Uniform motion problems?
476201
1= 8mph\n2= 6 hrs walked 4hrs run\n3= 60 miles to the farm
372422
If each bug moves towards its target at constant rate r inches per second, then I think "infinitesimally speaking" because each target is moving at right angles to the direction of its seeker that the distance between seeker and target will lessen at the same constant rate of r inches per second.\n\nThus it will take 10 inches / (r inches per second) = 10/r seconds until all four meet in the center, during which time each will have traveled 10/r * r = 10 inches.
985616
All are true. For the last one,\nsuppose acceleration = a.\nThen the velocity is at + c\nand the position is at^2/2 + ct + d,\nwhere c and d are constants.
974284
I certainly hope you have visited the link below for visual and mathematical insights into this problem.\n\nHowever, the only materials that come to mind which lend themselves to a physical model of this transformation are:\n\n o thin-but-rigid springs or "lead" wire(s) as supports/frames\n o plastic wrap as a surface substrate\n o flexible, rubberized, quick-drying paint as a protective layer\n\nI regret that my imagination is limited with respect to this problem, but good luck with your project!
986259
(a) Displacement is given by the equation:\n\nx = 6 cos[3π(t+(1/9))], so at t = 2 sec\n\nx = 6 cos[3π(2+(1/9))] = 3 m\n\n(b) Velocity is v = dx/dt = -18πsin[3π(t+(1/9))], so at t = 2 sec\n\nv = -18πsin[3π(2+(1/9))] = -48.97 m/s (in the negative x direction)\n\n(c) Acceleration is a = dv/dt = -54π^2cos[3π(t+(1/9))]\n\nso at t = 2 sec, a =  -266.48 m/s^2\n\n(d) Phase of the motion is θ = 3π(t+(1/9)), for t = 2\n\nθ = 19.89 radians\n\n(e) The frequency = ω/2π = 3π/2π = 1.5 revolutions per sec\n\n(f) Period = 2π/ω = 2/3 sec
311411
Math model of problem:\n\nD2S/Dt^2 = 2.1  \n\nIntegrating:\n\nDS/Dt = 2.1*t + c1\n\nIntegrating again:\n\nS = 2.1*t^2/2 + c1*t + c2\n\nFrom the statement of the problem:\nWhen t=0 DS/Dt=0 and S=0\nTherefore c1 and c2 both are zero\n\nSo S = 1.05*t^2\nSo t = square root of S/1.05\nThat is t = square root of 70/1.05\nthus t = 8.165 secs.\n\nSo boat accelerates for 8.165 secs in travelling from rest over a distance of 70 metres
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
497479
When calculating with error, remember to MAXIMIZE error wherever possible. For example, since V1 and V2 are in the denominator of their respective terms, error should be subtracted to maximize the overall terms. Check me on this, but I found an average velocity of (19.5 ± 0.5) m/s.
217864
J=Iw, w=v/r (v = tangental velocity) I is diffrent for each object, or group of objects.  so J=Iv/r I=Jr/v\n\nI for uniform objects does = sigma MR^2.
1075715
Here are two sites that look into the problem:\n\nhttp://www.kw.igs.net/~jackord/bp/f6.html\nhttp://math.fullerton.edu/mathews/N310/projects/e6.htm
267038
You might try reading Whitehead and Russell's Principia Mathematica.  I can't promise it will help, as I haven't read it, but I've been told that this issue is addressed.
901641
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
788625
This is a momentum question.  If the end result is comming to a halt, then then each vehicles' momentum was equal and opposite.  Momentun (M) = mass (m) * velocity (v),  therefore you are left needing to solve the equation m1*v1 = m2*v2...you can do the rest I hope....
500368
The links below have good info. The first briefly addresses your first question, and the second is a good brief history of the problem which provides some insight as to the difficulties encountered. I actually played around with this problem a little on an elementary level and it is interesting. Cheers.\n\nhttp://www.sciam.com/print_version.cfm?articleID=000DBA55-B698-1C71-9EB7809EC588F2D7\n\nhttp://www.geocities.com/fermatnow/flt/
617356
Something about beam misalignment\n\nhttp://www.reference.com/search?db=web&q=jogged%20beam
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs