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

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An object in free fall accelerates at 10 m/s², which means its speed increases over time. After 5 seconds, the object reaches a speed of 50 m/s, but the average speed during this time is 25 m/s since it started from rest. The distance covered can be calculated using the formula d = 1/2 gt², resulting in 125 meters after 5 seconds. This discrepancy arises because acceleration affects speed, not the distance directly. Understanding the relationship between acceleration and average speed clarifies why the object covers more distance than initially expected.
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Hi,

As I understand, the distance covered by an object in free fall is described as d = \frac{1}{2}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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Accelerating at 10m/s2 for 5 seconds means you reach 50m/s. But you started off stationary, so your average speed has been 25m/s. In 5 seconds at and average of 25m/s you cover 125 m.
 
AbsoluteZer0 said:
As I understand, the distance covered by an object in free fall is described as d = \frac{1}{2}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.
 
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