Solve Free Falling Motion Homework: Traveled in 3rd Sec of Fall

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An object released from a height of 50m falls under gravity, and the distance traveled during the 3rd second is often misunderstood. The correct interpretation of the problem is to calculate the distance covered specifically between the end of the 2nd second and the end of the 3rd second, not the total distance fallen after 3 seconds. The formula d = Vi*T + 1/2at^2 is valid, but users initially misapplied it by calculating total distance instead of the distance for the specified interval. The provided answer of 24.5m is accurate when the problem is correctly understood. Clarifying the wording resolves the confusion surrounding the calculations.
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Homework Statement


An object is released from a height of 50m. How far has the object traveled during the 3rd second of fall?

Homework Equations


d = Vi*T + 1/2at^2

Doesn't work though - answer provided is 24.5m.

The Attempt at a Solution


I did a simple distance at 3 seconds with 9.81 m/s^2 as acceleration, got 29.43m. Using the formula above I got 44.45m - again, not 24.5m as provided for answer.Thanks a lot!
 
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wheeljeeper said:

Homework Statement


An object is released from a height of 50m. How far has the object traveled during the 3rd second of fall?


Homework Equations


d = Vi*T + 1/2at^2

Doesn't work though - answer provided is 24.5m.


The Attempt at a Solution


I did a simple distance at 3 seconds with 9.81 m/s^2 as acceleration, got 29.43m.
you have calculated it's speed after 3 seconds, in m/s, not its distance.
Using the formula above I got 44.45m - again, not 24.5m as provided for answer.


Thanks a lot!
This formula is the correct one to use, but the problem is asking, somewhat vaguely, not for the distance traveled after 3 seconds, but rather, the distance traveled between the 2nd and 3rd second of travel.
 
Got it.
Yep, wording was the only confusing part. Thanks a lot!
 
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