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Acceleration problem

  • Thread starter mslena79
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  • #1
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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
 

Answers and Replies

  • #2
learningphysics
Homework Helper
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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?
 
  • #3
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velocity at end of stage 1: v=(3.0m/s^2*t)?
 
  • #4
learningphysics
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velocity at end of stage 1: v=(3.0m/s^2*t)?
Yup. Now using the variable a2... what's the velocity at the end of stage 2?
 
  • #5
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a2=2.5v-v/t
 
  • #6
learningphysics
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a2=2.5v-v/t
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.
 
  • #7
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comes out to 4.5 m/s^2 still, right?
 
  • #9
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thanks
 

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