i need two examples of rectilinear motion.?
91916
a car trvelling in a straight line\na beam of light
855447
if it is equations of motion then \nv=u+ft\ns=ut+1/2ft^2\nv^2=u^2+2fs\nhope this is what you wanted
74552
these are data from end fixing of the structure you should introduce to solve the system of equations, otherwise you would get infinite solutions or (rigide body motion)-Good Luck
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
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!
398633
Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
905757
This is uniform motion, if it is going in a straight line at constant speed.\n\nIf there are no forces acting upon a body, and therefore no acceleration (remember F=ma) or change in direction, nor any change in angular momentum, the motion is uniform. If it were changing direction, accelerating or going round in circles then it is non uniform.
895462
Above answers are right on. Here are some examples.\nVelocity is vectoral. It has both speed and direction implied.\nWhereas temperature is purley scalar, only quantity implied.\nOther examples of vectoral quantities are acceleration ,\nmomentum and force. Other scalar examples are energy,\nmass,volume and area.
268070
Check out this site. It gives you a few examples. It was helpful for my cousin when she was leaning how to do them.\nhttp://www.purplemath.com/modules/solvelin.htm
822161
1. when b = 0\n2. only if it has zero slope\n3. it must have a y-intercept. it can have no x-intercept (for example y=x²+1)
268125
just investigate for example ODE y"=-y and then add to the left part increasingly small non-linearity +c*y^2 and find solution of the first order of c - you will see how different modes interact with each other.
771568
Basically you have the right idea. Use momentum (mass x velocity) for the motion and the sudden change in speed and/or direction (acceleration) and you get force x time = kinetic energy trying to rip you apart or squash you like a bug.
12859
The Shaum's Outline Series book will show you how to do the problems.\n\nhttp://books.mcgraw-hill.com/search.php?keyword=fluid+mechanics&template=&subjectarea=ALL&search=Go
834761
Lets call the distance of the first part S1 and the total distance S2, and the corresponding times T1 and T2.  We know:\nS1 = 40\nS2 = 100\nT2 = 9.85\n\nIf V is the maximum velocity we know S1 = 0.5VT1 (const acc.)\nFor the second part we know S2-S1 = V(T2-T1) = VT2 - VT1\n\nVT1 = 2S1 so S2-S1 = VT2 - 2S1 or V = (S2+S1)/T2 = 140/9.85\n\nSo a) - V = 14.21 m/s\n\nWe know V^2 = 2AS1 where A is acceleration so A = V^2./2S1\nor A = 14.21^2 /(2x40)\n\nSo b) A = 2.525 m/s^2\n\nWe know V = AT1 so T1 = V/A = 14.21/2.525 = 5.62 s\n\nSo c) the two parts are 5.62s and 4.23s
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!
837607
You can find the math related to this topic in "rotational motion or Circular motion" in any physics text book.\n\nYou need to search a good high standars book  for it..\n\nIf it contains above mentioned chapters, you will find some questions related to it..\n\nHope this helps you..\nGood Luck!
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj