travel independent of used speed. There's a third way of solving this, by calculating both crosspoints using derivative maths... This would be the sure way, but I can't be arsed to read up on that right now.
