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Find Velocity and Time from an Acceleration vs Time Graph

  1. Jan 17, 2016 #1

    CGI

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    1. The problem statement, all variables and given/known data
    The rocket car is subjected to a constant acceleration of Ac = 6 m/s^2 until T1 = 15 s.
    The brakes are then applied which causes a deceleration at the rate shown until the car stops. Determine the maximum speed of the car and the time T when the car stops.

    2. Relevant equations
    Area of Rectangle = L x W
    Area of a Triangle = 1/2(b)(h)

    3. The attempt at a solution
    So I know that the area under the curve for an acceleration vs time graph would represent the velocity, but without having what the final time T was or what the final acceleration was, I'm not sure how to go about it. I know that at 15 seconds, the velocity is 90 m/s and that's about it.

    IMG_0825.JPG
    Any help would really be appreciated!
     
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  3. Jan 17, 2016 #2

    TSny

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    Are you asking about the first part of the question: getting the maximum speed? If so, at what time does maximum speed occur? Before the brakes are applied, after the brakes are applied, or at the instant the brakes are applied?
     
  4. Jan 17, 2016 #3

    CGI

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    I would say before the brakes, so that must mean that the 90 m/s IS the maximum speed that car attains.

    Is that correct?
     
  5. Jan 17, 2016 #4

    TSny

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    Yes, the car is picking up speed all the way until the brakes are applied. So, maximum speed occurs at t = 15 s.
     
  6. Jan 17, 2016 #5

    CGI

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    Okay that just leaves me with finding the time at which it stops. So that means the velocity should equal zero which means that the area of the triangle should also be equal to 90 which I think might help, how exactly would I go about finding the time?
     
  7. Jan 17, 2016 #6

    TSny

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    Good!
    Note the little triangle with sides of ratio 1 to 2.
     
  8. Jan 17, 2016 #7

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    Oh I forgot to ask about that! Unfortunately, I'm not entirely sure about what that means..
     
  9. Jan 17, 2016 #8

    TSny

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    It allows you to get the slope of this part of the acceleration graph.
     
  10. Jan 17, 2016 #9

    CGI

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    Okay, so the slope is rise over run, which would mean that the slope here is 1/2 right? But I'm not sure how I would use that to get the time :confused:
     
  11. Jan 17, 2016 #10

    TSny

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    Try to write an expression for the area of the triangle of the graph in terms of the time interval Δt (between t1 and t).
     
  12. Jan 17, 2016 #11

    CGI

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    Hmmm.. it seems like the only thing I can think of is the integral from t1 to t of the area of the triangle is equal to the velocity. Is that in the right direction?
     
  13. Jan 17, 2016 #12

    TSny

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    Yes. And the integral is just the area of the triangle. You are ultimately looking for the base of that triangle. Can you express the height of the triangle in terms of the base? You can use similar triangles with one triangle being the little triangle with sides 1 and 2.
     
  14. Jan 17, 2016 #13

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    Would the height just be equal to half the base?
     
  15. Jan 17, 2016 #14

    TSny

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    Yes.
     
  16. Jan 17, 2016 #15

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    Okay, so the integral would be:

    90 = integral of b^2/4dt from 15 to t? Which becomes b^3/12t?
     
  17. Jan 17, 2016 #16

    TSny

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    b^2/4dt is the integral (i.e., the area of the triangle).
     
  18. Jan 17, 2016 #17

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    Okay so something more like this?

    90 = (t^2 - 225)/4 in which case, t = 24.2?
     
  19. Jan 17, 2016 #18

    TSny

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    I should have said that b2/4 is the area (instead of b2/4 dt). You know what this area must equal. So, what do you get for the base b?
     
  20. Jan 17, 2016 #19

    CGI

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    I get 18.97 for the b. Which would be the answer for t right?
     
  21. Jan 17, 2016 #20

    TSny

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    I agree with your answer for b (assuming you're still using the second as a unit of time). But I don't agree with your answer for the instant of time t when the car comes to rest.
     
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