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Acceleration problem

  1. Aug 31, 2007 #1
    1. An automobile starts from rest and accelerates to a final velocity in two stages along a straight road. Each stage occupies the same amount of time. In stage 1, the magnitude of the car's acceleration is 3.0m/s^2. The magnitude of the car's velocity at the end of stage 2 is 2.5 times greater than it is at the end of stage 1. Find the magnitude of the acceleration in stage 2.



    2. I'm having some trouble, because it seems like there is missing info. Do I just use the velocities?



    3. ((2.5)(3.0m/s^2))-3.0 m/s^2 = 7.5m/s^2 - 3.0m/s^2 = 4.5 m/s^2
     
  2. jcsd
  3. Aug 31, 2007 #2

    learningphysics

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    Let t equal the time that each stage takes... let a2 be the acceleration in the second stage.

    What is the velocity at the end of stage 1 in terms of t?

    What is the velocity at the end of stage 2 in terms of a2 and t?
     
  4. Aug 31, 2007 #3
    velocity at end of stage 1: v=(3.0m/s^2*t)?
     
  5. Aug 31, 2007 #4

    learningphysics

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    Yup. Now using the variable a2... what's the velocity at the end of stage 2?
     
  6. Aug 31, 2007 #5
    a2=2.5v-v/t
     
  7. Aug 31, 2007 #6

    learningphysics

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    Yup a2 = (2.5v - v)/t ...(you need the parentheses)

    Now you can substitute from the previous equation into this one to solve for a2.
     
  8. Aug 31, 2007 #7
    comes out to 4.5 m/s^2 still, right?
     
  9. Aug 31, 2007 #8

    learningphysics

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    Yup.
     
  10. Aug 31, 2007 #9
    thanks
     
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