Suppose a ball is dropped from the top of a skyscraper. Assume we are able to measure its descent by counting the number of floors that the ball crosses in each second. We will find that in the first second, the ball covers approximately 5 meters, in the second second, it will cover 15 meters, in the third second, it will cover 25 metres approximately and so on. We find that the ball is accelerating, that is its speed (velocity in a given direction, in this case, downwards) is increasing and the distance covered in each second is increasing more and more. This is an example of an uniform acceleration, due to earth's gravity, which is 9.8 m/second^2 or approximately 10 m/sec^2.
While the velocity (speed) of an object is expressed in meters/second, the acceleration is defined as the change in velocity/sec and is expressed in metres/second/second or m/sec^2.
The velocity v is given by v = u + a.t where u is the initial velocity if any and a is the acceleration and t is the time.
Distance travelled s = ut + 1/2 . a. t^2
v, u and s are related by the equation, v^2 - u^2 = 2.a.s
2007-04-03 02:27:43
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answer #1
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answered by Swamy 7
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Uniform acceleration is a constant rate of change of velocity. So you can write the velocity with respect to time as v=at+b, where a & b are constants. Now to get position as a function of time we can integrate;
x=integral over t of v=at^2/2+bt+c, where c is any constant. This has the form of a parabola.
One easy example to think of: If you drop something, it accelerates uniformly towards the ground (with a/2=g=9.81 ms^(-2).
2007-04-03 02:30:35
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answer #2
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answered by Adam B 2
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A body has a uniformly accelerated motion if it travels in a straight line and its velocity increases by equal amounts in equal interval of time
2007-04-03 02:47:26
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answer #3
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answered by Crystal 4
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Gravity
2007-04-03 02:38:19
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answer #4
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answered by JOHNNIE B 7
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acceleration is constant as a function of time.
2007-04-03 02:22:14
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answer #5
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answered by Anonymous
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