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Help Passing Equation

  1. Sep 12, 2006 #1
    passing equation - help please

    I am really confused :yuck: with this stuff, old timer needs some help....

    A vehicle moving at a speed of 50 mph is slowing traffic on a two-lane highway. What passing sight distance is necessary, in order for a passing maneuver to be carried out safely?

    Assume that the following variables have the values given:

    Passing vehicle driver's perception/reaction time = 2.5 sec
    Passing vehicle's acceleration rate = 1.47 mph/sec
    Initial speed of passing vehicle = 50 mph
    Passing speed of passing vehicle = 60 mph
    Speed of slow vehicle = 50 mph
    Speed of opposing vehicle = 60 mph
    Length of passing vehicle = 22 ft
    Length of slow vehicle = 22 ft
    Clearance distance between passing and slow vehicles at lane change = 20 ft
    Clearance distance between passing and slow vehicles at lane re-entry = 20 ft

    You should also assume that the passing vehicle accelerates to passing speed before moving into the left lane.

    The distance travelled by slow vehicle ??
    The distance traveled by passing vehicle ??
    The distance travelled by opposing vehicle ??
    The time taken for passing vehicle to complete the pass ??
    The relative distance ??
    The clearance distance between opposing vehicle and passing vehicle upon lane re entry ??
     
  2. jcsd
  3. Sep 12, 2006 #2

    berkeman

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    Staff: Mentor

    Welcome to PF, Probie. You need to show some of your work before we can be of much help. I'd recommend that you make a plot of the two vehicles' position versus time to help you figure this out. Draw time on the horizontal axis and distance on the vertical axis. Then Draw a straight line for the vehicle that is getting passed -- he's going a constant 50mph. Then start to draw the line for the passing vehicle, given some of the initial conditions. That should help you to get started.
     
  4. Sep 12, 2006 #3
    Okay I drew to lines like you said, and all I get is two black lines on my paper staring back at me. I do not know if this is right but this is as far as I can get, as I said old timer needs help (50). I do not kow how to find the time I think it is t = D/v but I am not sure, and if it is how do I find the distance? Call me stupid....maybe just to old to learn.

    Relative distance = 84ft

    So = speed of opposing veh

    So = 60 * 1.46
    So = 87.6 f/s/s

    Ss = speed of slow veh
    Ss = 50 * 1.46
    Ss = 73 f/s/s

    diff = 87.6- 73
    diff = 14.6 f/s/s
     
  5. Sep 12, 2006 #4

    berkeman

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    Staff: Mentor

    Well first off, I'm about the same age, and I learn new stuff all the time. So no, you don't get to use that excuse.

    Next, the question is a little hard to understand, but I'm guessing it's something like this:

    -- Driver in back sees that the road is clear at time t=0, but takes 2.5 seconds to start doing anything.

    -- At t=2.5s, rear driver accelerates at 1.47mph/s, and reaches 60mph 20ft behind the slower car. How long does this take, and how far back from the 20ft did the rear driver have to start?

    -- Faster driver now travelling 60mph travels the 20ft + 22ft + 22ft + 20ft distance and pulls back in, completing the pass. (quiz question: Why 22ft twice?)

    You should be able to start adding up the numbers and distances, and start answering some of the questions. You should also be able to start drawing the plot of the passing vehicle on your graph. Since he starts behind the slower car and going the same speed, for t=0s to t=2.5s, the passing vehicle's line will be below (behind) the other vehicle's line on the graph, and parallel to it. Do you see why?

    EDIT -- Fixed a typo
     
  6. Sep 12, 2006 #5
    You are reading the question properly, the time of 2.5 sec is called percetion/reaction time.
    The two vehs are on the highway and when the way is clear to rear veh accelerates and passes.
    But there is veh coming toward the passing veh at a speed of 60 mph.

    Why 22 ft twice, because you must count the length of the veh being passed plus the length of the passing veh. (for instance what if the passing veh is 40 ft long, it would then take longer to clear the slower veh and also it would close the distance between the passing veh and the opposing veh sooner).

    If I am doing this right so far I have come up with a time of 6.80 sec to reach a speed of 60 mph. from 50 mph.
    60 - 50 = 10 / 1.47 = 6.80sec
     
  7. Sep 13, 2006 #6
    Okay here is what I get (no laughing allowed):

    A = acceleration V1 = rear veh V2 = lead veh D = difference in speed

    D = V1 - v2
    D = 60 - 50
    D = 10 mph

    T = D / A

    T = (V1 - V2) / A
    T = (60 - 50 ) / 1.46

    T = 6.80 sec.

    So the rear vehicle travelled started 147 ft back of the lead veh. Then at 20 ft behind the lead veh he pulled out to pass covering a distance of 88 ft in 5.98 sec. then pulled back into his own lane in front of the veh he passed.

    diff. in speed of two veh = 10 mph
    14.70 f/s/s = 10 * 1.47 f/s/s
    D = 14.70 * 10
    D = 147 ft.

    The passing veh covered a distance of 599.76 feet during the passing manuver.
    D = V * t
    D = 88.2 f/s/s * 6.80 sec
    D = 599.76


    The opposing veh covered a distance of 599.76 ft at a speed of 60 mph.
    So if the distance between the passing veh and the opposing veh was 1800 ft then the distance between the passing veh and the opposing veh on completion of the passing manuver was 453.38 ft

    The distance travelled by slow vehicle = 499.80 ft
    The distance traveled by passing vehicle = 746.76 ft
    The distance travelled by opposing vehicle = 599.76
    The time taken for passing vehicle to complete the pass = 6.8 sec
    The relative distance = 88 ft
    The clearance distance between opposing vehicle and passing vehicle upon lane re entry = 453.48 ft
     
    Last edited: Sep 13, 2006
  8. Sep 15, 2006 #7
    Could someone please check my above post to see if I did this right. Thanks
     
  9. Sep 16, 2006 #8
    Okay here is my work...someone please see if I am correct with this.

    A = acceleration V1 = rear veh V2 = lead veh D = difference in speed

    D = V1 - v2
    D = 60 - 50
    D = 10 mph

    T = D / A

    T = (V1 - V2) / A
    T = (60 - 50 ) / 1.46

    T = 6.80 sec.

    So the rear vehicle travelled started 147 ft back of the lead veh. Then at 20 ft behind the lead veh he pulled out to pass covering a distance of 88 ft in 5.98 sec. then pulled back into his own lane in front of the veh he passed.

    diff. in speed of two veh = 10 mph
    14.70 f/s/s = 10 * 1.47 f/s/s
    D = 14.70 * 10
    D = 147 ft.

    The passing veh covered a distance of 599.76 feet during the passing manuver.
    D = V * t
    D = 88.2 f/s/s * 6.80 sec
    D = 599.76


    The opposing veh covered a distance of 599.76 ft at a speed of 60 mph.
    So if the distance between the passing veh and the opposing veh was 1800 ft then the distance between the passing veh and the opposing veh on completion of the passing manuver was 453.38 ft

    The distance travelled by slow vehicle = 499.80 ft
    The distance traveled by passing vehicle = 746.76 ft
    The distance travelled by opposing vehicle = 599.76
    The time taken for passing vehicle to complete the pass = 6.8 sec
    The relative distance = 88 ft
    The clearance distance between opposing vehicle and passing vehicle upon lane re entry = 453.48 ft
     
  10. Sep 17, 2006 #9

    berkeman

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    Staff: Mentor

    > covering a distance of 88 ft in 5.98 sec.

    Basically it looks right, but why 88 ft. instead of 84 ft?
     
  11. Sep 19, 2006 #10
    Typo ...sorry about that.
     
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