draw the diagrams s/t, v/t and a/t for a uniform motion and a uniform accelerated motion?
602635
For uniform motion:\n\ns/t is a steadily increasing line [straight diagonal across your page]\nv/t is a flat line at a non-zero level [horizontal]\na/t is a flat line at zero [horizontal]\n\nFor uniformly accelerated motion:\n\ns/t is an increasing curve [like the right-hand side of y=x^2]\nv/t is a steadily increasing line [straight diagonal]\na/t is a flat line at a non-zero level [horizontal]
238922
Howstuffworks has good diagrams and explanations that may help you.  Good luck.
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
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F = d/dt(mv) [the time deravitive of the momentum] = dp/dt  where p is the momentum\n\nWork = Energy = F x distance\n\nPower = Energy/time\n\nPower = (F x distance)/time\n\nPower = [(dp/dt) x distance]/t\n\nremember dp/dt = d/dt(mv) = m dv/dt + v dm/dt
728641
acceleration can be found using\na=(m1-m2)g/(m1+m2) = 3*9.8/5 = 5.88 m/s^2
614293
Hertzsprung-Russell  Diagram.  See the wikipedia article for more info.
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a) it is always same speed.\nb) Vf=at+Vi\nVf= -8m/s (because it is going downward)\nvi= 8m8s\na= -9.8m/s^2\n\n-8m/s=(-9.8m/s^2)t+8m/s\n-16m/s=(-9.8m/s^2)t\nt=1.63 s
382082
use d=V1*t + [(a*t*t)/2]\nd=distance=490m\nV1=initial velocity = 0m/s\nt=time = ?\na=acceleration due to gravity = 9.81m/s*s\n\nusing equation above, t will come up to be:\nt=(square root)(2*d/a)\nt=(square root)(2*490m / 9.81m/s*s)\nt=9.99 sec or 10 sec.
150092
t=to\n_____\n√1-v2/c2
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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
985616
All are true. For the last one,\nsuppose acceleration = a.\nThen the velocity is at + c\nand the position is at^2/2 + ct + d,\nwhere c and d are constants.
268064
acceleration = change in speed / time taken\n\nThe units are metres per second per second\n\nFor example: \nIf a car starts at 0 m/s and over a period of 10 seconds increases its speed to 25 m/s then its acceleration is 25/10 = 2.5 m/s^2.\n\n-------------------------------------------\n\nAcceleration is the rate of change of velocity, hence the units.  If you were to plot a graph of velocity against time the gradient would be the acceleration.\n\nJust as velocity can be written v=ds/dt acceleration can be written a=dv/dt.
1186433
Well, I think that for the first oneyou can get the acceleration by deriving it twice, then you get the integral from 1 to 5, let's see...\n\nds/dt= 6t^2+3=v\ndv/dt= 12t=a\n\nfor a\n\nI think you should do it this way\n\nintegral of vdt/ integral of dt from 1 to 5\n\nso you will get\n\n(2t^3+3t)/t from 1 to 5\n\nso it will be\n\n2t^2+3 from 1 to 5\n\nevaluating you'll get\n\n2*(25)+3-2-3\n\n=50-2\n\naverage speed = 48\n\nfor b\n\nyou just substitute t=4 to the first derivative\n\nV=6*(4)^2+3=99ft/min
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
716223
The change in velocity of an object undergoing uniform acceleration, a, for a given time, t, is given by:\nv = at\nwhere v is the change in velocity of the object.\n\nIf the object is accelerating at a constant rate of 10 m/s^2 for 5 seconds, the change in velocity the object experiences is 50 m/s.\nSince (we assume) the object started from rest, the objects final velocity after 5 seconds is 50 m/s.\n\nTo find the average velocity, divide the total displacement of the object by the total time.\nIn this case, the object has traveled 125 meters in the 5 seconds it has been accelerating (d = 1/2 at^2), so dividing this by 5 seconds and we get an average velocity of 25 m/s.
834149
The period T is 0.9s\nT = 1/f = 2pi/w so w = 2pi/T\n\nAs the movement is circular, the total of the forces is in this plan and to the centre and the value is mw²R.\n\nSo P + S = mw²R where P is the weight and S the tension of the string and R the radius and a the angle between the string and the vertical\nIn projection, we have :\nmw²R = S sin a\nmg = S cos a\n\nDividing the 2 equations :\ntg a = w²R/g\nand R = L sin a\n\nso w² = g/(Lcos a)\nso cos a = g/Lw² = g T²/(L 4 pi²) \n\nso cos a= 9.8 0.9²/ (0.3 4 3.14²) = 0.6702\n so a = 47.91°  or a =0.83 rad\n\nBut this angle is the angle between the string and the vertical so angle between the string and the horizontal is 90-a = 42.09°
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ok biker bike, here we go.\n\nacc.= g = 9.8 m/s2, (any object moving in air will have this acc. or retardation)\n\nb) we apply formula\ns=ut + 1/2at^2\nu=20\nt=4\ns=20*4 + 1/2 *9.8 *4*4\n  =158.4 m\n\nc) v=u + at\n  therefore, t= v-u/a\n                   =50-20/9.8\n                   =3 seconds (approx)