In a position function, when is the particle stopped?
357903
When V(t)=0 the particle stops.\nIt can have a(t)=0, but still moving (constant speed)
457512
It stopes before comes down
788532
Is it x*(e-2) or x^(e-2)?\n\nI don't think it matters anyway, because the bottom fluctuates between 3 and 5 while the top increases indefinitely, i.e. approaches "infinity" if you like.  So there's no limit.
856832
(a) The particle is held stationary between the plates by balancing the weight of the particle with the force due to the electric field, which is expressed as the charge times the electric field (F = q*E). Since the top plate is negative and the bottom plate is positive, the sphere must be positively charged (deficient of electrons) in order for the electric field to balance the particle's weight.\n\n(b) The balanced position means that the net force on the sphere is zero so that the force due to the applied field balances the weight of the sphere\n\nq*E = m*g\n\nFor the parallel plate geomtery, the electric field is expressed as\n\nE = V/d\n\nwhere V is the applied potential and d is the separation. Solving for q in terms of V and d yields\n\nq = m*g*d/V\n\n(c) The charge deficit, in this case, is then\n\nq/e\n\nwhere e = 1.602 x 10^-19 C is the value of the elementary charge.
140359
That sounds like a "Robert's Rule" question.  Try http://www.robertsrules.com/.  They have a forum specifically for this type question.\n\nMy understanding would be that once a motion carries its permanent unless and until another motion to reverse or suspend was passed.
863361
x = -11 + 12t + 3t^2 + 4t^3 - 4t^4\n\nTaking the 1st derivative gives the velocity.\nv = dx/dt = 12 + 6t + 12t^2 - 16t^3\n\nTaking the 2nd derivative gives the acceleration.\na = dv/dt = 6 + 24t - 48t^2\n\nTo find when the acceleration is maximum, we have to take the derivative of the acceleration equation and set it equal to zero.\n\nSo, da/dt = 24 - 96t = 0.\nThus, t = 1/4.\n\nThe location of the particle at this time is then found by substituting t = 1/4 into the location equation.\n\nx = -11 + 12(1/4) + 3(1/4)^2 + 4(1/4)^3 - 4(1/4)^4\n= -11 + 3 + 3/16 + 1/16 - 1/64\n= -497 / 64\n\nThe velocity at this time is found by substituting t = 1/4 into the velocity equation.\n\nv = 12 + 6(1/4) + 12(1/4)^2 - 16(1/4)^3\n= 12 + 3/2 + 3/4 - 1/4\n= 14
425296
Instantaneous rate of change of a function (7 words),\n\nOR,\n\nSlope of the tangent line (5 words). :P
991201
One of two things will happen:\n\n1) The so-called unmovable object will be moved.\n2) The so-called unstoppable oject will be stopped.\n\nThis existance of an unstoppable object precludes the existance of an unmovable one, and vice versa. Of course, it's likely that neither one exists anyway so the question is pointless.
52711
shift or change\nfixed or immobile\nvector
415527
First of all, the sum is incompleted.  Is it for maximum height, or for range?  Secondly, let us assume you are asking for maximum height.  Then differentiating the given equation of motion, d = -5(t+1)^2 + 258 with respect to 't', we get dt/dt = -10(t+1).  At maximum height, the velocity is zero. \nTherefore, -10(t+1)=0, or t = -1.  This is absurd time can not become negative once the motion is initiated.  Check up there is something wrong in the problem.
723707
s'(t) =(t-3)(-2t^{-3}) + t^{-2}\n=( -2t+6 +t)/ t^{-3}\n=(6-t) / t^{-3}\nso the velocity is 0 when the top =0\ni.e 6-t-0\nt=6\ns
502439
An object in motion will remain in motion until acted upon by an external force. An object at rest will remain at rest until acted upon by an external force.
796769
The point P moves in such a way that its Cartesian coordinates with respect to an origin O at time t is given by x= e^(-t) , y= 2te^(-t). the distance OP is denoted by r and the angle between OP and the x-axis by θ.\n(a)Using an analytical method, find the value of t when r^2 is stationary. Determine the nature of the stationary point.\n\nr² = Δx² + Δy²\n= (e^(-t) - 0)² + (2te^(-t) - 0)²\n= e^(-2t) + 4t²e^(-2t)\n= e^(-2t)(1 + 4t²)\nd(r²)/dt = -2e^(-2t)(1 + 4t²) +8te^(-2t)\n= 2e^(-2t)(1 - 4t + 4t²)\n= 0 for stationary points\nSo 1 - 4t + 4t² = 0 as e^(-2t) > 0 for all t\nie (1 - 2t)² = 0, ie t = ½, ½\nWhen t <½, d(r²)/dt > 0 and when t >½, d(r²)/dt > 0\nThus when t = ½ there is a stationary inflexion\n\n(b) Find, in terms of t, the rate of change of θ with respect to\n\ntanθ = Δy/Δx\n= 2te^(-t)/e^(-t)\n=2t\n\nsec²θ * dθ/dt = 2\nie\n\ndθ/dt = 2/sec²θ\n= 2/(1 + tan²θ)\n= 2/(1 + 4t²)\n\nCan you show the non analytical method for part (a) too!\n\nx = e^(-t) → -t = lnx ie t = -lnx\n\ny = 2te^(-t) \n= -2xlnx\n\nx: .. 0 ...... 0.25 ... 0.5 .... 0.75 . 1.0 .. 1.25 .. 1.5\ny: undef . -0.15 . -0.36 . -0.22 ... 0 .... 0.28 . 0.61\n... (→0)\n\nLim x → 0 (x lnx) = 0 and OP increases with x \n(limiting min value of OP = 0)\n\nMin value of xlnx occurs when lnx + 1 = 0 ie x = -1/e and at this point OP² will show its inflexion as it cannot be a minimum and nor can it be a maximum so all it can do is slow its rate of growth.
861527
Stop, stop, stop. If it was f(x)=2logx, f(x) cannot be = to (1/log2)*logx, as  1/log2 is not = 2.\nDon’t hurry, put your question correctly with necessary details and don’t save on ()-s in formulas.
240568
Conservation of momentum
121355
well if your taking the derivative, then there should be no h's left over at the end, since they go to zero with the limit.  The derivative of that function is [5(5x+2)+25x]/(5x+2)^2.  I'll leave it to you to simplify.
142433
you havn't closed "void determinecost (...)" function with curly braces " } "\n\nin c++, you have to close any block of code with } that has been opened with {