Linear Motion: 0.525kg Ball Travels 200m in 10s

AI Thread Summary
A 0.525kg ball rolls down a hill with uniform acceleration, covering 200m during the second 10 seconds of its motion. The discussion revolves around calculating the acceleration and distance traveled in the first 5 seconds. Users suggest using the formula a = 2d/t² to find acceleration and then applying d = ut + ½ at² for distance. Confusion arises regarding the interpretation of "during the second 10.0 s" and the correct application of equations. Ultimately, the calculations yield varying results, indicating a need for clarification on the problem's parameters.
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Homework Statement



A 0.525kg ball starts from rest and rolls down a hill with uniform acceleration, traveling 200m during the second 10.0 s of its motion.

Homework Equations



xf= xi + vi +1/2a(t)^2
x= vi + at

The Attempt at a Solution



a= 4 m/s^2

x=50 mits wrong
 
Last edited:
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A 0.525kg ball starts from rest and rolls down a hill with uniform acceleration, traveling 200m during the second 10.0 \rm s of its motion.

So, are you trying to solve for acceleration?
what does this part mean, "during the second 10.0 \rm s of its motion" is it a typo?
 
solve the distance it travels @ 5 seconds.
 
So the question states that the ball travels during the second 10s and you are trying to find the distance after the first 5 seconds?

Use a = 2d / t² to find acceleration where t will be your 10 seconds
Then use the equation d = ut + ½ at² where t will be your 5 seconds and solve for distance...

I also got x = 50... are your units correct?

Edit:
Maybe it is 16.63 m? (edited again)
a=200m/150 s² = 1.33m/s²
d = ut + ½ at²
d = 0 + ½ (1.33)(5)²
 
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