Finding Acceleration of a Ball in Constant Motion

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A ball initially at rest accelerates and covers 10 meters during its third second of motion. To find the acceleration, the distances covered in the first two seconds are calculated using the formula s = ut + 1/2*a*t^2. By determining the distances for 2 seconds and 3 seconds, the difference is set to 10 meters. The solution reveals that the acceleration is 4 m/s², confirming the calculations. This approach effectively utilizes kinematic equations to solve for acceleration in uniformly accelerated motion.
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Homework Statement


A ball initially at rest and constantly accelerating covers a distance of 10 metres during its third second of motion. What is its acceleration?


Homework Equations


No idea which equations I need. I'm guessing I need
a = (v-u)/t

The Attempt at a Solution


I've got no idea how to even apporach this. I'm thinking:
s = 3 s
u = 0 m/s
v =
a =
t =
 
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Welcome to PF!

Hi ride4life! Welcome to PF! :smile:

Hint: call the acceleration a, and find the distances after 2 seconds and after 3 seconds, and find the value of a that makes their difference equal 10 metres :wink:
 
What formula should I use to figure the distanes covered in each second?

0 - 1 = ?
1 - 2 = ?
2 - 3 = 10m = 10m/s/s

I was thinking in the 1st second it traveled 2.5m and in the 2nd second it traveled 5m. I think that turns out to be 2.5m/s/s. That's just a guess. That's all I can figure out. :(
 
You can use, displacement s = ut +1/2*a*t^2.
Find s for 2s and 3s. Then find the difference which is equal to 10m.
 
So for
2 - 3 = 10m

s = 10m
u = ?
a = ?
t = 1s
 
Ball starts from rest. So u = 0.
S2 = 1/2*a*(2)^2...(1)
S3 = 1/2*a*(3)^2...(2)
Find S3 - S2, equate it to 10m, and solve for a.
 
Thanks for the help, I got 4m/s/s which is the right answer. :D
 
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