If it takes 3 hours to travel 24 miles downstream and 6 hours to travel downstream, how fast is the current...
337691
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.
639920
its simple.\nwhen boat travels downstream the two velocities add up to give the answer 15+3=18km/hr.\nwhen boat travels upstream the velocity of river gets subtracted from velocity of boat which gives the result as 15-3=12km/hr.
607556
Everyone is wrong, it is 48MPH, think of it this way.  Lets say the distance is 60 miles, it takes 1.5 hours at 40, and 1 hour at 60, so a total of 2.5 hours, divide 120 by 2.5 and it is 48mph.  This will always be the answer reguardless of distance, whether 1 mile in each direction or a million miles in each direction.
1175600
They will meet when:\n\n4mph x Time = 3miles - 5mph x Time\n\nSo the answer is 0.33333 hours \n\nOr just 20 minutes.\n\n(not 27)
954922
440=Vr(t+30)\n(Vr+Vb)t=440+(Vb-Vr)30\nTotal time elasped is 1hour\nRate of stream Vr=440yards per hour
631391
let T1 and V1 be velocity and time to lake\nand T2 and V2 be velocity and time to ocean\n \nT1 x V1 = 600\nT2 x V2 = 240\nalso\nT1 = 2T2\nV1= V2 + 6\nhence\n2T2 x V1 = 600\nsubstituting again V1 = V2 + 6\n2T2 x(V2+6) = 600\nT2 x V2     = 240\nor T2       = 240/V2\nsubstituing in equation:\n2 x (240/V2) ( V2 + 6) = 600\nx V2 both sides\n2x 240 (V2+6)= 600 V2\n480 V2 + (6x480) = 600 V2\nsubtract 480 V2 from both sides\n6x 480 = 120 V2\ndivide by 120 each side\n6 x 480/120 = V2\nV2 = 24\nV1 = 30 \nT2 = 10 hours\nt1 = 20 hours \nchecked in top equations
905252
Lets say its 3:25 now. \n\n5 minutes ago it was 3:20 which makes (4 times 5 many minutes) past 3.\n\nNow if it is 3:25 now then its 155 minutes to 6\n(35 min 3:25 to 4 + 60 min 4 to 5 + 60 min 5 to 6).
1022304
Point A and Point B are 600 miles apart. A train leaves Point A travelling toward Point B at 5:30AM at a constant speed of 65 kph. Another train leaves Point B heading toward Point A at 6:20AM at a constant speed of 40 kph. What time do the two trains meet?1 kilometer = 0.6214 miles\n\nx=kilometers     5/6(50 minutes/60 minutes)*65= 54.166...km\nx=.6214 (600 miles)\nx=372.84 kilometers apart\n5/6(50 minutes/60 minutes)*65= 54.166...km\n372.84-54.17= 318.67km between the 2 trains at 6:20\nthats as far as I can get (I don't remember how to do this... hope this helps!)
694439
D(istance) =  S(peed)   x   T(ime)\nor   D / T   =  S\n12 minutes =  .20 hr  \n11km /  .20h  =    55 km/h
143101
speed is distance divided by time. So work out the distance the water needs to travel and divide by the time taken. Add more details if you want to know a more complex way to work out the speed.
1071909
roughly 11 meters per second, or 24 miles an hour.\n;o)
392981
Is the speed at which it travels.
915373
Each crest travels 3m in 6 s, so the wave's speed is 3/6 = 1/2 m/s.
355718
I have walked 24 miles. It took me about 5 hours.
100012
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?)
786356
at 3 sec r = 3*8 = 24m\n\na = pi*r^2\n\nda/dt = d(pi*r^2)/dt \nda/dt = 2*pi*r*(dr/dt)\nda/dt = 2*pi*24*8\nda/dt = 384*pi\nda/dt = 1206.4 meters^2 per sec
866341
Trip time and trip current.