related rates problem?
923704
Draw a diagram to represent the positions of the two ships at time t.  This will be a right triangle.  Side A will be the position of ship A and side B will be the position of ship B.  The distance between the ships will be the hypotenuse.  Let's call this S.  By the Pythagorean theorem,\n     S^2 = A^2 + B^2\nDifferentiate with respect to time, t.:\n\n2 S dS/dt = 2 A dA/dt + 2 B dB/dt\n Simplify by dividing the entire equation by 2\n\nS dS/dt = A dA/dt + B dB/dt\nSolve for dS/dt:\n\ndS/dt = 1/S (A dA/dt + B dB/dt)\n\ndA/dt is constant at 24 knots\ndB/dt is constant at 18 knots\n\nAt 5 PM,    A = 30 + 24(5) = 150\n                  B = 18(5) = 90\n                  S = SQRT( A^2 + B^2) = SQRT(150^2 + 90^2)\n                     = 174.9\nTherefore dS/dt = (1/174.9)(150(24) + 90(18)) = 29.8 knots.\n\nIt's just a coincidence that spoonish's answer is about the same as mine.  Only differential calculus can solve this type of problem.
371210
3 * 10^8 m/s (approx.- this is used for the purpose of calculations)\nthe exact value is hard to calculate but the closest approximation is 299,792,458 m/s
49314
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
965285
rate b = rate a / sqrt (mwtb/mwta)\n          = [rate a * sqrt mwta]/sqrt mwtb\n          = 0.0235 * sqrt64 / sqrt 352\n          = 0.188/18.7624 = 0.01 = 10 x 10^-3...aprrox the same answer\n she got that 9.88 by taking m wt UF6  as 362....lol\nplug it in to see...\nbtw, your algebra was wrong....
794278
The distance between them is calculated by:\n\nD = √((50t)^2 + (60t)^2) = √(6100t^2) = 10√(61)t\n\nD' = 10√(61) (a constant rate of change)\n\nSo I agree with you!
596087
This is very straightforward. Which step are you having trouble with? The same goes for your other question.
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
232038
Distance = rate x time\n\nif you're going 8 feet per second for 4 seconds, you've gone 32 feet...
732600
SRY!! Matt is going to a basketball game out of town. If he could travel 75mph, He would arrive one hour early. If he could travel 45mph he arrives one hour too Late. How far is Matt's trip\n\ndis=75*(time-1)\ndis=45*(time+1)\n\nthen\ndis= 225mile,    time=4hour
93876
A quick guess without having reviewed your math, try adding g = 9.8 m/sec^2 to your answer and see if that is the answer that is needed.
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
1007024
distance equals rate times time\nThat is,\nd = rt\nd = 117 and t = 2.25\nSo we have\n117 = r(2.25) or\nr = 117/2.25\nr = 52 mph
194533
Let the second man's rate be x.  We are given the first man's rate is x+0.5.  Finally, let y be the distance the first man can walk in 5 hours.  We know that this is the same distance the second man can walk in 6 hours.\n\nSo you get two equations, and two unknows.  We know the first man takes 5 hours to walk the distance y:\n\ny/(x+0.5) = 5\n\nWe know the second man takes 6 hours:\n\ny/x=6\n\nSo solve for x, and that gives you the second man's rate.  The first man's rate will be x+0.5
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
735022
There will be a term proportional to distance added to a constant delay term, so the form will be t=kD + c. Solving for k and c:\nt1=.152=100k + c\nt2=.229=170k + c\nt2-t1=.077=70k\nThen k=.0011\nt1=.152=.11 + c --> c=.042\nt2=.229=.187 + c --> c=.042\nResult is\nt=.0011*D + .042
425296
Instantaneous rate of change of a function (7 words),\n\nOR,\n\nSlope of the tangent line (5 words). :P