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Find Velocitiy from an Acceleration vs. Displacement Graph

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  1. Jan 16, 2016 #1

    CGI

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    1. The problem statement, all variables and given/known data
    The a-s graph for a rocket moving along a straight horizontal track has been experimentally determined. If the rocket starts from rest, determine its speed at the instants s = 20m, 50m, and 70m, respectively. On the graph, S1 = 50m, a1 = 10 m/s^2, S2 = 70m, and a2 = 20 m/s^2

    2. Relevant equations
    ads = vdv



    3. The attempt at a solution
    I know that the solution for the velocity at S2 = 70m is V = 40 m/s, but I'm not sure how to get there.
    I know we talked about the relationship for ads = vdv so I tried taking the integral on both sides to calculate for the velocity.

    I tried to solve for the velocity at S2

    The integral from 0 to v of vdv = the integral from 0 to 20 of ads
    So, V^2/2 = 20s which means V^2 = 40s where S = 70 meters.
    So I get V = √(40)(70) = 52.9, which is not correct.

    20160116_194713.jpg Any help would be much appreciated!
     
  2. jcsd
  3. Jan 16, 2016 #2

    TSny

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    What is the graphical way to interpret the integral of ads from s = 0 to s = 70 m?
     
  4. Jan 16, 2016 #3

    CGI

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    My best guess is the area of that section from 0 to 70? Is that in the right direction?
     
  5. Jan 16, 2016 #4

    TSny

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    Yes.
     
  6. Jan 16, 2016 #5

    CGI

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    Okay, so I got 800 for the total area underneath, but I'm not sure what the next step is
     
  7. Jan 16, 2016 #6

    TSny

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    Good.

    How is the integral of ads related to the integral of vdv?
     
  8. Jan 16, 2016 #7

    CGI

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    They're equal to each other? So I could set v^2/2 = 800 which in the end makes v = 40.

    Did I think about that correctly?
     
  9. Jan 17, 2016 #8

    TSny

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    Yes.
     
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