It's the acceleration due to gravity on Earth. A stationary object dropped will be moving at 32 ft/s after one sec. and 64 ft/s after another sec.
2007-01-16 05:07:21
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answer #1
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answered by scruffy 5
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That is the unit of acceleration due to gravity. It means: for each second of time, the falling body increases its speed by 32 ft/sec. So, when something falls off the top of a building, at the first instant (zero seconds), it's going 0 ft/sec. At 1 second, it is going 32 ft/sec (accelerated from 0 to 32 ft/sec). At 2 seconds, it's now going 64 ft/sec (having accelerated from 32 to 64 ft/sec). At second 3, it's now going 96 ft/sec (accelerating from 64 to 96 ft/sec), and so forth. By the way, this is true of bodies falling in a vacuum under ideal conditions. In real life, air resistance starts pushing back on the falling object, and thus slowing it down. So, for example, a feather or a piece of paper does not follow this formula.....except it would if in a vacuum.
2007-01-16 05:13:50
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answer #2
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answered by august51944 2
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Explain 32 feet/sec/sec?
These are the units of acceleration. It means that the speed of an object is increasing 32 feet per second every second.
If you had said -32 feet feet/sec/sec, then the object would be decellerating.
2007-01-16 05:10:43
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answer #3
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answered by ironduke8159 7
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This is the acceleration due to gravity. It is how fast something falling in freefall speeds up. If you jump out of an airplane, you start out at 0 feet per second, the the next second you are travelling 32 feet per second, then the next second you are travelling 64 feet per second. So you speed up an extra 32 feet per second...every second. Of course, while falling you are experiencing no air resistance or anything. So you fell from a plane in a vacuum, but the plane could not be flying in a vacuum, so....you jumped from a building...and you had a space suit on.
2007-01-16 05:08:41
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answer #4
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answered by Anonymous
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It is the acceleration due to gravity. Acceleration is the change in velocity, where the units of velocity are ft/sec. Hence, the change in velocity will have (ft/sec)/sec = ft/sec^2. It means for a free falling object, with no wind resistance, the velocity will change by 32 ft/sec every second.
Bozo
2007-01-16 05:08:22
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answer #5
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answered by bozo 4
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That is the acceleration due to gravity (in english units) at the surface of the earth. 9.8 meters per second squared in metric units.
So If I drop an object from rest. One second later, it's going down at 32 feet per sec (9.8 meters per sec). Two seconds later, it's going 64 feet per sec. And it keeps accelerating like that until something like the ground or air stops or slows it.
2007-01-16 05:08:27
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answer #6
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answered by Anonymous
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It is the acceleration of gravity. When an object is dropped, its speed is 32 feet per second for the first second, 64 feet per second for the next second, 96 feet per second for the third second and increases by 32 feet per second for every second it drops until it hits something.
If you drop an object that provides wind resistance such as a feather or an open parachute, the rate of acceleration is slowed.
2007-01-16 05:12:24
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answer #7
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answered by Forward Kindness 3
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it's the acceleration (ft/sec)/sec or ft/sec^2
mathematically you get acceleration from a velocity v by:
dv/dt
and you already know that velocity is from
ds/dt where s is a distance and t is time
2007-01-16 05:10:28
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answer #8
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answered by Anonymous
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Like, a box goes thirty-two feet a second, every second.
That's acceleration.
The unit of force, the Newton, is a kilogram moving a meter per second, every second (1m/s/s or 1m/s^2).
2007-01-16 05:07:52
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answer #9
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answered by ZZ 4
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This is the acceleration due to gravity at sea level on Earth.
2007-01-16 05:08:33
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answer #10
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answered by gebobs 6
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