Constant acceleration

  • Thread starter Doraneli
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  • #1
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1. A boulder initially at rest rolls down a hill with an acceleration of 3.69m/s^2. If it accelerates for 7.72 seconds, how far will it move?


2. d=v0(initial velocity)t+.5at^2



3. d=(0)(7.72)+.5(-9.8)(3.27)^2
-39.24+-52.4
-91.64m

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  • #2
HallsofIvy
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1. A boulder initially at rest rolls down a hill with an acceleration of 3.69m/s^2. If it accelerates for 7.72 seconds, how far will it move?


2. d=v0(initial velocity)t+.5at^2



3. d=(0)(7.72)+.5(-9.8)(3.27)^2

Why is t= 7.72 in the first term and 3.27 in the second?

-39.24+-52.4
Where did the "-39.24" come from? And why the "+-" for the second term?

-91.64m

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #3
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You are using the correct equation, but your values for 'a' and 't' are wrong.
The acceleration of the ball is given to you in the question, so you don't need to use 9.81m/s². And also the time stated in the question is 7.72s.
 

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