Boat A moves in  straight line at 20 km/h and boat B moves in a straight line at 32 km/h.?
539310
After 2.5 hrs, Boat A has gone 50 km, and Boat B 80 km. Let the distances of A and B be two sides of a triangle, a 70 degree angle connecting them.\n\nThus we have a triangle with two known sides and one known angle.\n  \nLabel it A, B and C starting from the bottom left angle. Label each angle's opposite side its respective lowercase (a,b, and c). Fill in for b and c, and angle A. We don't know if this is a right triangle or not, so we're going to need to use the law of cosines:\n\na^2 = b^2 + c^2 - 2*b*c*cos(a)\nSubstitute for each known variable:\na^2 = (50)^2 + (80)^2 - 2(50)(80)cos(70 degrees)\na^2 = 2500 + 6400 - 8000 * cos (70 degrees)\na^2 = 8900 - 8000 * .34202\na^2 = 6163.84\na = 78.5101\n\nThus, using the law of cosines, the distance (a) from the two boats is approximately 78.5101 km. It is not exact as there is no exact value for cos (70 degrees).
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
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.
274949
Here is a solution.  Let x be the distance the motorist traveled at 20km/h on city road, and let y be the distance the motorist traveled at 50km/h on the highway.  We are looking for y.\n\nFirst, we know that the total distance is 155km\n\nx+y=155     (eq1)\n\nSecond we know that if you travel x km at 20km/h, it will take you x/20 hours.  Since the total hours is 4, then:\n\nx/20 + y/50 = 4      (eq2)\n\nThat is two equations, two unknowns, so we can solve the problem.  We'll rearrange eq1 to be:\n\ny=155-x     (eq3)\n\nThen substituting this into eq4 yields\n\nx/20 + (155-x)/50 = 4 \n\nor\n\n5x/100 + 2(155-x)/100 = 4\n5x+2(155)-2x=400\n3x=90\n\nx=30\nThat gives us x, and we can solve for y from (eq 3)\n\ny=155-30\n\ny=125\n\nAnswer: 125 km
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.
104070
Heather leaves a town at 2 p.m., going 56 mph, traveling east. Erin leaves the same town at the same time going 64 mph, traveling in the same direction. What time of day will it be when Erin is 72 miles ahead of Heather?\n============================================================\n\nHeather's distance = 56x\nErin's distance = 64x\nx = hours traveled\n\nThe distance between them at any time can be expressed as a simple equation:\n\nd = 64x - 56x\nd = 8x\n\nYou want to find "x" when d = 72 miles:\n\n8x = 72\nx = 9 hours\n\n2:00 p.m. + 9 hours = 11:00 p.m.
48038
distance = rate x time, \n\nLet x be the distance she can go out on the lake, and t the time it takes to get there.\nGoing out she paddles a 3km/h so d = 3 x t\nComing back, the distance is the same, the rate is 2 km/h\nand the time she has is whatever is left of the 2.5 hrs.\nShe use t amount of time going out so (2.5 -t)\n\nGoing out   d = 3 x t\nComing back d = 2 x ( 2.5 - t)\nThe distance is the same both ways, so \n3 x t = 2 x (2.5 - t), now solve for t and then figure out d.
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.
1007306
Silly bequalming,\n\nIf the angle is 90 degrees, the kayaker has reached the opposite shore.  Fortunately these kayakers are tough, and never ever slow down.  This really comes down to a ratio, between the kayakers progress moment to moment, and the rate of the current.  I think we all know that this is possible, depending on the different rates.\n\nFirst, some definitions.\nT = time\nT0 = time at the beginning = 0\nTc = time at completion\nT* = some time between T0 and Tc\ntheta(T) = angle of kayaker, which is dependent on time\n\nand notation:\n\nintegral(T*..0) [f] = integral of function f, with upper bound T* and lower bound 0.\n\nlim(T*->Tc) is the limit as T* goes to Tc\n\nLets pick a hypothetical point half way across (time = T*).  The kayaker's distance from the opposite shore is the total distance minus his progress (cos(theta)P), his position downstream is the total effect of the current (W*T) minus what the kayaker has made up (sin(theta)P)..\n\nD - integral(T*..0) [cos(theta(T*))P] = distance to travel\nWT* - integral(T*..0)[sin(theta(T*))P] = distance downstream\n\nNow we can form a function of time to describe the angle theta(T) at time T*\n\ntan(theta(T*)) = opp/adj = \n    D - integral(T*..0) [cos(theta(T*))P] / \n    WT* - integral(T*..0)[sin(theta(T*))P] = \n\nsimplifying:\n   tan(theta(T*) = \n     [D - sin(theta(T*))P] / [WT* - (-cos(theta(T*)P - P)]\n\nso \n\ntheta(T*) = \n   arctan{ [D - sin(theta(T*))P] / [WT* - (-cos(theta(T*)P - P)] }\n\nugly, but manageable.\n\nSo now to the real questions, how long to cross?  The cross is complete when their distance from the opposite shore is zero, i.e.\nwe know that \n\nD - integral(Tc..0) [cos(theta(Tc))P] = 0\nor\nintegral(Tc..0) [cos(theta(Tc))P] = D\nor\n[sin(theta(Tc))P] = D\n\nWe just need to solve for Tc, so we try a limit\n\nlim(T*->Tc)[sin(theta(T*))P] = D\n\nsub in our function for theta(T*) to get\n\nlim(T*->Tc)[sin(arctan{ [D - sin(theta(T*))P] / \n         [WT* - (-cos(theta(T*)P - P)] })P] = D\n\nNumercially i would start with small values of T, and increase them until I get T value for which this is true.  That would require values of W, D and P.\n\nOk, I've wasted an hour.  the rest is up to you.  no guarantees on the quality, by the way.  check my math.
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.
981265
Call the speed the kayaker would make in still water x (mph).\n\nWhen he paddled with the current, the actual speed of the kayak was x+6.\n\nWhen he paddled against the current, the actual speed of the kayak was x-6.\n\nSince the lengths of both trips are the same, we have:\n2*(x+6) = 3*(x-6) <-> 2x + 12 = 3x - 18 <-> x = 30.\n\nHe would manage 30mph in still water (wow, what a beast :)))
32272
well Bus B travels less distance in the same time, meaning the velocity is greater for bus B and since velocity is the slope on a distance time graph, the slope of the graph would be steeper for bus B than for bus A.\n\nif you need a numerical answer, the slope for bus A would be:\nv= d/t= 300/12= 25 m/s\nand the slope for bus B would be:\nv= d/t= 200/12= 16.67 m/s
331284
boat A because the other one sunk before the finish
977084
a.) newtons\nb.) km/h\nc.) f(x) = 0\nd.) x = f(0) + c
888670
First, draw a picture. Let C be Chris, B be Blaire, and A be the ball at some point. Thats a right angled triangle, where side BA is how far the ball has dropped; CB = 100 feet, and CA = the fishing line.\nNow, we need to know how fast the ball is dropping. Since its distance is 16t^2, its velocity is 32t (differentiating). So side BA is changing at 32t feet per second at any time.\nWe want to know the rate at which side CA is changing. But we know that CA^2 = CB^2 + AB^2, so CA = sqrt(CB^2 + AB^2) = sqrt(100^2 + AB^2).\nNow, we need to differentiate both sides with respect to time.\nThus, d/dt(CA) = d/dt(sqrt(100^2 + AB^2)).\nWe need the chain rule for the RHS: sqrt(something) differentiates to 0.5(something)^-0.5, then we multiply by the inside differentiated. AB^2 differentiated (with respect to t) is 2*AB *d/dt (AB) = 2*AB*32t.  So the overall equation becomes:\nd/dt(CA) = 0.5(100^2 + AB^2)^-0.5 * 2 * AB * 32t.\nNow, we know t = 3 and AB = 16*3^2, so substitute that in and you'll get the answer of 78.9 feet/second.\n\nI'm likely to have made a mistake, but hopefully that gives you the idea of what to do.
125621
B) Increase\n\nThe difference in the two objects' velocities would be constant since that are both accelerating at the same rate, but the first object already had a higher initial velocity (from Object B' point of view).\n\nThe equation relating the distance traveled with respect to initial velocity, accelertion and time is,\nd = v_i * t + 1/2 a*t^2\nSo for both objects, after B is released the 1/2 a*t^2 part of the equation is the same, so we can ignore that part and focus on the v_i * t part.\nIf we start t=0 at the moment object B is released, object A's velocity is some non-zero number while B has no initial velocity.  So as time goes on, the distance between the 2 objects inceases.
892284
1 in water = 248.8 N/m2= 0.0361 lb/in2 = 25.4 kg/m2 = 0.0739 in mercury = 1.8664 Torr = 2.4884 millibars\n\nInches W.G. are generally used when working with air handling systems, where pressure differences are normally quite small.