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Avoid Car Crash - Algebra Based

  1. Sep 9, 2008 #1

    Dig

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    1. The problem statement, all variables and given/known data
    35. Speedy Sue, driving at 30.0 m/s,
    enters a one-lane tunnel. She then
    observes a slow‐moving van 155 m
    ahead traveling at 5.00 m/s. Sue
    applies her brakes but can accelerate
    only at ‐2.00m/s2 because the road is
    wet. Will there be a collision? State
    how you decide. If yes, determine
    how far into the tunnel and at what
    time the collision occurs. If no,
    determine the distance of closest
    approach between Sue’s car and the
    van.

    2. Relevant equations
    s = v0(t1 - t0) + 1/2a(t1 - t0)^2


    3. The attempt at a solution
    I have reached an answer to this problem, but I am not sure that it is correct.

    I'll copy my notes as I wrote them:

    We need to find displacement in the amount of time decelerating at -2.00m/s^2.
    v0 = 30m/s
    a = -2.00m/s^2

    So, I need time to solve for the displacement, so I used t=v/a to get 15seconds. This was (30m/s)/(-2.00m/s^2) which gave me -15s. I am not sure if I went astray here because I interpreted the -15s as the time it would take for the vehicle to stop.

    I plugged this all in to the displacement equation to get s=225m

    So, if Suzie is going 30m/s and decelerating at -2.00m/s^2, she will not stop until 225m after 15 seconds.


    THE VAN
    The van would need to travel 70+ meters in 15 seconds in order to avoid collision. I came to this conclusion since the van is already 155m ahead of Suzie and she ends up at 225m.

    The velocity of the van is 5m/s, so in 15 seconds, the van will travel 75 meters. Adding this distance to the distance between Suzie's vehicle and the van gives us 230 meters which is clear of the 225 meters that Suzie's vehicle will travel.

    Distance of closest approach
    I am not entirely sure about this one, but I would imagine that the distance of closest approach would be when Suzie's vehicle comes to a halt at 225m. This would make the closest approach 5m.


    Any help is very much appreciated.
    Thanks :)
     
  2. jcsd
  3. Sep 9, 2008 #2

    Doc Al

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    Staff: Mentor

    Figuring out how long it would take for Sue to come to rest (assuming no collision) is a good first step. That's the 15 seconds you calculated.

    Now do this. Write expressions for the position of Sue as a function of time and the position of the van as a function of time. Using those expressions, how would you solve for the time of the collision (or prove that there isn't one)?
     
  4. Sep 9, 2008 #3

    Dig

    User Avatar

    Thank you for your reply.

    Here are my attempts.

    Sues' Car: f(t) = v0(t) + (1/2)a(t)^2
    For example: f(3) = 30(3) + (1/2)-2.00(3)^2
    Which gives 81meters out of 225meters which is the total she traveled in 15 seconds.

    The Van: f(t) = vt
    For example: f(4) = 5.00(4)
    Which gives us 20meters out of the total 75meters that she traveled in 15 seconds.

    I am not entirely sure how I would go about solving this besides finding and comparing the positions of both of the vehicles at each second interval up to the 15 seconds when Sues' vehicle comes to a complete halt.

    I came to my original conclusion by comparing the final distances of both of the vehicles.

    Any help is appreciated :).
     
    Last edited: Sep 9, 2008
  5. Sep 9, 2008 #4

    Dig

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    I have found the answer manually using the functions I created and my calculator. I would still like to know if there is an easier way to do it. I simply did the calculations and wrote down the results for each second up to 15. For the van's calculations, I added each result to 155meters which was the gain it had on Susie's car.

    It appears that Susie's vehicle catches up to and hits the van at 12 seconds. The calculation for Susie's vehicle at 12 seconds is 216m while the van is at 215m. The van continues to be hit by Susie's vehicle during the 13'th second and then escapes it's clutches at second 14 by 1m.

    I am so glad I asked you guys for help on this as I was pretty sure of myself. Thank you so much :).
     
  6. Sep 10, 2008 #5

    Doc Al

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    Staff: Mentor

    Good. I would write the position of Sue's car like this:

    [tex]x_1 = v_0t + 1/2at^2 = 30t -t^2[/tex]

    Since the van has a 155 m head start, I'd write its position (measured from the starting point of Sue's car) as:

    [tex]x_2 = x_0 + v_0t= 155 + 5t[/tex]

    That's OK, but it's the hard way. Instead, let the equations do the work. In order for a collision, car and van need to be at the same position. Set up an equation and solve for the time when [itex]x_1 = x_2[/itex].
     
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