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formula of resultant displacement

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Question:WHAT IS THE FORMULA FOR DISPLACEMENT? I have to find the displacements for the following values. I have a graph of velocity versus time for motion with variable acceleration. All that is given is the time in seconds. I have to obtain the velocity, acceleration and displacement and here is what I have gotten so far. Maybe you guys can draw this table it would help u understand my question better I cant make a table but heres the closest I got. Time(s)Velocity (m/s)Acceleration (m/s^2)Displacement (m) 0.01.0 1.02.500.45? 3.03.750.15625? 5.03.500.25? 6.02.500.50? 9.03.751.125? 10.05.501.50? the question marks are for the displacemnt which i dont know how to get. DO THESE VALUES MAKE SENSE? Please give me formulas and a few examples as to how u got the answer for each value. Thanx!

Answers:Integrate the velocities using the trapezoid method to find the displacements: http://en.wikipedia.org/wiki/Trapezoidal_rule

Question:It seems as if I've forgotten this formula (we must have found it in my physics class though). I have a starting acceleration, a final acceleration, a displacement, but no time. I need to find an acceleration.. What's the equation?

Answers:I should think that if you have an initial and a final accel, and you need to find an accel, that you simply subtract a = a - a Now, this accel could be effected a number of ways: 1) a force (like gravity or friction) acting on a mass, a = F/m 2) a spring acting on a mass, a = kx/m 3) electromagnetic forces 4) etc. EDIT: I suspect that you meant initial and final VELOCITY, and a displacement. That's tricky. I don't know of such a formula. If there is one, I would think that it would have to be something like a = kv /x, where k is some dimensionless constant. v /x will give you the dimensions of acceleration. Or t = k(x/v) will yield seconds. Maybe t = x / v ? Try graphing velocity on the y axis and t on the x axis. (I know we don't know t, but some time has elapsed; don't use units.) If we assume UNIFORM acceleration, the line from (0 , v ) to (t , v ) will be a straight line. The area under the curve will be displacement. Maybe you can figure out how to make the area match your known displacement; that will give you t = t and you can go from there. That's all I've got. EDIT #2: I've been thinking about this, at the expense of some sleep. The graph described above i basically a rectangle with a right triangle on top (I assumed linear acceleration). The area is readily given as s = (v + v )t, so t = 2s / (v + v ) For uniform acceleration, a = v / t = (v - v )/t = (v - v )(v + v ) / 2s your formula! a < 0 if v < v Voila! FINAL EDIT: I was working another problem and I stumbled upon a more direct answer to your question. If you multiply the numerator terms in the eqn above, you get a = ( (v ) - (v ) ) / 2s which can be rearranged as (v ) = (v ) + 2as which is Torricelli's equation.

Question:What's the formula to get the total displacement? (a) Start from rest with constant accelera- tion of 2.45 m/s2 for 17.3 s; (b) Constant velocity for the next 0.895 min; (c) Constant negative acceleration of 8.86 m/s2 for 4.47 s. What was the total displacement x for the complete trip? Answer in units of m.

Answers:For (a) s1 = 0.5*a1*t1 let Vi be the velocity at the end of part (a); Vi = a1*t1 for (b) s2 = Vi*t2 = a1*t1*t2 for (c) s3 = Vi*t3 - 0.5*a3*t3 Total is s1 + s2 + s3

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Darla Dimple's Rampage of Displacement and Volume :A black hole, according to the general theory of relativity, is a region of space from which nothing, including light, can escape. It is the result of the deformation of spacetime caused by a very compact mass. Around a black hole there is an undetectable surface which marks the point of no return, called an event horizon. It is called "black" because it absorbs all the light that hits it, reflecting nothing, just like a perfect black body in thermodynamics.[1] Under the theory of quantum mechanics, black holes possess a temperature and emit Hawking radiation, but for black holes of stellar mass or larger this temperature is much lower than that of the cosmic background radiation. Despite its invisible interior, a black hole can be observed through its interaction with other matter. A black hole can be inferred by tracking the movement of a group of stars that orbit a region in space. Alternatively, when gas falls into a stellar black hole from a companion star, the gas spirals inward, heating to very high temperatures and emitting large amounts of radiation that can be detected from earthbound and Earth-orbiting telescopes. Astronomers have identified numerous stellar black hole candidates, and have also found evidence of supermassive black holes at the center of galaxies. In 1998, astronomers found compelling evidence that a supermassive black hole of more than 2 million solar masses is located near the Sagittarius A* region in the center of the Milky Way galaxy, and ...

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