Why Does an Object in Free Fall Cover More Distance Than Expected at 10 m/s²?

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AbsoluteZer0
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Hi,

As I understand, the distance covered by an object in free fall is described as d = [itex]\frac{1}{2}[/itex]gt2 or d = 5t2 on earth. Objects accelerate at 10 m/s2.

Using the first equation, if an object has fallen for 5 seconds then it has covered a distance of 125 meters. If objects, however, accelerate at 10 m/s2, then why hasn't the object fallen 50 meters?

Thanks,
 
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AbsoluteZer0 said:
As I understand, the distance covered by an object in free fall is described as d = [itex]\frac{1}{2}[/itex]gt2 or d = 5t2 on earth. Objects accelerate at 10 m/s2.

Using the first equation, if an object has fallen for 5 seconds then it has covered a distance of 125 meters. If objects, however, accelerate at 10 m/s2, then why hasn't the object fallen 50 meters?

If an object accelerates at 10 m/s2, this means that the speed after 5 seconds is 50 m/s (if the initial speed was zero). Acceleration tells you how fast the speed changes, not how fast the position changes.