IF YOU WERE A SHIPMASTER AND YOUR SHIP WAS GOING 22 NAUTICAL MILES...HOW FAST WAS HIS SHIP TRAVELING AT SEA LEVEL IN MILES PER HOUR? EASY!
123273
25 mph\n\nhttp://www.disastercenter.com/convert.htm
267498
Get in touch with me, buong, I am, and I will be leaving in 4 days. My name here is Glalien, I come from Pleedeeze, it takes 8 hours to arrive here from our planet, via time warping. Our ships are made of nothing you have on this planet. Our propulsion is of magnetics. You will not believe me if I tell you why we are here. Look me up.
106197
SIne they travel at right angles to each other, the distance between them at any point of time will be the hypoteneuse of a right angle triangle. The path of each boat from the dock will form the other two sides.\n\nLet teh speed of the slower boat be x mph\nHence, the speed of the faster boat = (x+1) mph\n\nUsing Pythagoras' Theorem\nx^2 + (x+1)^2 = 5^2\n\nSolving the equation,\nx^2 + x^2 + 2x + 1 = 25\n\n2x^2 + 2x = 24\n\nx(x+1) = 12  (divided by 2 throughout)\n\nx^2 + x - 12 = 0\n\nx^2 + 4x - 3x - 12 = 0 (FInd two numbers whose sum is 1 and product is -12)\n\nHence, \nx(x+4) -3(x+4) = 0\n\n=> (x-3)(X+4) = 0\n\n=> x = 3 or x=-4\n\nSpeed cannot be negative. Hence, x = 3\n=> x+1 = 3+1 = 4\n\nHence,\nSpeed of slower boat = 3 mph\nSpeed of faster boat = 4 mph
11962
The answer is A. You mean to tell me you couldn't figure this out for yourself?\n\nHere's how it's done.\n\n1. The unit of measure is feet per hour (ft/h).\n\n2. There are 60 minutes in an hour.\n\n3. 30 minutes x 2 = 60 minutes, which is an hour, as mentioned in #2 above.\n\n4. Since you multiplied 30 minutes by 2, you should likewise multiply 1000 feet by 2. 1000 feet x 2 = 2000 feet. There's your answer. Got it?
26849
...traditional merchant ships were accustomed to average speeds of under 5 knots (9 km/h), clippers aimed at 9 knots (17 km/h) or better. Sometimes these ships could reach 20 knots (37 km/h).
415867
They travelled a total of 476 miles in 4 hours.  So together they are travelling (476/4) mph or 119 mph.\n\nSo you have two equations:\nJ + S = 119\nJ - S = 5\n\n2J = 124\nJ = 62 mph\nS = 57 mph
123648
Average rate per hour = total distance / total time\n=( x + x )/ total time          (1)\nTime required to travel x distance at the speed of 10 miles\nper hour = x / 10 hours.\nTime required to travel x distance at the speed of 8 miles\nper hour = x / 8 hours.\nSo the total time is\n=(x / 10) + (x / 8)\n=18x/80 \n=\nPutiing this value in the denominator of (1)\nWe get\nAverage rate per hour = total distance / total time\n= 2x/[18x/80]\n=80/9\n=8.88\n=8.9 miles/hr\n\n( In general if the speeds are 'a' & 'b', the the average \nis given by their harmonic mean viz. 2*a*b/(a+b) )
402576
try: \n\nF=m*a =>     0.3*M*g = M*a    =>    a=0.3*g\n\na*t = 60 miles/hr   =>    \nt = 60miles/hr / (0.3 g) = (60 miles per hour) / (0.3 * 9.8 ((m / s) / s)) = 9.12 seconds\n\ndistance\nD=1/2*a*t^2 = ((0.5 * 0.3 * 9.8 * (m / s)) / s) * ((9.12 sec)^2) = 122.266368 meters
103497
Depends on the level of elevation.  The higher elevations have less air pressure.  At sea level it's about fifteen pounds for every square inch.\n\nConversion to mph is about 31.07 mph or 546.83 inches per second.\n\nHere are some drag formulas\nhttp://carini.physics.indiana.edu/E105/drag-force.html
257741
If I'm reading your description right, I think your first boat is heading up and to the right ("direction N.68 degrees E") making an angle of 68 degrees with north, or 22 degrees with east.  And I think your second boat is headed directly east ("S.90 degrees E").  So I'm assuming that they form a triangle, with one angle being 22 degrees and another one being a 90 degree angle, since we're looking for the point where one boat will be directly south of the other.\n\nThe first boat traveled 12t miles so far.  The second boat left an hour later and traveled 16(t-1) miles.\n\nThe first boat's path turns out to be the hypotenuse of the triangle, and the second boat's path will be one of the adjacent legs to the 22 degree angle.  So we can set up an equation with trig identities:\n\ncos 22 = 16(t-1)/12t\n0.9272 = 4(t-1)/3t\n2.7816t = 4t - 4\n4 = 1.2184t\nt = 3.3 hours\n\nSo after 3.3 hours, one boat will be directly south of the other.\n\nFinding that distance can be done one of several ways: using the Pythagorean theorem, sine relationships, law of cosines, etc.  I'll use sine, since we're already doing trig.  After 3.3 hours, the northernmost boat will have traveled 39.6 miles.  The distance between the boats becomes our side opposite the 22 degree angle.  So,\n\nsin 22 = distance between/39.6\ndistance = 39.6sin 22\n= 14.83 miles\n\nNow if I read the angle wrong, then simply go back and change all the "22" angles to the correct one and recalculate.
658118
The width of the river is 1 mile =  1760yards\n\nYes, let A be the faster boat leaving from shore A, and B the slower boat leaving from shore B. A will a travel a distance (720+x) yards from shore A to where they first meet while B travels 720 yards from shore B. A reaches shore B first and waits10min and then starts back, while B reaches shore A later and does the same thing and they meet 400 yards from shore A. Since they wait the same time and keep their speed constant we must have 720-400 = 320 = x. So the river = 720+720+320 = 1760 yards.\n\nIt's easy to see if you eliminate the wait time and imagine A turns around as soon as he reaches shore B.  A will be distance x from shore B on the return trip when B is distance x from reaching shore A. So they must still travel the width of the river to meet 400 yards from shore A. So we must have x+400 = 720 as the distance B travels to reach shore A and then return to the 2nd meeting point. So x = 320 and the river is a mile wide,
180239
I'd say 17000-19000 miles per hour.\n\nIts at a inclined geo-stationary orbit probably.
400545
28 hours..6.5 mph\n\nnot difficult..10 miles the first hour\n                   3 miles the next hour\n\n13 miles in 2 hours therefore 6.5mph\n\n182/6.5=28
262089
what type of ship? A merchant ship will do about 20 miles an hour..(30 kph).......a sailboat will average 4 to 6 mph (6-9 kph)....so, do the math...........if youre asking can you take a sailboat from Germany to Galveston in 3 months, sure...and have time to stop along the way.
384910
To solve this problem, you need to find the time the bus needed to travel the 200 miles.  You should divide this problem into 3 intervals:\n\n1) 150 MPH for first 50 miles\n2) 125 MPH for 2 minutes\n3) 150 MPH for reaminder of trip\n\nFor interval 1, we need to caclulate the time taken to travel 50 miles:\n\n(50 miles)(1 hour / 150 miles) = 1/3 hour = 20 min\n\nFor interval 2, we needed to calculate the distance traveled in order to solve for interval 3:\n\n(125 miles / 1 hour)(2/60 hour) = 25/6 miles\n\nNow, we can find the # of miles for interval 3:\n\n# miles = 200 - (50 + 25/6) = 875/6 miles \n\nNow we can find the time needed to travel this distance:\n\n(875/6 miles)(1 hour / 150 miles) = 35/36 hour (around58 min)\n\n---------\n\nTo find average speed, you need to take [total distance/ time].  therefore,\n\navg. speed = [200 miles]/[(1/3) + (1/30) + (35/36) hr] ------->\n\n149.38 miles per hour\n\n------------\n\nHope this helps
607518
i am no genius but i did try. here's how i went about it.\nspeed = distance/time.\n(1/30)s = 525 lines, therefore time for 1 line = ((1/30)/525)\n=(1/15750)s\n20in = 1 line.\nresweeping the line, therefore distance = 40in.\n40in = 3.33ft (1ft = 12in)\nagain, speed = distance/time, so\nspeed = (3.33ft)/(1/15750s) = 52500 (ft/sec)\nhopefully, this is the right answer. please let me know. thanks.
199491
You would have to travel the last mile at an infinite velocity to average 60mph over the total 2 miles\n\nV = 1/(2/60 - 1/30) = 1/0 = infinity\n\nEDIT- more descriptive version of the equation\n\nDt/Vt = D1/V1 + D2/V2\nhence\nV2 = D2/(Dt/Vt - D1/V1)