When we say that an object is under constant acceleration,
we mean that it is accelerating uniformly over time. The most commonly
discussed example of this is a dropped object.
When you drop it, for a very short time it seems to barely
move. Then as time passes, the object gains speed and continues gaining
speed.
NOTE: The air here on Earth complicates
things because it hampers falling acceleration, so we will pretend that
it doesn't exist.
Each second, the object moves 9.8m/s
faster. What does the physicist say? They
say, "The acceleration of an object dropped near the Earth's surface is
9.8m/s/s."]
How fast would you be falling if
you fell for 3 seconds?
END OF SECOND 1==>9.8m/s
END OF SECOND 2==>9.8+9.8=19.6m/s
END OF SECOND 3==>9.8+9.8+9.8=29.4m/s
Pretty easy, huh?
How far would you fall?
Hmm... since average velocity equals
displacement/time AND it equals Vf+Vo/2, then
displacement=(time(Vf+Vo))/2==>3sec(29.4m/s-0m/s)/2==>44.1m