displacement vs time ; velocity vs time ;  and acc vs time graphs are quite common in kinematics.\n\nBut I am looking for graphs such as acc vs velocity and velocity vs displacement\n\nCan anyone tell me where can i find thoses graphs?
199785
what you are looking for is called Phase Space or Phase Portrait in analytical mechanics.\n\njust an example:\nhttp://www.cds.caltech.edu/~shawn/LCS-tutorial/motivation.html
546189
All your answers are correct. Here is my take:\nx(t) = 1 - 10 cos t*pi. Therefore v(t) = 10 pi sin t*pi and a(t) = 10 pi^2 cos t*pi, are the velocity and acceleration, which are the first and second derivatives of x(t). \nv(t) is maximum in one direction when t*pi = pi/2 + 2*pi*k, k=0,1,2,..\nThis leads to  t= .5 s,  2.5s, 4.5s and a speed of 10 pi = 31.42 cm/sec. But v(t) is maximum in the *other* direction when t*pi = 3*pi/2 + pi*k, k=0,1,2,.. leading to t= 1.5s, 3.5s,....\nIn both cases when the sine in the v(t) is maximal, the cosine in a(t) is zero.
504239
Ok, iam assuming you know the equations of motion: \nv= u + at\ns=ut + 0.5at^2\nv^2=u^2 + 2as\ns=((u+v)/2)t\n\nWe need to break this motion up into seperate sections.\n\nNote however that all the units in this question are in SI units, so they are consistent with each other. You don't have kilometres which you would need to convert into metres for example, before using these equations. In that case I won't put in the units by the numbers in my calculations everytime, just to make it look clearer. But they are there and they are consistent.\n\n\nFirst section is from rest (u=0) accelerating at 2.6m/s^2 for 18s. We will find the displacement travelled here.\n\nWe know u(initial velocity)=0, we know a(acceleration)= 2.6 and t=18 and we want s(displacement)\n\nthe equation connecting these four is s=ut + 0.5at^2 \n(we don't need v and this equation does not have v)\n\nso s = (0 x 18)  +  (0.5 x 2.6 x 18^2)\n       = 421.2m << note this displacement, we're going to use it at the end\n\n\nThe next section, we have the automobile continuing its motion at a constant velocity that it has gained after accelerating at 2.6m/s^2 for 18s.\n\nHere, we need to find what this constant velocity is, which is easy. \n\nSince the body is accelerating at 2.6 metres per second every second it is gaining a velocity of 2.6 metres per second, every second.\n\nSo the velocity it has gained after it has accelerated for 18s is just \n2.6m/s^2 multiplied by 18s \n= 46.8m/s^1\n\nIt's important you notice that I am using 18s because this is how much time it has accelerated for to reach that constant velocity; it stopped accelerating after 18s. 15s is the time it travels at that speed it has gained after accelerating for 18s.\n\nNow this velocity is constant so every second it travels a displacement of 46.8m, the displacement it travels here is just 46.8m/s^1 multiplied by 15s\n= 702m << note this as well.\n\n\nWe have one last section now. Here, it is accelerating at -2.9m/s^2 from a velocity of 46.8m/s^1 (the constant velocity we calculated remember) to a velocity of 0m/s^1 (rest). So we need one of the equations of motion here.\n\nWe know u=46.8m/s^1, a= -2.9m/s^2 and v must = 0m/s^2 \nwe want s. No t is involved here so the equation that doesn't have t is \nv^2=u^2 + 2as\n\nThis equation isn't giving us displacement in this form though. We need to rearrange to make s the subject by leaving it alone on one side of the equal sign.\n\nFirst lets put u^2 to the other side of the equal sign (iam taking it away from both sides of the equal sign to do this). \n\nThat gives us v^2 - u^2=2as\nNow we just need to divide both sides by 2a to leave s on its own.\nThat gives us \n\n(v^2 - u^2)/2a = s\n\nNow referring back to our values, we just subsitute them in,\nwhich gives us\n\n(0^2 - 46.8^2) / (2 x -2.9) = 377.6275862m << note this lastly\n\nRemember all them displacements that we noted, now all we have to do now is add them together. To find the total displacement travelled. That is \n\n421.2m + 702m + 377.6275862m = 1500.827586m\nWhich is your answer\n\nA few notes here:\nThe equations we used gives us displacement only, not distance travelled. That's why I always referred to them as displacements. And same for velocity. The equations only deal with velocity, not speed. The difference is that speed doesn't care about it's direction, velocity does. However, the automobile just travels in a straight line and straight forward in this case; there isn't any case where it travels backwards or anything. It just travels in one direction, so displacement automatically gives us distance travelled anyway and velocities in this case is the same for speed.\nAlso I didn't round up the any numbers (especially that long 377.62... number) becuase I don't want to round numbers until I get to the final answer, becuase that will introduce inaccuracies. I can however round up now at the final answer. \nI leave tha
396034
You draw it under the curve\n\nThe tangent line will only touch the curve at one point for this type of smooth upward bending curve.(At least for the part of the graph shown).\n\nYour graph plots total displacement verses time. If the graph was a straight line then this would indicate a constant "velocity" but since the curve bends upwards you are dealing with a positive acceleration.\n\nPractically, from your graph, suppose you wanted to know the approximate slope of the tangent line at t=5. Use the two closest data points: t=4, d=8.4 and t=6, d=16.5. Then the slope is about (16.5-8.4)/(6-4) = 4.05 at t=5. it will be slightly greater than this. You can use a straight edge to make a visual adjustment and then you can find the slope by dividing the displacement value at t=5 by (5 minus the value of t where the straight edge crosses the time axis).\n\nThe other way is to use the data to deduce, a closely as possible, the formula of the function being graphed. For example if the displacement was directly proportional to the square of the time, d = k t^2 for a determined value of k, then the slope of the tangent line at any time, t, would simply be 2 k t which is the time derivatiive of the function.
505057
They all involve friction, but I think you mean which has the most force due to friction, right? :-&#92;  If so, I believe the answer is that they are all the same.\n\nIn simple situations, the frictional force created by sliding two surfaces against eachother is given by:\n\nf = u * N\n\nWhere u (mu) is the frictional coefficient (an experimentally-determined constant) and N is the normal force, which is mg (mass times accelleration of gravity) if the surface isn't tilted.\n\nSo, assuming the frictional coefficient is the same for all sides of the book, there is no reason why changing the sides would make a difference.
601120
Mean is the average of the data.\nMedian is the middle number (if placed in ascending/descending order)\nMode is the number that appears the most.\n\nExamples:\n\nLet's say you are trying to discover the force of gravity by measuring the time it takes a car to roll down a ramp.  You know the length of the ramp and the angle of the ramp and therefore, need only a good approximation of the time that it takes for the car to go down the ramp (your number will be low do to friction).\n\nLet's say that these are the times that you got from 10 trials:\n\n\n9.3\n9.3\n9.9\n8.8\n9.5\n9.7\n15.4  (the car came off the track and was slowed down a lot)\n9.6\n8.9\n9.2\n\nThen you would have had a mean of 9.96\na median of  9.4 (line them up in ascending oder 8.9, 8.9, 9.2, 9.3, 9.3, 9.5, 9.6, 9.7, 9.9, 15.4 and since there is an even number of data points, take the mean of the 5th and 6th)\nand a mode of 9.3\n\nNow 15.4 is obviously a bad point.  Removing it we get:\na mean of about 9.7\na median of  9.3\nand a mode of 9.3\n\nSo only the mean really changes, this is because the mode and median are designed to get rid of "bad" data points.
602962
To find the average velocity, you find the ratio of the position difference to the total time elapsed.  \n\nSo, average velocity \n= [r_vec(2)-r_vec(0)] / 2\n= [(13.8,10.4)-(3.8,0)] / 2\n= (10,10.4) / 2\n= (5,5.2)\n\nTo find the instantaneous velocity at time=a, you find the derivative of r_vec with respect to t for both vector components and then plug in a.\n\nThe derivative of r_vec with respect to t is:\n(5t, 5.2)\n    \nAt t=0, the instantaneous velocity is:\n(0, 5.2)\n\nAt t=1, the instantaneous velocity is:\n(5, 5.2)\n\nAt t=2, the instantaneous velocity is:\n(10, 5.2)\n\nThese units are all in cm, as given in the problem.
478352
It seems that the men are all streaking at a constant velocity of 17 miles per hour.  This is a very simple problem.  All you need to know is the relationship between velocity, displacement and time:\n\nv = dx/dt\n\ndt = dx/v\n\ndt = 3/17\n\ndt = 10.6 minutes or 0.18 hours
38737
Code:\nx'' - 2nd derivative of x with respect to time (t) (acceleration in x)\n\nx' - 1st derivative of x wrt time (velocity in x)\n\nAnd the same system for the y coordinate.\n\nx''=1.2 then by integration wrt time gives x'=1.2t+constant. Boundary conditions are x'=4750 m/s at t=0, so the constant = 4750 m/s. Equation is now Vx = x' = 1.2t + 4750. Substitute in t=743s, answer is Vx=5641 m/s. \n\nSame method for y'' with acceleration of 8.5 m/s^2, and boundary conditions of y'=0 at t=0 gives Vy=6315.5 m/s.\n\nFinal answers:\nVx=5641.6 m/s\nVy=6315.5 m/s
375475
the final velocity of an acceleratiing system is equal to the initial veloctiy plus the acceleration times the time over which it is accelerating\n\nso \nv(final) = v(initial) + a t\nyou know the final 12 rad/s\nyou know the time 3 sec\nyou kow the inital 0 rad/sec\n\nso 12 rad/s = a *(3 sec)\nso a = 4 rad/s^2\n\n\n\nNow for part b\n\nthe distance traved in an accelerating system is equal to the initial velocity time the time plus half the acceleration time time squared\n\nor \nx = vt + 0.5 a t^2\n\nt = 3 s\nv = 0 rad/s\na = 4 rad/s^2\n\nx = 0.5 * 4rad/s^2 * (3 sec) ^2\n\nx = 18 rad\n\nit is still just the equations of motions.
466786
Great Question! I have the answer!\n\nImagine you are moving an object from left to right. You will see there is a displacement and a definite time taken to move from one place to the other. \nImagine you are moving faster. Notice that the time reduces.\nNow you are moving in ultrafast speed such that the time taken is nearly zero. \nWhat does this mean? There has been a displacement with no time. \nThis is what is called Yoga! Ancient Saints use to travel in time space exactly by controlling their displacement at no interval of time.\nSo do not confuse yourself with the things which will not happen with our life.\n\nTimelessness would be felt in time machines, where in the the machine travels at a speed equal to infinity. Or atleast close to infinity!\n\nOk! is it possible to have a timelessness?\n\nThis looks like an imaginery thing. Just in fictions!
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Hey! I just finished learning about active transport! The teacher confused me soooo bad! I had to have a friend in my class explain it to me. I would suggest going to google and typing in "active transport" and clicking on the news tab on the top. good luck with your project!
369454
Velocity is basically speed, normally a constant speed used when measuring extremely fast objects such as bullets. Acceleration is the rate at which an object increases speed. For example a car can accelerate from a standstill to 60mph in ten seconds
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Torque = Force * Distance(from fulcrum)\ntherefore\nTorque =(mass * acceleration) * D\nThe higher the torque, the greater the acceleration, but you'll need a duration (time) or distance travelled to determine a velocity.\n\nMore torque = more acceleration.
627010
velocity is totally different from speed, speed accounts for total distance covered per unit time whereas velocity depends on displacement, it differs by change in displacement per unit time, i.e. higher the displacement in short time more would be the velocity, if the person after travelling a distance returns back to the original position then the it accounts for zero displacement, therefore it would be zero velocity no matter what the direction is.
228480
Eggman is good except for one point: velocity is a vector not a scalar.  He was right that velocity was a speed and direction.  He just used the wrong term.\n\nAnother example would be that of a pendulum.  When the pendulum bob is at the peaks on either side, it is experiencing acceleration, but it has zero velocity for the instant it is at either peak.\n\nThis is actually the same as throwing a ball up in the air, and the ball stops for an instant at the very top of its path.  The only difference is the pendulum keeps going.  The point is, the peaks of a pendulum, and the peak of a ball being thrown straight up into the air are really the same thing happening.
406941
Acceleration and Force are directly related. So the question is the same as asking for the position that gives 1/2 max force.\n\nSpring force and displacement are directly related. So the question is the same as asking for the displacement that is1/2 max displacement.