The velocity of a particle moving along the x-axis is given by v(t)= e^t +t e^(2t) . What is average velocity?
354866
The average value of a function is given by\n\n(1/(b-a)) S f(x) dx, evaluated from a to b (I'm using S as the integration symbol).\n\nSo for this question, the average will be given by\n\nVavg = (1/6-0) S e^t + te^(2t) dt\n= (1/6){e^t + [(e^2t)/2][t - (1/2)]}\n= (e^t)/6 + te^(2t)/12 - e^(2t)/24\n\nEvaluating from 0 to 6,\n= (e^6)/6 + 6e^(2(6))/12 - e^(2(6))/24 - {(e^0)/6 + 0e^(2(0))/12 - e^(2(0))/24}\n= (e^6)/6 + e^(12)/2 - e^(12)/24 - (1/6 - 1/24)\n= 74663\n\nI've been looking at your answers and the equation and either there's something missing in your equation or there's an answer missing somewhere.
900660
You must compute the slope at which the particle is moving at\nthat point.  Then the ratio of the absolute speed to the horizontal \nspeed is equal to the ratio of the hypotenuse of the slope to adjacent side.  And of course the slope is computed by the derivative's value at x=4.  So\n  y=x^1/2\n  dy/dx = 1/2(x^(-1/2)= 1/(2x^1/2)   \n  dy/dx=1/4    at x=4  So  the ratio of the  vertical speed to the horizontal speed is 1/4 and the proportional hypotenuse  in the direction of the particle(i.,e. the curve) is\n (1^2 + 4^2)^1/2 =  17^1/2\n\n   So    (17^(1/2))/4 = v/3cm/s   where v is the velocity in direction\nof the curve is\n\n       v=(3/4)(17^(1/2))\n\nStuart's approach is slightly different and fascinating to me\nbut also right.\n\n  Garipopki is also right but a bit work intensive I would say.\n\n Garipopki's answer is closer to mine and I'm backing off\nfrom supporting stuart. Work intensive or not, correctess\nsupercedes all.\n\n\nAt this point I'm responding to garypopki's comment that I'm\ncomputing the velocity along the curve and not the velocity\naway from thr origin. Good point. So the velocity away from\nthe origin is cos of the angle between the two directions times\nthe absolute velocity of the particle which is in the direction of the curve.\n    Angle of the curve at the point (4,2) is  α   where\n  tanα = 1/4,  so α=14d (degrees, this is physics so I won't use\nradians)\n\n   and the angle from the origin from which the particle is receding\n is  β where tanβ = 1/2,  so β=26d   \n\n   26d-14d=12d \n\n   So the component of velocity away from the origin is the absolute velocity times the cos of the angle between them.\n\n    Absolute velocity is  3.0923 so velocity from the origin\nis cos(12)3.0923 = \n    3.024cm/s  \n\nwhich I have to admit is\nawfully close to the 3.018 that you'd get from garypopki's\n(27/20)5^(1/2)
303805
I don't see how components are necessary.  \n\n"r" and "v" are just shorthand for r(t) and v(t), assuming we are looking at position (r) and velocity (v) as time varying "vectors".  And v(t) is just the time derivative of r(t) (by DEFINITION), that is, v(t) is DEFINED as r'(t).\n\nSo there isn't any actual CONTENT in expanding them into components. We are just choosing different ways to symbolize the quantities involved and there isn't anything to prove.
448267
the average orbital velocity of earth is 29.783 km/s.\n\nhttp://en.wikipedia.org/wiki/Earth
901648
put x=2 directly into the given equation to get value of x at t=2...the answer u get is the displacement\n\ndifferentiate the equation wrt time to get velocity equation and then differentiate the velocity equation to get acceleration equation\n\nin both cases put t=2 to get velocity and acceleration\n\nconvert the given equation in terms of sin\ncomparing the new equation with x=asin(wt+@)\nwhere a=amplitude\n          w=angular frequency\n          @=phase\n\nalso use equations w=2 x pi x f and T=1/f\nhere x denotes multiplication \nf=frequency \nT=period
466161
time average means sum of hte probability density over a certain time interval divided by the time interval. and it will be a function of position since the wave function is a function of position AND time.\nso u have W(x,t) where W is the wave function and x is position t is time. you integrate it with respect to time from a starting time to an ending time (not that hard, its just integral of e^-iwt, i think) and then divide by the time interval. and ya. i dunno.\n\nhope that helps. let meknow if u need more help. later
821342
THe VdW equation is P= nRT/(V-nb) - a(n/V)^2, just plug and chug.  But dont forget to switch your T into Kalvin
294130
You should be more specific with your question. A lot of people don't know what a v-t graph is (velocity-time).\n\nAnyway, it's a horizontal line. Velocity remains constant while time keeps moving forward.
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.
46701
v=u+at\nv^2-u^2=2as\ns=ut+1/2at^2\ns=disance traveled\nu=initial vel.\nv=final vel.\na=accelaration\nt=time
743381
Velocity(at time t) = Velocity (at time = 0) + Acceleration x Time\n\nV(t) = Vi + At\nV(t) = 32 ft/sec + (-32 ft/sec/sec)(t)\nV(1 sec) = 32 - 32 = 0 ft/sec\n\ny(t) = y(initial) + (V x t) + 1/2(A)(t^2)\ny(1 sec) = 0 + (32) +  (-16) = 16 feet\n\nTo hit ground, y = 0, so (V x t) = 1/2(A)(t^2)\n32t = 16t^2\n2 = t, t = 2 seconds\n\ny(max) occurs at zero velocity, which is at t = 1 sec, x = 16 feet
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
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB
931347
x =70m, t = 7s, Vf = 15m/s\na.)\nx = 1/2(Vi + Vf)t\nVi = initial velocity\nVf = final velocity\n\n70m = 1/2(Vi + 15m/s)(7s)\n10m/s = 1/2Vi + 7.5m/s\n2.5m/s = 1/2Vi\nVi = 5m/s\nb.)\nVf = Vo + at\n15m/s = 5m/s + 7s(a)\n10m/s = 7s(a)\na = 1.43m/s/s\nbasic kinematics\nyou can't use distance = rate x time because there is acceleration and the speed is not constant.
888567
a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)
763881
Take the x-axis as the line of action of the moving ball. and suppose the second ball moved at an angle theta wrt to the line of action with certain velocity that u need to find out. Now resolve the velocity components of both balls along x-axis and y-axis before and after collision. Now conserve the momentum along both x-axis and y-axis before and after collision. solve the equations and u will know all the unknowns.\n\nFor the second question find 'e' i.e. co-effecient of restitution and depending on the value it will be elastic or inelastic.
1158387
as you can see that velocity vector as a function of time is given \nv(t) = 5 t^2 - 200 t - 150;\nthe acceleration is given,\nso using the vector calculus (vector differentiation)\nwe differentiate the velocity vector,we have,\na(t) = 10 t - 200;\nso in the given problem the acceleration of the particle is given to be 100 m/s/s;\nthere fore equating both we have,\n10 t - 200 = 100;\nwhich gives us t = 30 s.\nto know the distance through which the particle traveled we have to find the displacement vector first.\nto get the displacement vector again we take the help of vector calculus this time vector integration\nv(t) = 5 t^2 - 200 t + 150;\nint [v(t)] = s(t) = 5 t^3 / 2  - 200 t^2 / 2 + 150t;\nand substituting t = 30s will give the answer as -40,500m;\nthe answer is negative indicating that the particle has travelled along the negative direction of the axis taken into consideration.