A man travels 86 miles in 2 hours. What is his average speed in miles per hour?\n\nPS: not homework.
132140
speed=distance/timetaken\ntherefore the average speed=86/2=43miles per hour
39489
A would take 22 minutes to get to 15 miles and 32 minutes to get to 20 miles. Unfortunately, there's insufficient information to be precise on B's journey. Did he spin off before the start?.. or at which point en route? Guessing it, I would say A arrived at 15 miles and at the 20 miles point first. NB. I have had to make certain assumptions in your question to arrive at this answer (eg I asumed A slows down to 40mph after 4 miles?)
457809
You need to convert feet to miles and seconds to hours. There are 5280 feet in a mile and 3600 seconds in an hour.\n\nx ft/s * (1 mi/5280 ft) * (3600s/hr) = x * 22/15 mph\n\nSo you multiply the speed in feet per second with 22/15, or 1.4666666666666666...
367643
135 miles / (60 miles / hour)\n\n= 135 miles * (1 hour/60 miles) = 2.25 hours\n\nThe way to work these kinds of problems out is to pay attention to the units.  \n\nRemember that "miles per hour" means that number of miles divided by one hour.  You are given miles and miles per hour, so what operation between "m" and "m/h" yields only time (hours or h) as a result?  The answer is m divided by (m/h).
427384
5 hours X 40 miles + 30 miles= 230 \n\n230/5= 46 miles per hour...is that right? yeah i think so
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)
439060
1.time taken=distance/speed=26/43 hours\n 2.time taken=distance/speed=68/54\n=34/27hr\n=1 7/27 hr
462243
Average speed = (8+8.5)/2=16.5/2 = 8.25 m/s\nTotal distance = 8.25x99=816.75m\nTime taken for 1st Lap = 816.75/(2x8) = 51 sec\nTime taken for 2nd Lap = 816.75/(2x8.5) = 48 sec
351135
its right.......... average speed= distance/time\n\nso average speed= 10*3/1\n=>30km/hr
484841
well if the car needs to go around twice and average 60 mph then it would need to travel 120 mph total, 60*2.  If it went once around at 30 mph, 120-30=90. Therefore it would have to go 90mph for the second lap.\n\nAnother way to look at it would be to use the D=RT formula for distance = rate times time.  Solve it for the time meaning T=D/R so the time it would take for two laps to average 60mph would be 10miles/60mph or 1/6 of an hour, or 10 mintues.  Now one lap of 5 miles at 30mph would take T=5m/30mph or 1/6 hour = 10 minutes.  This leaves no time for the second lap to take place.\n\nThis is a paradoxical question.  If the second lap was at 90 mph then the average would be 60mph (30mph+90mph)/2=\n120mph/2=60mph.
239957
He has to drive two miles (one mile up and one mile down) in two minutes.  He already took two minutes to get up the hill, so it is impossible for him to average 60 mph.
514801
Something's wrong with the writeup for the first question.  I'm guessing it should have said THR = RHR + 0.65(MHR-RHR), because otherwise you're gonna get a THR of, like, 10000 beats per minute.\n\nProceeding from that assumption, the THR is RHR + 0.65(MHR-RHR).  You're told that the RHR is 54 bpm.  The MHR is 220 - the age(43), so it's 177.  Therefore the THR is 54 + 0.65(177-54) = 54 + 0.65(123) = 54 + 79.95 = 133.95 bpm.\n\nIn the second question, you didn't provide the "following equations," but think of it this way.  The airplane flew for a total of 15 hours at an average speed of 350 miles per hour.  That means it went 5250 miles.  In the final 7 hours it went 2275 miles.  So in the first 8 hours it had to go 2975 miles.  2975 miles/8 hours = 371.875 mph.  I don't exactly know how to write that in an equation, but now that you understand the reasoning, hopefully you'll be able to recognize the correct equation.  If all else fails, plug the data into each equation and see which one returns 371.875.
560757
Let x = speed going into the country, then\nx + 5 = speed returning\nd = rt\nd = x(5) for the trip into the country\nd = (x+5)(4) returning from the country\nThe two distances must be equal.\nTherefore,\nx(5) = (x+5)(4)\n5x = 4x + 20\nx = 20\n20mph into the country and 25mph returning.
488396
D.\n\nHe would have to run 9 mph 2 catch up with/ overtake him. cuz Joe can run a mile every 10 minutes, so hes a mile a head of Ken before Ken even starts. So Ken has 20 minutes to run 3 miles. There is 3 sets of 20 mins in an hour and 3 miles per 20 minutes = 9mph! lol
259698
As you can tell, there's not really enough info within the problem itself to solve it.  You'll need to get a map, atlas, or some internet source to determine how much of the 481 miles will be driven in Oregon, and how much in California.  I won't look that up for you, but let's pretend that it's 121 miles in Oregon, and the other 481-121=360 miles in Cali.\n\nAssuming you constantly maintain the speeds indicated, you can calculate the time it takes you for each leg of the trip by using the formula:\n\n   distance = speed x time\n\nOr, equivalently,\n\n   time = distance / speed\n\nSo, for the 121 miles in Oregon, over which you will be traveling 65 mph,\n\n   time = 121 miles / 65 mph = (approximately) 1.86 hours\n\nand for the 360 miles in California, over which you will be traveling 70 mph,\n\n   time = 360 miles / 70 mph = (approx.) 5.14 hours\n\nAltogether, then, you'd be traveling 1.86 + 5.14 = (approx.) 7.00 hours.\n\nIf you want to write that out as one long problem, you could write it this way:\n\nTotal Time = Time1 + Time 2\n           = Dist1/Speed1 + Dist2/Speed2\n           = 121 mi/65 mph + 360 mi/70 mph\n           ~ 1.86 hr + 5.14 hr\n           ~ 7.00 hr\n\nNow, don't forget, I picked the distance values, 121 and 360 miles, out of thin air.  You'll need to look them up and do the problem yourself.  But hopefully that should give you an idea of how.  Good luck!
144302
if the question is: If it takes 3 hours to travel 24 miles downstream and 6 hours to travel UPstream, how fast is the current? then,\nlets call the travelling speed Vt and the current speed Vc\ndownstream we have Vt+Vc=24miles/3hours=8mph\nupstream we have Vt-Vc=24/6=4mph\nto solve the system of two equations we can add the equations and get 2Vt=12, Vt=6mph\nVc=8-6=6-4=2mph\nThe current is 2mph fast.
50817
If you start with the times from A to B is t1 and the time from B to C is t2, then the distance from A to C is t1*10 + t2*50, and the total time is t1+t2, so the average speed is:\n\n    (t1*10 + t2*50)/(t1+t2)\n\nThat's the weighted mean of 10 and 50, with weights t1 and t2.\n\nIn general, to average a bunch of averages, you need to compute a weighted average.  For example, if you have ten rocks averaging one pound apiece, and I have five rocks averaging four pounds apiece, then to determine the average of all the rocks, you would take the weighted average of 1 (with weight ten) and 4 (with weight 5.)  That is:\n\n    (1*10 + 4*5) / (10+5) = 30/15 = 2