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Finding velocity as a function of position, constant acceleration

  1. Mar 6, 2014 #1
    This isn't actually a homework question, it's for a vehicle safety project at work, but I think it fits in this forum.

    1. The problem statement, all variables and given/known data

    A truck is traveling down the road at 60 mph, and begins to decelerate at a rate of 0.8 g.

    I'd like to find the total distance it takes the truck to stop, and come up with a formula for the velocity as a function of displacement from the original braking point.

    2. Relevant equations

    I think that the relevant equation is v2 = 2a(s2-s1)+v12


    3. The attempt at a solution

    Well, I know the truck is traveling at 88 fps, and slowing at a rate of 0.8 * 32fps2, so it will take

    88 fps / (32.2 fps2 * 0.8 g) * 44 fps = 150 feet to stop

    But when I plug those numbers into the above equation, I get

    v2 = (2 * (0.8 * 32.2 * 60 / 88) * (-x) + (60)2)0.5

    That has the right y-intercept (65 mph), but the wrong x-intercept, around 104'

    If I change the "2" on the right side of the equation to 1.38, it looks right, but I can't work out why it would be 1.38 instead of two.

    Argh! Any help much appreciated!
     
  2. jcsd
  3. Mar 6, 2014 #2

    PeroK

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    The time it takes a truck to stop is v/a. The average speed during this time is v/2. So the distance travelled when the truck stops is [tex]s = v^2/2a[/tex]
    And, if the truck is still travelling at speed v1, then we have:
    [tex]s_1 = (v^2-v_1^2)/2a[/tex]
    Where a is the magnitude of the deceleration and v is the initial speed.
     
    Last edited: Mar 6, 2014
  4. Mar 6, 2014 #3
    Thanks for your response. However, I understand that part already--what I'm trying to work out is the equation for velocity as a function of position.
     
    Last edited: Mar 6, 2014
  5. Mar 6, 2014 #4

    PeroK

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    You mean you understood all that but yet you couldn't rearrange the equation for yourself:

    [tex]s_1 = (v^2-v_1^2)/2a[/tex]
    [tex]v_1^2 = v^2 - 2as_1[/tex]
    [tex]v_1 = \sqrt{v^2 - 2as_1}[/tex]
     
  6. Mar 6, 2014 #5
    Hi--thanks for the response. Well, I worked my way to the final equation you posted (it matches up to the final equation in my original post--may not have been clear, I didn't realize I could write in LaTeX). The problem is that I calculate a stopping distance of 150' by using average speed * time, but a "displacement when velocity equals zero" of 120' when I use the last of the equations you just posted.

    Maybe I'm doing something else wrong? Here's the equation I'm using, with variables.

    [tex]v_1 = \sqrt{(65mph)^2 - 2(0.8 g \cdot 32.2fps^2 \cdot 60mph/80fps) \cdot x}[/tex]
     
    Last edited: Mar 6, 2014
  7. Mar 6, 2014 #6

    PeroK

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    Are you trying to find speed as a function of position, or initial speed as a function of total stopping distance?
     
  8. Mar 6, 2014 #7
    Speed as a function of position. I'd like to know how fast the truck is going at a given displacement.
     
  9. Mar 6, 2014 #8

    PeroK

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    You need all your quantities expressed in the same units: fps. Do that before you plug anything into your equation.
     
  10. Mar 6, 2014 #9
    Well, everything is expressed in mph.
     
  11. Mar 6, 2014 #10
    Oh my gosh, how did that work!

    Expressed in fps, it works fine.

    y-intercept 88 fps, x intercept 150'

    So to change the y axis to mph, I just multiply the right side of the equation by 60mph/88fps, right?

    Sheesh, didn't see that one coming.
     
  12. Mar 6, 2014 #11

    PeroK

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    1 mph = 5280/3600 fps = 88/60 fps, so yes.
     
  13. Mar 6, 2014 #12
    Thanks so much for your help on this. Graphs are coming out perfectly.
     
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