YES! but not permanently ,only momentarily. To explain, say that u throw a ball at velocity V upwards , the earth's acceleration is -G. consider T=0 at the moment of throwing. V+(-G)T is the velocity of the ball at any instant T (after throwing). At the time T=V/(+G) *we have the ball's velocity 0 but the ball persistently has -G acceleration* all along duration of flight until it hits ground . *Its because the object has (-G) acceleration even at velocity 0 that the ball starts falling and finally comes down to hit the ground*.
2007-07-25 06:24:00
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answer #1
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answered by Neatest Inbox Holder 2
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It's a false premise for a problem. It depends on "with reference to what". If you are talking about yourself sitting at rest in a chair with reference to the earth, then you have zero velocity and you are experiencing an acceleration of 1 g, pointed away from the center of mass of the earth. The 'reference frame' you choose will determine whether you are considering acceleration as rate of change of velocity of an object relative to some other viewpoint, or as the FORCE experienced by that object. This difference explains why attempts have been made to explain gravity by saying it distorts space and time. Because when you are in a gravitational field, you are either experiencing an acceleration by change of velocity over time (falling), or as a force (sitting). Zero G acceleration is free fall, one G vertical acceleration is stationary, as measured by force experienced near the earth's surface. All other change in motion relative to the earth, when measured as a force, gets vectored in with the inescapable gravitational acceleration. And you never truly escape gravity, you can only go places where there is less or more of it. For a fun demonstration of acceleration and gravity relative to 'reference frames', go out and do some loops in an airplane. Or figure out why an airplane doing a 60 degree banked turn without losing or gaining altitude or speed subjects the passengers to 2 G's. So, reference frame is everything. Even in geostationary orbit, if you look down, you might figure you have zero velocity and zero acceleration. But then if you look at the universe spinning around you, you would have to conclude that you are at the very least accelerating toward the earth just enough to counteract your tendency to fly off into the universe in a straight line. You can't escape the vortex!
2016-03-15 23:02:38
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answer #2
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answered by Anonymous
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The ball thrown in the air is the perfect example. The acceleration is always g, but at the highest point the velocity is zero.
A car starting from rest, at the instant it begins to move, has zero velocity and non zero acceleration. If it had zero acceleration it would never begin to move.
2007-07-25 06:21:53
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answer #3
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answered by Dr D 7
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Yep!
Throw a ball up in the air. It rises, slows down, and stops. Now it has zero velocity. But it does not have zero accelertion as evidenced by the fact that its velocity continues to change as the ball now begins to fall back to the earth, faster and faster. During the whole process, the acceleration was constant: about 9.81 m/s^2 (32.2 ft/s^2) downward.
2007-07-25 06:17:10
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answer #4
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answered by anobium625 6
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Yes it can. For example, if you are rolling backward in your car, your velocity is v < 0, you are going in a negative direction. So now you put the car in gear and gun the engine; so your backward, negative velocity changes to a foward, positive velocity.
To change from negative to positive velocity, you must go through v = 0, for an instant you'd be standing still. But, even so, you are accelerating because, by definition, a change in velocity is acceleration.
2007-07-25 07:00:44
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answer #5
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answered by oldprof 7
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I see what I did wrong
velocity = (dx/dt) + constant change in distance wrt time
accel = d(dx/dt)/dt change in velocity wrt time
now the velocity can be zero (dx/dt + c = 0)
and the acceleration d(dx/dt +c)/dt != 0
Sorry
2007-07-25 06:13:37
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answer #6
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answered by Anonymous
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