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Homework Help: Will there be a collision?

  1. Sep 16, 2004 #1
    Speedy Sue, driving at 30 m/s, enters a one-lane tunnel. She then observes a slow moving van 155 m ahead travelling at 5 m/s. She applies her brakes but can accelerate only at -2.0 m/s^2 because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and what time the collision occurs. If no, determine the distance of closest approach between sues car and the van.

    here's my greatest guess....

    i'm going to assume they collide...(if i can do that)

    S----------- 155 m -----------V--------x---------C

    S = sues starting position
    V = the vans starting position
    x = the distance travelled by the van
    c = collision

    d = vt + 1/2at^{2}
    155+y = 30t - t^{2}
    y = 30t - t^{2} - 155

    d = vt + 1/2at^{2}
    y = 5t


    30t - t^{2} - 155 = 5t
    t^{2} - 25t + 155 = 0

    using the quadratic formula, i got

    t = 13.62 and t = 11.38 seconds (approx.)

    now which one do i accept? am i doing this right or the smartest way? thanks!
  2. jcsd
  3. Sep 16, 2004 #2
    my mistake...i defined the variable as x, but used y in my solution....

    S----------- 155 m -----------V--------y---------C

    S = sues starting position
    V = the vans starting position
    y = the distance travelled by the van
    c = collision

    any suggestions/help would be greatly appreciated...
  4. Sep 16, 2004 #3
    Position = A/2t^2 + Vi*t + Pi

    A = acceleration
    Vi = velocity initial
    P = position initial
    t = time

    Sue’s position equation is: -2/2t^2 + 30t + 0
    The van’s position (assuming the van has a constant acceleration) equation is: 0/2t^2 + 5t + 155

    The question you are being asked is when their positions are equal, to get that you need to set their positions equations equal to each other. -2/2t^2 + 30t = 5t + 155

    Are there any real solutions to that quadratic equation? And if they are, what do those solutions signify?
  5. Sep 16, 2004 #4
    isn't that what i did? the answer i got was t = 13.62 and t = 11.38 seconds (approx.). but how does that make any sense? how can they intersect twice? shouldn't i get a negative answer that i could discard? thanks for the quick reply joh
  6. Sep 16, 2004 #5


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    Homework Helper

    In my opinion the best way to do this is to calculate the distance Sue will cover when she accelerates negatively, and then check the distance covered by the truck, and compare both distances. I mean if Sue covers a distance D [D1+D2] and the truck covers a distance [D2] and Sue distance is > the distance of the truck + 155, then it hit the truck, don't you agree?

    If you like my idea, here's a hint:
    What speed Sue at least should have to avoid collision?
    Last edited: Sep 16, 2004
  7. Sep 17, 2004 #6
    The equation assumes constant acceleration while in reality she is going to stop accelerating once she hits the truck.

    Basically that equation works like they are driving next to each other. In that case they would have equal distances twice, she would pass him, and then he would pass her.

    But in your scenario that isn’t the case once they collide they stop. So the smaller time value.
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