lets say one hour on each, average speed
217871
treadmill would lkely burn 30% more calories for average users, but this would depend on intensity.\nIf boht users achieved similar target heart rates, and maintained them , the calorie consumption would be similar.
235275
average speed is total_distance/total_time.\nAssume distance is x, so it the average speed is: \n2x/(x/20+x/30) = 2/(1/20+1/30)\n = 2*60 / (3+2) \n = 24
109818
average speed: total distance/total time\ntotal distance: 60 miles\ntotal time: 100 minutes = 1,6666666666666667 hour\naverage speed = 60 miles/1,6666666666666667 hour\naverage speed = 36 mph\n---------------------------------------------------\nvelocity is zero because man does not change of position
11497
20m/s.\nbecause the total time is 5s. and by defn of average speed it is the ratio of total distance travelled to total time taken i.e 100/5 = 20m/s
403179
East bound speed=x, in 5 hours 5x\nWest                     =x+6,             5x+30\n10x+30=240\nx=21 miles per hour.\nEast=21\nwest=27\nVR
282369
An hour for each shot.
91143
No, just 7-8 hours average
609913
4:00\n\nEST = CST + 1 hour
648028
very easy 45km/hr   avarge speed depend on both avarge speed\ntotal trip avarge = avarge of here to ther+ avg of there to here / 2  \n\ntrip avg= 50+40/2    = 45
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
560127
53.3 km/hr\n\ntrip takes 1.5 hours so 80/1.5 gives you km/h.
33259
213.75169013 mile/hour
441201
I believe its 150/min
597418
15 min a least the average time about 20
334789
1 hour or 60 mins or 3600 seconds :-) is the time it takes but it will still be 12:30 :-p
531764
440=Vr(t+30)\n(Vr+Vb)t=440+(Vb-Vr)30\nTotal time elasped is 1hour\nRate of stream Vr=440yards per hour
363861
OK what you need to remember about any problem like this is the formula Distance Traveled=Rate of Travel * Length of Time Traveled, (D=R * T).  As long as you have two of those values, you can calculate the third.  You know the distance, and you know the rate or travel.  To calculate the Time, you get it by itself on one side of the equation.  Divide both sides of the equation by R (rate), like this:\n      D = R * T\ndivide both sides by R\n     D / R = (R * T) /R\nthe R's on the right side will divide out and leave you with\n     D / R = T\nNow you plug in the values for D and R\nFor train A\n     D = 1000 (miles)\n     R = 125 (mph)\nso  T= 1000/125 = 8 hours\nYou know the time is expressed in hours, because your units divide out just like numbers do.  Your are taking miles and dividing them by miles per hour, and that leaves you with hours.\nIf you add in the stops the train makes (2 stops, 15 minutes each) you would add 30 minutes, or 0.5 hours to the total time, it takes Train A 8.5 hours to go from Station 1 to Station 2.\n\nDoing the same thing with Train B...\nT = D / R\nT = 1000 / 100  (Train B is travelling the same distance, 1000 miles, and travelling at 100mph.)\nSolve the equation and you get T = 10 hours, and adding in one stop of 15 minutes (or .25 hours), you get 10.25 hours.\n\nSo, Train A gets there first.  Even if Train B doesn't make it's stop until near the very end of the run, Train A gets to the station in 8.5 hours, and even without any stops, at 100 mph, Train B is still 1.75 hours out of the station at that point.\n\nIt's a useful formula to learn and understand, you can apply in many different situations.\n\nGood luck.