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What is required to just barely avoid a collision.

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data

    A car (car1) is traveling in the positive x direction with a speed v when the driver notices a car (car2) that is stopped in front of him a distance d. In order to avoid a collision, he immediately applies the brakes resulting in an acceleration of –a1. At the same time, the parked car starts to accelerate in the positive x direction with an acceleration of a2. Determine the distance d in terms of v, a1 and a2 if the cars are to just barely avoid a collision.


    2. The attempt at a solution

    This is my first physics class ever and I'm completely lost. But I know that the cars will not collide if traveling in the same direction with the same velocity. So, does that mean that I have to find the distance at which the velocity of car2 equals that of car1?
    I think the kinematics equation you would use is (vf)^2=(vi)^2 + 2ad
     
  2. jcsd
  3. Feb 23, 2012 #2
    Say that the positions of car 1 and car 2 are [itex]x_1[/itex] and [itex]x_2[/itex] respectively. Imagine plotting [itex]x_1[/itex] and [itex]x_2[/itex] as a function of time on the same graph. What will the shapes of the curves be? What can you say about the number of times we want the curves to intersect?

    Hint: [itex]x_1(t=0) = 0[/itex] and [itex]x_2(t=0) = d[/itex]
     
  4. Feb 27, 2012 #3
    Sorry it took so long to reply! But thank you so much for helping.

    So on a position graph, if the x-axis is the origin, then the car1 would be traveling toward the origin with a constant negative acceleration. and car2 would be traveling away from the origin with a constant positive acceleration. We do not want the two lines to intersect because that would mean that they are in the same position at the same time, correct?
     
    Last edited: Feb 27, 2012
  5. Feb 27, 2012 #4
    Okay, here's the thing. In physics questions, "just barely avoid" often means that the event would actually happen in real life, but for the purposes of the problem you can consider it as having been avoided. In other words, when you solve this question and find a [itex]d[/itex], if you actually used that [itex]d[/itex] in real life you would hit the car. The reason they say "just barely avoid" is that you could theoretically pick some arbitrarily large [itex]d[/itex] so that the cars would definitely avoid the collision, but then you wouldn't really have done any physics work. The upshot of all of this is that you want to find a [itex]d[/itex] such that the paths of the cars intersect only once.

    The paths of the cars are not lines. Remember that we're plotting their positions as a function of time.
     
  6. Feb 27, 2012 #5
    Okay, sadly I have been looking at this for a couple of hours. But I think I am finally onto something.

    If, in the strange world of physics, we want to find the instant where the cars intersect only once then I believe we would want to find where the position of Car1 equals that of Car2.

    So for Car1: x=xi+vit-(1/2)at2 where the initial position (xi) is equal to zero.

    For Car2: x=xi+vit+(1/2)at2 where the initial position would equal the distance d from Car1 and the initial velocity would equal zero.

    This is only as far as I've gotten. Is everything looking alright so far?
     
  7. Feb 27, 2012 #6
    Yes, all of that is good! Summarizing what you just said, we have the following:

    [tex]x_{1}=vt - \frac{1}{2}a_{1}t^2[/tex]
    [tex]x_{2}=d + \frac{1}{2}a_{2}t^2[/tex]

    You still need to tell me what they look like though :)
     
    Last edited: Feb 27, 2012
  8. Feb 27, 2012 #7
    What they look like? Well I would imagine that Car1 is an acid yellow 1936 Desoto Airstream Coupe and Car2 is a 1938 Fiat Topolino Street Rod -- Fire truck red with yellow flames of course.
     
  9. Feb 27, 2012 #8
    :rofl:

    I meant the shapes of the graphs, but that's fine too. In fact, the curves are parabolas due to the [itex]t^2[/itex] terms. There are three possible ways that two parabolas can intersect; once, twice or never. We already know that we want them to intersect once, so what are you going to do with the two equations?
     
  10. Feb 27, 2012 #9
    Oh! It makes more sense that you meant the shapes of the graphs. Sorry about that.
    Well, to get back on topic, what I meant by "line" before was actually a curve. Car1 would have a negative parabola and Car2 would of course have a positive parabola. So to find where they intersect, x1=x2

    So v1t-(1/2)a1t=d+(1/2)a2t2

    Solving for d you get: d=v1t-(1/2)a1t2-(1/2)a2t2
    and you can simplify to get d=v1t-(1/2)t2(a1-a2)

    but now I'm stuck... what do I do to get rid of t?
     
    Last edited: Feb 27, 2012
  11. Feb 27, 2012 #10
    Note that you've set the positions equal to each other and since the positions are functions of time, this must be a quadratic equation for the times when the positions are equal. Collecting like terms gives:

    [tex]0 = (1/2)(a_{1}+a_{2})t^2 + (-v)t + d[/tex]

    Now what can you do with this?
     
  12. Feb 27, 2012 #11
    Okay so I got

    t= (v±√(v2-4(a1+a2)(d)))/(2(a1+a2))
     
  13. Feb 27, 2012 #12
    Okay that equation did not turn out the way I expected. But basically I just did the quadratic equation
     
  14. Feb 27, 2012 #13
    Okay, fair enough. You should've used [itex](1/2)(a_{1}+a_{2})[/itex] instead of just [itex](a_{1}+a_{2})[/itex] for the coefficient of [itex]t^2[/itex] but the idea is there. The main point though to notice is that there are in fact two solutions because of the plus or minus sign. Thus we need to pick our value of [itex]d[/itex] such that there's only one and now we have a way of doing that. Note that if the argument of the square root is zero, then the plus or minus sign becomes useless and we have only one solution. Now you need to find a [itex]d[/itex] that satisfies this. Also note that this is equivalent to setting the discrimanent of the quadratic [itex]t[/itex] equation equal to zero, so you don't even really need to solve it.
     
  15. Feb 27, 2012 #14
    Oh ya, I forgot about the 1/2. Well because the prompt specifically states that it is moving in the positive x direction, does that mean that we pick which ever equation would come out positive?
     
  16. Feb 27, 2012 #15
    Simply eliminating one root (ie. the negative root) because a negative time doesn't make any sense is usually a good idea, but here the negativity of an answer is totally arbitrary. You could put the origin anywhere you want and even get both roots to be negative. All you know is that one car is [itex]d[/itex] in front of the other one. However, we do know that we only want one solution. Thus, we don't eliminate one root, we shrewdly pick a [itex]d[/itex] that forces the plus or minus sign to become useless. The only way this is going to be true is if you have [itex]\pm 0[/itex]. Thus, solve the following for [itex]d[/itex]:

    [tex]\sqrt{[-v]^2-4[(1/2)(a_{1}+a_{2})(d)]}=0[/tex]
     
  17. Feb 28, 2012 #16
    Okay, I see! This answer was actually do a couple days ago and I got it wrong haha, but I still wanted to understand how to do it.
    Thank you for all of your help! I couldn't have done it without you :) I hope you have a fantastic night!
     
  18. Feb 28, 2012 #17
    Glad to help! :biggrin:
     
  19. Feb 28, 2012 #18
    Hey sorry to bother you again since we kind of already finished this conversation. But I just wanted to clarify, do you assume that the ±v is 0?
     
  20. Feb 28, 2012 #19
    No, [itex]v[/itex] can be anything. The roots are:

    [tex]t = \frac{v\pm \sqrt{v^2-2d(a_{1}+a_{2})}}{a_{1}+a_{2}}[/tex]

    Then we picked a [itex]d[/itex] to make the square root [itex]0[/itex] so that:

    [tex]t = \frac{v\pm 0}{a_{1}+a_{2}}[/tex]
    [tex]t = \frac{v}{a_{1}+a_{2}}[/tex]
     
  21. Feb 28, 2012 #20
    Oh but I think the prompt called for us to find distance in terms of velocity and acceleration, instead of finding time
     
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