Constant Acceleration

  • Thread starter Manh
  • Start date
  • #26
billy_joule
Science Advisor
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To change velocity, acceleration must occur. You've already found that change in velocity on page one.

Constant does not mean zero, it means it doesn't change with time. That is, the acceleration is the same for the entire 15 seconds.
 
  • #27
62
0
vf^2 = vi^2 + 2a(delta x)
22.22^2 = 11.11^2 + 2a(250)

Is this the right way to find a?
 
  • #28
billy_joule
Science Advisor
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The equation you used in post #23 is fine. Though, that equation (in #27) will work it is unnecessarily complicated.

The problem is that you used 16.67 m/s in post #23, that is the average velocity, you need to use the change in velocity.

Acceleration = Change in velocity / change in time
 
  • #29
62
0
a = 11.11 m/s / 15 s
a = 0.74 m/s^2 ?
 
  • #30
This would be it.Also,nice work on not forgetting to put the units of measurement after your results.
 
  • #31
62
0
Thank you everyone (billy_joule, LittleMrsMonkey, and Tallus Bryne) for helping me solve the problem. I really appreciate your time to work with me!
 

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