tag:blogger.com,1999:blog-15628310.post358564363502394180..comments2021-04-22T08:14:35.383-04:00Comments on Question of the day: But How Fuel Efficient Are They?Anonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-15628310.post-37557605088008991272011-01-05T13:54:51.296-05:002011-01-05T13:54:51.296-05:00Can we visualise this problem as an analogue clock...Can we visualise this problem as an analogue clock? I.e. every time the minute hand laps the hour hand, the point of overlap increases by a constant unit? So, if the first overlap on the racetrack is N, the second overlap point will be 2N?Carl Cookhttps://www.blogger.com/profile/18413104231588611923noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-71484603886050990422009-11-11T08:59:18.304-05:002009-11-11T08:59:18.304-05:00Good points all. 300 feet from point A is the ans...Good points all. 300 feet from point A is the answer. <br /><br />USAF Pilot, that was impressive work!Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-74786763047162415212009-11-10T20:28:42.768-05:002009-11-10T20:28:42.768-05:00Yeah, whether it passes 150 feet in front of point...Yeah, whether it passes 150 feet <i>in front of</i> point A or 150 feet <i>behind</i> point A, the 2nd pass will occur 300 feet from point A.jeredhttps://www.blogger.com/profile/12528386127450345049noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-42649820844239253352009-11-10T13:46:37.345-05:002009-11-10T13:46:37.345-05:00I just had to look at the math. So I got a little...I just had to look at the math. So I got a little creative. <br /><br />The track is over a mile long, so lets call it exactly 1.5 miles. Hence, when car 1 passed car 2, car 1 had traveled 1.5 miles + 1.5 miles - 150 feet (2.97 miles). (But when you subtract the 1.5 miles car 1 did on his first lap, they would be at the same place) Car 2 had traveled 1.5 miles - 150 (1.47 miles). Since the time for both was the same, let's call it 3 min after the start. So we get a speed for car 1 of 59.43182 mph and a speed of car 2 of 29.43182 mph.<br /><br />Now if you do some simple D = S * T formulas, you can find that at exactly 6 minutes, car 1 (faster car) had traveled 5.94318 miles. Car 2 (slower) had traveled 2.94318 miles. But since car 1 had traveled an additional 3 miles, while it was lapping car 2, they both reach 1.44318 miles around the track at exactly 6 minutes. 1.44318 / 1.5 * 5280 (feet in a mile) = 300 feet short of point A.USAF Pilothttps://www.blogger.com/profile/01326339691636594880noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-76594859817957449802009-11-09T15:38:31.021-05:002009-11-09T15:38:31.021-05:00300 yards from point A.300 yards from point A.Andyhttps://www.blogger.com/profile/14117563810484999900noreply@blogger.com