what 2 jobs requires accurate calculations of velocity?
139933
Probably Rocket scientist, and balistics expert.  Other ones would be physicist or physical scientist.  Oh and if your talking equations, well anything with force.
923403
Start off by drawing the picture. You have a right triangle, with a hypotenuse of 25, a base (b), and height (h). \n\nDon't get jumbled down by numbers yet, ignore them. Visualize what you are calculating.\n\nVelocity / speed is the derivitive of position. (v = delta x / delta t). So all you need to do is find an eqation relating the geometry using known and unknown quantities and then differentiate with respect to time. Sounds confusing but all you have to do is....\n\nYou know, a^2+b^2 = c^2 for a right triangle. In our problem,\n\nb^2+h^2 = 25^2\n\nThis is our position equation. Now differentiate with respect to time.  DO NOT forget the CHAIN RULE! <------ VERY important\n\n2*b*db/dt +2*h*dh/dt = 0.\n\nFrom here just use algebra; where,\n\nb=7 ft, and db/dt = 2 ft/sc.\n\nWhich leaves two equations and two unknowns.\n\nThe important thing to remember is to take this step by step and try to visualize what you are calculating. Also, looking at the units of the given quatitities can give you some idea of what you will be required to do. \n\nAlso, just remeber that speed / velocity is the derivative of position with respect to time (the units reflect this).
478314
If you have a numerical approximation, your integration can not be more accurate than approximate, right?  And that may be close enough.  Integration is often the process of finding the area (integrating) under a curve, although the "curve" may be linear.  If you can draw your curve on a graph where for every value of x there is a single value of y, you can divide the graph into vertical slices where the bottom of each slice is the x axis and the top of each slice is a short horizontal line at the "eyeball" average height of the y axis in the slice.  If you divide the x axis into say five slices you then calculate the area in the five slices (A1 = x1 x y1, A2 = etc.).  You can increase the accuracy by using say 10 or 20 slices.  Of course calculus uses an infinite number of slices rather than approximation.\n\nAs a simple example, you can plot a "curve" of velocity vs time, where the velocity is Y and time is X.   In case one, velocity is constant (60 mph) and time increases to say 5 hours.  Y is a straight horizontal line at Y = 60 MPH from X = 0 to X = 5 Hr.  The "area" under the curve is a rectangle; 60 MPH x 5 Hr = 300 Miles.  So the integration represents distance traveled.\n\nIf in case two velocity increased linearly from Y = 0 to Y = 60 as time increased from X = 0 to X = 5, the area under the curve would be a triangle and can be calculated as 1/2 base time height or 1/2 (5 - 0) x (60 - 0) = 150 (miles).\n\nWhen the function is a real curve, you still need to estimate the area under the curve (with slices?) to integrate it.  You could have used slices in the above two examples to get an approximate answer but calculating area of a rectangle or triangle was easier and more accurate.  Good luck.
769161
Its a pain to do this with X and Y components, but there is an easier way....\n\nRange Formula - used when start and end heights are the same\nQ = Theta\nd = ((Vo^2)(sin 2Q))/g\n\n301.5 = ((Vo^2)(sin 2(25)))/9.81\n\nVo = 62.13 m/s\n\nNow that u have initial velocity, find the y component\n\nVo Sin Q\n\n62.13 sin 25 = 26.26 m/s\n\nThen No Time Formula\n\nVf^2 = Vo^2 + 2gd\n\n0 = (26.26^2) + 2(-9.81)d\n\nd = 35.15 m\n\nHope this helps
428624
You need to be a bit more specific about which quantities you are trying to determine.  "Flow" is not a quantity to be calculated.
844762
There are jthree  formulae cncerning velocity.These are v=u+at , V^2=u^2+2as ,and s=ut +(1/2) at^2 ,where u= initial velocity,v=final velocity ,a=accelaration.,s= distance travelled and t= time.Now how to decide which formula is to be used?in the problem,just see what they want you to find out and note down what is are already given in the problem, if they give you u,a and t and ask you to find out v,apply formula no.1.But if they gdonot give you time but give the distance or displacement and asks you to find v,donot hesitate to apply formula no,2 and so on.if you have any difficulty to understand.please do not hesitate to write to me through yahoo answersand i shall be glad to assist you
927051
The given final speed doesn't seem to compute. I get 14.8 m/s. \nMa*Va + Mb*Vb = (Ma+Mb)*Vf \nis the formula that applies to show conservation of momentum in a completely inelastic collision, which is what you have. The only explanation I could suspect is some cargo was spilled. \n\nTo continue the assignment - kinetic energy is given by\n(1/2)*m*v^2\nThere is no requirement that all the original kinetic energy be measurable as kinetic energy after a completely inelastic collision. But you should be able to point to where the missing energy went. Some of the energy probably went into cart damage. Perhaps the cargo slid around.
984947
W=F*L – work done by F=145N on distance L=7.2m,\nU=k*m*g*cos(a)*L, work wasted on friction, where k*m*g*cos(a) is force of friction, while\nk=0.31, a=11.7deg, g=9.8m/s^2, m=10.3kg.\nY=m*g*L*sin(a) work done to hoist m to height of L*sin(a) meters.\nE0=m*v0^2/2 initial kinetic energy of m, v0=1.3m/s \nNow energy balance: E0+W=U+Y+E1, E1 is final kinetic energy of m E1=m*V^2/2, V is final speed of m. Thus kinetic energy change dE=E1-E0=W-U-Y.\nAs E1=E0+W-U-Y=m*V^2/2, then V=sqrt[2*(E0+W-U-Y)/m], \nI hope you cope with numeric calculations, all are SI units
898274
Yes, you can do it, but it will require iterations.  \n\n1) Convert all units to standard engineering units (feet, pounds, seconds).\n\n2) Set up the mechanical energy equation:\n(p2 - p1)/gamma = z2 - z1 + (f*L*v^2)/(2*D*g)\nwhere p2 -p1 is the pressure across the pipe (20 psi, but convert to lb/ft^2), z2 - z1 is the rise in the pipe (if any), gamma is the specific weight of water (62.4 lb/ft^3), L is the length of the pipe (50 ft), D is the I.D. of the pipe (ft), and g is the acceleration of gravity.  \n\n3) Now you have a problem: v is the velocity of the water, and f is the friction factor, which depends on v.  So, you guess a value for v, and calculate f from it.  You look up f on a chart (it's called a "Moody chart").  But you'll need the Reynolds number and relative roughness of the pipe surface to do that.  Then you try another v, find f, etc.  Or, you can try this:\n\nhttp://www.fluidmech.net/jscalc/vmd03.htm\n\n4) Just keep plugging in different velocities until the head loss (right side of the equation) equals the left side.  Then convert to volumetric flow rate.\n\nHey, I get 14.2 ft/s, 0.0774 ft^3/s = 34.7 gpm
889350
to compute for the maximum height, the vertical component of the velocity is needed.\nSo Vy = 49sin30\n      Vy = 24.5m/s\n\nto solve for the max. height use the formula\nVy = sqrt(2*g*h)\nh=(V^2)/(2*g) = [(24.5m/s)^2]/(2*9.81m/s2)\nhmax = 30.594 m
69843
work = force times distance\nW=Fs
671105
almost exactly, assuming someone on earth is doing the measurement. someone moving very fast would get a different result.
82068
surveying does use it alot and so does desgining and architecture. Engineering, construction and factories and industries all use it. There are many jobs out there were trig can help. I hated it but it shocked me how much I have used it now.
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.
899180
In the direction of the velocity vector. That's the best that can be done, given the limited information you have provided.
573522
kinetic energy is 1/2 mv² so it depends on mass and velocity
744597
You have to keep your two vectors straight....you have one for horizontal motion and one for vertical motion.  To find the velocity on the desk (horizontal), you will need to know the time it took to land.  You can use the formula for gravity: d=1/2gt^2.\nd=1.33\ng=9.8\nt is the variable\n\n1.33=0.5(9.8)t^2\nt=square root of 1.33/4.9\n\nOK.  Now we know the time it took to land.  You can now find its horizontal velocity.  Velocity=Distance divided by time.\n\nVelocity=.25/square root of 1.33/4.9\nVelocity is approximately equal to 0.4799m/s.\nMore exact:0.479857435\nHope that helps.  Let me know if you're still confused.