How to derive 3rd equation of motion graphically?
454697
Which graph du you use? velocity verses time or velocity versus distance?
24383
here are some simple mechanics formulas dealing with velocity:\n\nv=d/t\na=v/t\nvf^2=vi^2+2ad\nd=vit + 1/2at^2\n\nwhere v is velocity, vi is initial velocity, vf is final velocity, a is acceleration, and t is time.  also the i and the f after the v are just subscripts, and ^2 just means sqared.
1084130
The projectile motion is the motion of a projector starting from origin (x=0 & y=0) with initial velocity (v) & inclination (a) where (a) is the angle between X-axis & (v) so the only accelerations affecting the projector are:\nx''=0 & y''=-g --------> (1) & (2)\n\nTo calculate velocity (x' & y'):\nfrom (1) & (2) we get:       \nx''=dx'/dt=0  \ntherfore: x'=constant=v cos(a) [from initial condition] --------> (3)\n \nfrom (3) we get: y''=dy'/dt=-g\ntherfore: y'=-gt+constant\n\nfrom initial condition:\nat t=0 ----> y'=v sin(a) & constant=0\ntherfore: y'=v sin(a)-gt ------> (4)\n\nThe parametric equation:\nfrom (3): dx/dt=v cos(a)\ntherfore: x=tv cos(a)+constant\n\nfrom initial condition:\nconstant=0\ntherfore: x=tv cos(a) -----> (5)\n\nfrom (4): dy/dt=v sin(a)-gt\ntherfore: y=tv sin(a)-0.5g(t^2)+constant\n\nfrom initial condition:\nt=0 therfore y=0 & constant=0\ntherfore y=tv sin(a)-0.5g(t^2) -----> (6)\n\nfrom (5): t=x/[v cos(a)]          sub in (6)\ntherfore\ny=[x tan(a)]-{ [g(x^2)]/[2(v^2)] }*[1+ tan^2 (a)] \nthis equation is called The Cartizian Equation\n\nTo find The Range ------> sub y=0 in The Cartizian Equation\ntherfor:\nR(range)=[2v sin(a) cos(a)]/g
1166133
Multiply both sides by R+r\nI(R+r) = E\nDistribute the I\nIR + Rr = E\nSubtract IR form each side\nRr = E - IR\nDivide both sides by R\nr = (E-IR)/R
84751
I presume you mean the third derivative of V(t,D) with respect to t, not D. The first derivative of V(t,D) with respect to t is acceleration a,\n \na=dV(t,D)/dt=-D/(t^2)\n\nThe second derivative of V(t,D) with respect to t is the boost B,\n\nB=da/dt=2D/(t^3)\n\nThe third derivative of V(t,D) with respect to t is\n\ndB/dt=-6D/(t^4)
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.
1001657
V= V(i) + AT Initial velocity plus acceleratin x Time\n\n\n   = V(i) + (F/m)T     Acceleration = Force/ Mass\n\nNote  use same system of units
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.
734266
Frequency, f = 440 Hz\nAmplitude, A = 2.24 mm or 0.00224 m\n\ny(t) = Asin(2πft), assuming that y(t=0)=0\n\nv(t) = dy(t)/dt = 2πfAcos(2πft)\n\na(t) = dv(t)/dt = -(2πf)²Asin(2πft)\n\nMax velocity: v = 2πfA = 2π(440)(0.00224) = 6.19 m/s\n\nMax acceleration: a = -(2πf)²A = -17120.37 m/s^2
861187
0.4*(100kg*g+F*sin(x)) = F*cos(x). ignoring any rotational component
897617
This is an equation with separated variables. Rewrite as \n\ndu/u^3 = (t^3-t)dt \n\nand integrate (find antideritive) of each side separately.
134530
First, generate 3 equations by inserting the values of x and y for each of the coordinates.\n\nEqn 1. For (-1,1): 1=a*-1^2+b*-1+c yields: 1=a-b+c\nEqn 2. For (1,3): 3=a*1^2+b*1+c yields: 3=a+b+c\nEqn 3. For (3,2): 2=a*3^2+b*3+c yields: 2=9a+3b+c\n\nStep 1: solve Eqn 1. for a:  a=1+b-c\nStep 2: insert a from Step 1 into Eqn 2.: 3=(1+b-c)+b+c ==> 2=2b ==> b=1\nStep 3: insert b=1 back into Step 1 and re-solve for a: a=1+1-c ==> a=2-c\nStep 4: insert the results from Step 2 for b and Step 3 for a into Eqn 3 to find c.\n  2=9*(2-c)+3*1+c ==> 2=18-9c+3+c ==> 19=8c ==> c=19/8\nStep 5. Insert the results from Step 4 and Step 2 into Eqn 1 to find a.\n  1=a-1+19/8 ==> a=-3/8\n\nTherefore:\n  a=-3/8\n  b=1\n  c=19/8\n\nStep 6: Verify for -3/8*x^2 + 1*x +19/8\n (-1,1): 1=-3/8*(-1^2)+1*-1+19/8\n (1,3): 3=-3/8*(1^2)+1*1+19/8\n (3,2): 2=-3/8*(3^2)+1*3+19/8
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
737967
Draw a free body diagram as shown below: \n\n                [m1]------> T<---[m2]------->89N\n              <--------            <---------\n                f1 =19.6N        f2=29.4N\n\n\nFirst solve for f1 and f2:\n\nf=ukR is the general formula where f is the force of friction, uk is the coefficient of friction, and R is the force normal to the surface. In this problem R is the weight of the block=mg. Now substitute known values\n\nf1=0.2*10*9.8\n   =19.6N to the left\nf2=0.1*30*9.8\n   =29.4N to the left\n\nLooking at the free body diagram:\n\nfor m1:\n\nnet force, f1=T-19.6\n                      \nFrom the formula, f=ma  where f is the net force, m the mass, and a is the acceleration.:\n\nf1=10a\n\nThus,\n\nT-19.6=10a\n         a=(T-19.6)/10\n\nfor m2:\n\nNet force, f2=89-29.4-T\n                  =59.6-T\n\nSimilarly as for m1:\n\n59.6-T=30a\na=(59.6-T)/30\n\nNow a for m1 is also the same a as for m2, therefore:\n\n(T-19.6)10=(59.6-T)/30\nThe zero on each side cancels out. \n3T-58.8=59.6-T\n4T=59.6+58.8\n4T=118.4\n  T=29.6N in the direction shown in the above free body diagram.
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
1044611
F = Gm1m2/ r^2\nF = ma  \na = v/t\nv = d/t\na =d/t^2 \nF = m d/t^2  = Gm1m2/ d^2\nm  = (m1 + m2)\nF = (m1 + m2) d/t^2 = Gm1m2 /d^2\nd/m2t^2 + d/m1t^2 = G/d^2\nd/m2 + d/m1 = Gt^2/d^2\nSQRT{ (1/G) d^2 (d/m2 +d/m1)} = t
929849
solve three body problems