calculus word problem..At noon, ship A is 30 nautical miles...?
918147
You have a right triangle with legs 106 and 88, a & b respectively. By the Pythagorean Theorem, the hypotenuse is square root of 18980. Now, use related rates. (dc/dt)2c=(da/dt)2a+(db/dt)2b. That becomes (dc/dt)*2*sq. rt of 18980=19(212)+22(176). dc/dt=28.67 knots per hour.
958527
You need to know the angle between "x" and "y", call it "a"\nLet initial position of "x" be "x0"\nLet the time from start of "x" be "t"\nLet "z" be the distance between "x" and "y" at time "t"\n\nThen:  \ny = 38 + 2t\nx = x0 + 11t   \n\nAnd from Pythagoras:\nz^2 = x^2 + y^2 - 2xycos(a)\n\nHopefully that gets you started....
550744
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.
49698
your variable m stays for "miles".\nYou thus just have to put 295 instead of m and do the exact reckoning: 0.05*295 + 30  14.75 + 30 = 44.75\n\nNow choose from the options the one which is closest to your number: 45\n\nIn this way you cannot do wrong
810640
See picture: http://mudandmuck.com/str2/BOAT.JPG\nShe must travel √(x^2 +4) miles by boat, and 6-x miles walking.\nTherefore, the time t is the sum of the distances divided by the rates, or\nt = √(x^2 +4) / 2 + (6-x)/5\n\nTake the derivitive of t with respect to x and set = 0 to find the minimum time.\n\nt = (1/2)(x^2 + 4)^(1/2) - x/5 + 6/5\ndt/dx = (1/2)(1/2)[(x^2+4)^(-1/2)](2x) - 1/5\n= x/[2√(x^2+4)] - 1/5 = 0\n5x = 2√(x^2+4)\n25x^2 = 4x^2 + 16\n21x^2 = 16\nx^2 = 16/21\nx = √(16/21) = 4√21/21 ≈ .87287156 miles from the point on shore nearest the boat.\n\n{Substituting this value back into the equation to find time, we see that the quickest time is ≈2.116515139 hours.}
496519
12 minutes (1/5 of an hour)\n\nDistance (d) = 10 miles\nRate (r) = 50 miles per hour\nTime (t) = Time in hours\nd = r * t\n\nSubstituting:\n10 = 50 * t\n\nt = 10/50\nt = 1/5 hours\n\n1/5 hour * 60 minutes/hour = 12 minutes\n\nThe answer is 12 minutes
113601
In order to solve this problem set up an equation\nLet X be the speed of the boat the first 3hours and X-5 the speed of the boat for the other 3 hours.\n \nThe formula for distance is D= Time x Speed, so:\n\n 3x +3(x-5)=57\n3x+3x-15=57\n6x=72\nx= 12\n\nSo the speed of the boat is 12 \nto find the distance for the fist 3hous multiply by 3:\n\n3x12=36    \n\n                        ANSWER; 36\n\nGood luck!!!
761202
LivingDownSouth's answer is not right.\nHe's tried to simplify it too much by simplifying the curve to a quarter of a circle.\n\nIf I have time later, I'll try to determine the actual curve ... and provide the answer for you.\n\nAdditional information entered later:\nI don't have a complete solution to the problem, but the following reduces it to a differential equation.\n\nFor simplicity, assume that the initial distance between the boats is 1 unit, that A is at (0,0) and B is at (-1,0), and that both travel at a speed of 1 (= 1 distance unit per time unit).\n\nThen A's position at any time t is (0,t).\nB's velocity consists of an x (eastward) component and a y (northward) component. these two velocities can be expressed as derivatives in the following way:\ndx/dt = -x/sqrt(x^2 + t^2)\n[Remember that x is a negative number between -1 and 0, so -x in the numerator is a positive number, indicating movement in the positive (eastward) direction.]\ndy/dt = t/sqrt(x^2 + t^2)\n\nIf you can solve these two equations (i.e., manipulate them and take the integrals in order to get rid of the derivative), then you will have expressions for x and y in terms of t. There will be a constant of integration, but you eliminate that by the initial condition (position at time 0 is (-1,0).\n\nTo find the eventual distance between the boats, take the expression t - y and determine what its value will be for a very large value of t. In this expression, t represents boat A's distance from the shore, and y (a function of t) represents boat B's distance from the shore. The difference is the distance separating the two boats.\n\nHope you can use this info.
677985
Let Bus A = travelling to Chicago (north)\nLet Bus B = travelling to New Orleans (south)\n\nGiven:\n- Bus A left at 5:00pm and Bus B at 5:30pm\n- Both Buses travel at 50mph\n\nSolution:\n- Find the distance (d) traveled by Bus A after 30 minutes (t).\n= d = r*t\n= d = 50mph * 0.5h\n= d = 25miles ~~~~~ This is the distance between both buses upon Bus B's departure.\n\n- Since both buses run at the speed, we can just split the remaining 300 miles (325-25)! So that would be 150miles per bus:\n= d = r*t\n= 150miles = 50mph * t\n= 150miles / 50mph = t ~~~~~ divide both sides by 50\n= 3hrs = t ~~~~~ This is the time needed by both bus to cover 150miles.\n\n- add 3 hrs to 5:30pm:\n= 5:30 + 3:00 = 8:30pm\n\nFINAL ANSWER: At 8:30pm, the two buses will precisely be 325miles away from each other.
303421
v=u+at\nor,30=10+a*20\nor,20a+10=30\nor,20a=20\nor,a=1 m/s^2\n\nF=ma\nor,F=3500*1\nor,f=3500 N
676280
given\nt=# of hours past midday(meaning 12:00pm I assume)\n =15:00-12:00=3\n\nD=3.5\n\nso\n\nthe simplied equation becomes\n1.5*A -.5*A*k=-2\nthere are alot of solutions to this problem...\nfor example \nif k=1, A=-2\nif k=2, A=-4... \nmore information is needed to narrow the solution to only two numbers.\nDoes it tell you the depth at midday(t=0, D=?)
371345
d=vt\n\nWhere d is distance, v is velocity, and t is time.\nV here is the speed of sound through water, which you'd have to look up, and t=2.9 seconds.  Remember when you solve that your "d" actually covers the distance from the ship to the ocean floor AND BACK, so you'll have to divide it in half to find the depth.
858533
C = 2 ii r\n300 = 2 ii r\nr = 300/2ii\nr = 47.74648 feet\nA = ii r^2\nA = 7,161.97244 feet^2\nAcres = 0.16442
115023
Imagine a `speed` x `time` chart. Plotting the data, we will have parallelogram whose area will the length the car walked between the two speeds.\n\nWe'll have, then, 30*((5.56+13.89)/2) = 291.75m
966246
d=r*t\nhighway= 60*highwaytime\ncity = 30*citytime\ncity = highway -  100\ncitytime =highway time - 5/6  50 min = 5/6 hour.\nsub in\n\nhighway - 100 = 30 (highwaytime- 5/6)\n60 * highwaytime - 100 = 30 * highwaytime - 30*5/6\nsimplify\n30* highwaytime = 75\nhighwaytime = 2 1/2 hours\nh  =2.5*60 = 150 miles\ncity =30* 1 2/3 = 50 miles\ntotal 200 miles
552943
by using cosine law, you can calculate easily the magnitude of the ship.\n(a^2+b^2)-2abcos(q)\n840^2+960^2-(2*840*960*0.81)\n=320832 lbs\nFor the direction, i need to see the question. It would be the opposite of the other's directions.
199496
If you find the latitude of the northernmost point and the latitude of a point directly south of that you can calculate it.  Multiply the difference in degrees by 60 to get the distance in nautical miles.  Then apply the conversion factors:  1 nautical mile = 1.152 statute miles and 1 statute mile = 1.609 kilometers.  Combined, that means to multiply the difference in longitude in degrees by 111.214.  (Use degrees and decimal degrees, not minutes and seconds).\n\nYou'll be pretty close with this method even though it assumes a spherical Earth.