velo90 0 #26 January 16, 2004 Doh! I wish I never posted this now It doesn't matter how long the track is. Quote Share this post Link to post Share on other sites
panzwami 0 #27 January 16, 2004 look at it like this.... Same race, same 90-mile-long track. He does the first lap at 90 mph, meaning that *he traveled 90 miles in one hour (miles per HOUR)*. If he then does the second lap at 270 mph, he will complete the 90-mile lap in 20 minutes (270mph * 1/3 hour = 90 miles). So, the total time that it took him to complete two laps, for a total of 180 miles driven, is one hour (first lap) + 20 minutes (second lap), for a total of 80 minutes driving time. 180 miles / 80 minutes = 135 mph average speed. The thing that you are not considering here is that the track itself is of fixed length. If the track was of variable length, then you would be correct. Consider: First lap, 90-mile-long track, 90 mph = 90 miles in one hour If the track were then tripled in length, to 270-miles-long, for the second lap, and he traveled at 270 mph, then = 270 miles in one hour. Thus, he would have driven 360 miles total (90+270) in two hours total (one hour for each lap), giving an overall average speed for the trip of 360 miles / 2 hours = 180mph. BUT, because the track is static at 90-miles-long, he would have to complete two laps (180 total miles) in one hour to average 180mph. But since the original question stated that he did the first lap at 90mph, we therefore know that he took one hour to do his first lap, and thus cannot make up the difference. Make sense? Matt ----- Quote Share this post Link to post Share on other sites
kallend 2,114 #28 January 16, 2004 It doesn't matter how long the lap is, it takes the same time to do 8 at 90mph as it does to do 16 at 180mph. Therefore he has already used up ALL his time in the first 8 laps. So there is no time left for the final 8 laps. Unless you're an economist working for the government, when you borrow it from your children.... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
kallend 2,114 #29 January 16, 2004 Quote Clever! Now solve this one. G W Bush has started the ball rolling so I am going to open up the first Martian skydiving school. The question is for student gear will I need larger parachutes than on earth due to the thinner atmosphere? Or will I need smaller parachutes due to weaker gravity? Quote Bigger - way, way bigger. By the way, an AFF course at the Martian school of skydiving cost $400 billion. Any takers?... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
panzwami 0 #30 January 16, 2004 Quote The answer will have to do with the proportion between the change of the atmospheric density, and that of the planetary gravity. If the gravitational decrease is only very slight, but the air is only 1% as dense as it is on Earth, much larger parachutes will be needed to react with the fewer air molecules. Conversely, if the atmosphere is only slightly thinner than on Earth, but the gravity is substantially weaker, smaller parachutes will be needed. I don't think there's a definitive answer to the question as given.... Matt ----- Quote Share this post Link to post Share on other sites
mr2mk1g 10 #31 January 16, 2004 180 miles per lap X 16 laps = 2880 miles in the race. first 8 laps = 1440 miles @ 90mph = 16 hours. Next 8 laps = 1440 miles @ 270mph = 5 hours. So it takes him 21 hours to complete the whole race track. 2880 mile of race track / 21 hours = 137... hmmm must go faster Now I getcha. If you want to do 2880 miles at 180mph average it takes you 16 hours but you've already spent 16 hours fannying round with a busted car. Quote Share this post Link to post Share on other sites
kallend 2,114 #32 January 16, 2004 Quote The answer will have to do with the proportion between the change of the atmospheric density, and that of the planetary gravity. If the gravitational decrease is only very slight, but the air is only 1% as dense as it is on Earth, much larger parachutes will be needed to react with the fewer air molecules. Conversely, if the atmosphere is only slightly thinner than on Earth, but the gravity is substantially weaker, smaller parachutes will be needed. I don't think there's a definitive answer to the question as given.... Matt ----- Of course there is - you just have to do the research.... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
kelel01 1 #33 January 16, 2004 Ok, 1st, it never said that the track was 90 miles long, but let's say that it is. 90 miles per hour, for 8 laps= 8 hours 270 miles per hour, for 8 laps= 2.666667 hours average speed = ((90*8) + (270*2.67))/10.67= 135-ish Ok, so while trying to prove you wrong, I proved you right. (DOH!) But at least now I'm satisfied with the answer. But there has to be a speed (albeit, not even remotely realistic) that would produce an average of 180. Kelly Quote Share this post Link to post Share on other sites
velo90 0 #34 January 16, 2004 Quote Of course there is - you just have to do the research. Sod the research, I'll go with 300 sq ft and use trail and error. If something goes wrong I'll post it in the incidents forum and get all the replys I'll ever need. Anyway, I have not got any takers yet. Dave Quote Share this post Link to post Share on other sites
panzwami 0 #35 January 16, 2004 Hence the "as given." It is, of course, possible to research Martian atmospheric density and gravitational strength, and then use those values to calculate the actual answer, but that information is not included in the original question. Thus, *as given*, I don't think there's an answer to the question. Matt ----- Quote Share this post Link to post Share on other sites
kallend 2,114 #36 January 16, 2004 Quote I get the feeling its one of those maths questions which means something in pure maths but doesnt have much application to the real world again. Still interesting though - love to broaden my horizons. Absolutely it means something. The same principle applies if it Shumacker on a circuit or Little Red Riding Hood driving to see her granny or you going to the DZ. You have 1 hour to get to the DZ with is 60 miles away. Due to construction, you only average 30mph for the first 30 miles. How fast do you have to drive the rest of the way in order to get there on time?... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
velo90 0 #37 January 16, 2004 DSo you think the $400 billion price tag for an AFF course is acceptable then? Dave Quote Share this post Link to post Share on other sites
kallend 2,114 #38 January 16, 2004 Quote Hence the "as given." It is, of course, possible to research Martian atmospheric density and gravitational strength, and then use those values to calculate the actual answer, but that information is not included in the original question. Thus, *as given*, I don't think there's an answer to the question. Matt Rubbish - the data have been all over the news for the last 2 weeks. -----... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
mr2mk1g 10 #39 January 16, 2004 its alright - I got it now. You can't get there on time as its already taken you the hour you alotted. Quote Share this post Link to post Share on other sites
skydivexxl 0 #40 January 16, 2004 Quote Can't be done. The answer isn't 270 Mph? Damn! Blog Clicky Quote Share this post Link to post Share on other sites
mr2mk1g 10 #41 January 16, 2004 Quote DSo you think the $400 billion price tag for an AFF course is acceptable then? plus beer - you'll have to buy lots of beer for the FIRST AFF course on Mars Quote Share this post Link to post Share on other sites
Darius11 12 #42 January 16, 2004 WOW should i have chinese food for lunch. I'd rather be hated for who I am, than loved for who I am not." - Kurt Cobain Quote Share this post Link to post Share on other sites
skydivexxl 0 #43 January 16, 2004 Quote Now assume he has spent the first 8 laps travelling at 90 mph. How long has he been driving? Simple subtraction will now give you the answer as to how long he has to complete the next 8 laps. The Track isn't 90 miles long... it's a track, more like 1 1/2 miles around. AND... it would be more like two minutes per lap @ 90 Mph, not hours... I still say it's 270 Mph... Blog Clicky Quote Share this post Link to post Share on other sites
kallend 2,114 #44 January 16, 2004 Quote its alright - I got it now. You can't get there on time as its already taken you the hour you alotted. The general rule is that your average speed on any journey is much more influenced by the slowest parts than by the fastest parts. Driving really fast (and risking a ticket) makes less difference than avoiding congested areas if you can. Traffic lights (which are an invention of Satan) have the worst possible effect on your trip time, because you are completely stationary while waiting, and the increased speed when you encounter a green one can never compensate for the time you waste at red ones.... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
kallend 2,114 #45 January 16, 2004 Quote Quote Now assume he has spent the first 8 laps travelling at 90 mph. How long has he been driving? Simple subtraction will now give you the answer as to how long he has to complete the next 8 laps. The Track isn't 90 miles long... it's a track, more like 1 1/2 miles around. AND... it would be more like two minutes per lap @ 90 Mph, not hours... I still say it's 270 Mph... Do you work for the Bush White House?... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
skydivexxl 0 #46 January 16, 2004 Quote Do you work for the Bush White House? Need a job? Hopefully that was not a joke about intelligence.... Blog Clicky Quote Share this post Link to post Share on other sites
Taylor610 0 #47 January 16, 2004 It doesn't matter how fast he drives, Jr. is gonna win anyway. Quote Share this post Link to post Share on other sites
panzwami 0 #48 January 16, 2004 nope, there's not. You're right, he never said the track was 90 miles long, but no matter what the length, it has to be *something.* It could be 1 mile or 1000 miles, but the math is the same. The key point here is the fact that he was going below his average speed for all of his time. He needed to complete the race in 8 hours, but drove below the average for all 8 hours. If at any point before 8 hours were up he were to fix his car, it would still technically be possible for him to achieve his desired average. Consider: Same race, 16 laps, 90-mile-long track. He wants to average 180mph, meaning he will have to complete the race (16 laps * 90 miles per lap = 1440 total miles) in eight hours (1440 miles / 180 mph = 8 hours). Say he drives 90mph for the first *seven* laps. He will then have completed 630 total miles (90-mile-track * 7 laps = 630 miles) in 7 hours. He wants to go 1440 miles in 8 hours, and he has already done 630 miles in 7 hours. SO.....he has one hour remaining, and 810 miles left to drive. He would have to drive 810mph for the last hour to make his goal (forget that a car can't do 810mph, this is just the math part of it). So, if he were to fix his car at any time before he used up his alloted amount of time, he would still technically be able to speed up and make his goal. The longer he drives below the average desired speed, the faster he will have to drive to make up the difference. As time T approaches 8 hours (meaning the amount of time he has left gets smaller and smaller), his required speed will get higher and higher, continuing to infinity. Matt ----- Quote Share this post Link to post Share on other sites
kelel01 1 #49 January 16, 2004 So, 810 miles an hour it is? Good. Done. Kelly Quote Share this post Link to post Share on other sites
skydivexxl 0 #50 January 16, 2004 Where is the 8 hours coming from? Blog Clicky Quote Share this post Link to post Share on other sites
