back to town. If the bus travels at a rate of nine miles per hour and he was to jog back to town at a rate of three miles per hour, how far would he ride so theat he'd be back in eight hours?
126949
let d=distance traveled in one direction(away from town or back)\ntime=distance/rate\n8hr= d/9 + d/3\nd= 18miles = answer
384368
Bob was travelling at 9.5 meters/sec when he started his jump. To gain maximum range he needed to jump at a perfect 45 degree angle. If he had done so, his initial velocity vertically would have been 9.5 /sqrt(2) = 6.72 meters/sec. His maximum height would occur at time t where 6.72-gt = 0.\n9.78t = 6.72\nt = 6.72/9.78 = 0.687 secs\n\nAfter that he would fall for another 0.687 secs till he hit the ground for a total of 2(0.697) = 1.37 secs\n\nIn this time, he would have traveled (6.72)(1.37) meters horizontally or 9.23 meters.\n\nSo his jump of 8.90 meters was 0.33 meters short of perfection.\n\nNote that air resistance (which is pretty important in reality) would have the effect of making the maximum achievable jump shorter than 9.23 meters. So his jump was better than it appeared..
674292
4 miles (give or take)\n\nyou basically work out how many miles the firetruck goes in 1 minute (which is .83 50/60) then multiply by the 5 mins travelling time. 5 x .83 = 4.16 miles...\n\nformula would be: (a/60) x b = c\n\nwhere:\n\na is the constant speed\nb is the time taken to arrive\nand\nc is the answer
526574
Joes' running time: 7:30 - 7:00 = 30 min.\n Distance = Speed x time.\n Distance = 6 mi/h x ½ h\n Distance = 3 mi\n\n Kens' running time: 7:30 - 7:10 = 20 min. (that is, a third of an hour).\n Distance = Speed x time.\n Speed = Distance / time.\n Speed = 3 mi / (1/3)\n Speed = 9 mi/h
551509
Assuming he doesn't get tired and can run at that pace:\n8.3 m/s x 60 secs/min x 60 min/hr x 10 hrs = 298,800 m\nObviously this is too fast. I believe you meant that the bear runs 8.3 m/hr. \nThen it would be:\n8.3 m/hr x 10 hrs = 83 m\nWhich is a little more realistic.
188412
At 15 mph, it will take Charlene 2 minutes to go around once (0.5 mi * 60 min/15 mph - the 60 min converts hours to minutes for calculations).  At 12 mph, it will take Chris 2.5 minutes (0.5 mi/12 mph).\n\nAt the 2 minute mark, Charlene has already gone around once and is now "behind" Chris.  Chris will have gone 0.4 mi by then (12 mph * 2 min/60 min).  So now we can set up an algrbra problem - how long will it take Charlene to catch up?\n\nCharlene is going 15 mph, and will travel 15t miles after t hours.  Chris is going 12 mph and will travel 12t miles after t hours.  But Chris has a 0.4 mi head start.  So your equation will be\n\n15t = 12t + 0.4\n3t = 0.4\nt = 2/15 hours, or 8 minutes.\n\nSo, since all of these calculations are based on a point in time 2 minutes after they both started riding, Charlene will pass Chris 10 minutes after they started.
390581
D = distance in feet\nt = time in seconds\nS = speed in miles per hour\n\n[D feet x (1 mile/5280 feet)] / [t seconds x (1 hour/3600 seconds)]\n= S miles per hour\n\nDoing the arithmetic, we get:\n0.6818 D / t = S mph\n\nIf D is 150 feet and t is 4 seconds, \n0.6818 (150)/ 4 = 25.56 mph\n\nIf the time is 8 seconds, the speed would be half as much, or \n12.78 mph
153796
M = Martina's distance from home at time t, in hour.\nM = 24t\n\nJ = John's distance from home at time t, in hour.\nJ = 48(t - 2) \n\nWhy (t - 2)? because, John started 2 hours later.\n\nThus, when " john catch up with Martina" means that their distance from home is equal. Thus, \n\n24t = 48(t - 2)\n  t = 2t - 4\nor t = 4.\n\nThis means 4 hours from the time Martina's leaves home, John, would have caught up with her. The time is just 4 hours after 9 am, i.e, 1 pm.\n\nRegards
564330
260 miles.\n\nFrom the half-hour difference we can find the length of the hill:\n\nt = d/v, t2-t1 = d (1/v2 - 1/v1)\nd= .5/(1/72-1/90) = 180\n\nThis takes 2 hours at 90 mph, so the other hour covers 80 miles.\n180+80=260
515038
Ok. 45 minutes is equal to 3/4 of an hour. To find the distance Travis has travelled at that point in time, you will need to multiply 55 (his rate) by 3/4. That is equal to 41.25 miles. Now we need to find if Roy will ever catch up to him. We are trying to find when their distances are equal, so time can be our variable. \n\n41.25 + 55t = 65 t \n41.25 = 10t\nt = 4.125 hours\nNow, if we plug that in to 41.25 + 55 t and it is greater than 450, then Roy can't reach Travis before Travis reaches Pecos. \n41.25 + 55 (4.125) = 268.125\n4.125 hours after Roy leaves Dallas, he will catch up to Travis 268.125 miles from Dallas.
373483
The second problem is easier than the first problem, so I'll tackle it first.\n\nLet's imagine that the trip is 120 miles. The distance doesn't matter if the average rates are as given above.\n\nIt takes 3 hours to travel 120 miles at 40 miles per hour.\nIt takes 2 hours to travel 120 miles at 60 miles per hour.\n\nSo it took 5 hours to travel a total of 240 miles.\n\nThis is an average rate of 48 miles per hour.\n\nI'll work on the first problem and hopefully I'll get back to you.\n\nUPDATE\n\nI think I have the first problem solved as well. It involves an infinite series! Call the people involved A and B.\n\nThe probability that A wins on the first toss is 1/2.\n\nIf A doesn't win on the first toss, then the probability that B wins on the second toss is also 1/2. But this only happens in 1/2 of all games. (In the other half, A won on the first toss.)\n\nIf B doesn't win on the second toss, the the probability that A wins on the third toss is also 1/2. But this only happens in 1/4 of all games (In half the other games, A won on the first toss. In half the remaining games - 1/4 of all games, B won on the second toss.\n\nFollowing this logic, A wins:\n\n1/2 + 1/4(1/2) + 1/16(1/2) + 1/64(1/2) + ...\n\nof all games.\n\nB wins:\n\n1/2(1/2) + 1/8(1/2) + 1/32(1/2) + ...\n\nof all games.\n\nLooking again at A, we have:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...)\n\nThe term in parentheses is an infinite series having initial term 1 and ratio 1/4.\n\nThe sum of an infinite series having ratio whose absolute value is less than 1 is equal to:\n\na/(1 - r)\n\nWith a = 1, and r = 1/4, the sum = 4/3. So:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3\n\nLooking again at B, we have:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...)\n\nThis sum should be half of the sum of the infinite series in A.\n\nWith a = 1/2 and r = 1/4, a/(1 -r) = 2/3\n\nSo:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...) = 1/2 * 2/3 = 1/3.\n\nA should win 2/3 of the time, and B should win 1/3 of the time. There is an advantage to going first!
94734
Distance = rate x time\n\nif you're going 8 feet per second for 4 seconds, you've gone 32 feet...
199345
speed=distance/time\nV=S/T\nS=VT (distance = speed x time)\n\nlet Vg be gary's speed\n    Vt be Tom's speed\n    S be the distance\n given that Gary completed the run in 20 min, we have \nS=Vg x 20min     (x means times)\ngiven that for time T=20min Tom completed 5/8 of the distance or St=5S/8=Vt x 20min\nsolve St=5S/8=Vt x 20min for S to get S=(8/5)Vt x 20min\n\nnow we have two equations for S, S=Vg x 20min and S=(8/5)Vt x 20min\nit means that Vg x 20min = (8/5)Vt x 20min   /divide both sides by 20min\nVg=(8/5)Vt\n\ngiven that Tom's average speed for the race was 75m/min less than Gary's, we have Vt=Vg-75\nsubstitute for Vt in Vg=(8/5)Vt\nVg=(8/5)(Vg-75)\n5Vg=8Vg-600\n3Vg=600\nVg=200 m/min\nVt=200-75 = 125 m/min (Tom's speed was 125 m/min)\nS=Vgx20min= 200x20=4000m (distance of the race was 4000m)
104064
well... man is jogging at 10 km/h... and yhe distance between A and B is 25 km.... then the man is jogging for\nt = (25 km)/(10 km/h) = 2.5 h\n-------\nNow the distance that bird makes in that time....\nd = (45 km/h)*(2.5 h) = 112.5 km
320123
The difference in rates is 1 km/hr. So if they are moving in the same direction, the faster walker will be 5 km ahead at the end of 5 hours.\n\nThe sum of the distances traveled per hour is 11 km. So at the end of 5 hours of travel in opposite directions, they will be 55 km apart.
471362
hope you find this useful....\nconsider instead of a round trip, he takes a trip in only one direction, from say A to B....i.e for 360 miles\nas you know, \nspeed = distance/time  or   time = distance/speed\n\nso , with constant speed say x, his time t  is 360/x.\n\nconsidering any point m between A and B, his total time  for the journey can also be written as the sum of the time taken from A to m, and time taken from m to B.\n\n t = 360/x = [m/x ]+ [( 360- m )/x ] ---------(1) \n\nthis is a general solution ,considering the trip as being in only one direction. but in our problem, the journey is up and down,so  the point m is midway (180),and his speed changes ( increases by 9)his total time changes(given as decreases by 2 hours).\n\nso the new equation is....\n\nt-2 = [m/x ]+ [(360 - m)/(x+9)]     ---------(2)\n\nsubtracting (2) from (1), we get,\n\nt - (t-2) =[ (360 -m)/x ] -  [ (360 -m)/(x+9)]-----------(3)\n\n 2 = [180/x] - [ 180/(x+9)]...........( since m = 180)\n\nsolve for x...\n\nthats the speed in the forward trip\nx+9 is the speed on the return trip.
498236
.8 *10= 8\n\nso the question is at a speed of 18 miles per hour how long would it take to run 8 miles. \n\n18 miles / 60 minutes = .3 miles per minute\n8 miles / .3 miles per minute = 26.667 minutes