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I need help with this problem can sum1 take a look ?

  1. Sep 22, 2004 #1
    I need help with this problem... can sum1 take a look plz?

    Two cars approach an intersection of two perpendicular roads. The velocities of the cars are v1 and v2. At the moment when car1 reaches the intersection, the separation distance between the two cars is d. What is the minimum separation distance between the cars during the motion?


    it might seem trivial, but i need help so someone who can do this plz show me how to do it?
    Thx
     
  2. jcsd
  3. Sep 22, 2004 #2

    Tide

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    Have you tried anything at all with the problem?

    Here's a nudge: take the position of one car moving along the x axis to be v1 t and the other one moving along the y axis at position -d + v2 t. Use Pythagoras to find the distance between them and calculus to find the minimum distance.
     
  4. Sep 22, 2004 #3
    Is this a trick question? If not, then it is not well-defined.


    Here are some of the possible interpretations

    1) The way the problem is written, you could simply say d (think about it)

    2) Do you take the length of the car into consideration? Presumably not.

    3) The separation distance is not defined as d>0. If it is 0 then the question turns into an insurance claim rather than physics problem.

    4) car2 can have already passed the intersection when car1 reaches it. There would be, however, a solution. Use v2 and calculate (=write down the formula) how long it took from the intersection point I to the point D where I and D are d apart. Once you have that time you only have to find the distance of car1 from I at that time (which is the speed of car1 times 'The time that it took car2 to travel the distance d'). If this question is for real, you can impress somebody with that answer.

    5) The most likely case is that the question does not take the length of the cars into account and assumes that car1 is at the intersection before car2 and the distance then, at that moment, is d. The question probably further asks what the ´continuous´ minimum distance is between the two cars when they are moving. I also assume that they use the word "minimum" in order to express that there is a straight line between the cars rather than a hill or something.

    So, in that case make a drawing of the situation and think of Pythagoras (by the way, the guy lived in a quite remote, but beautifully set village on the island of Samos, Greece -worth visiting). You will realise that car1 is always (v1*t) and car2 (v2*t + d) away from the point of intersection. There is one non-intuitive part: you have to think of t as negative time -t (the cars approached from the past) and hence use the modulus of t = |t|.

    The rest I leave to you since I think that this may be a homework question or something.

    Cheers,
    Roberth
     
  5. Sep 22, 2004 #4
    There are not numbers given in the question, plus why is it that the distance between car1 and the intersection always v1(t)? if t=1, distance=v1, t=2, distance=2(v1), so wouldn't that mean that as time goes by the car gets FURTHER away from the intersection instead of closer? Same goes for car2.
     
  6. Sep 23, 2004 #5
    You are almost there

    [It took me ages to find out how to reply to you on this forum. Anyway..]

    Yes, that is correct. However, that is the reason why I said you have to think of t as negative time -t.

    Start at t=-4, for example. At that point in time you have to think: the point of intersection is v*t away (t is now positive since we started at t=-4, think about it!). Sit down and make a table with t=-4, t=-3, t=-2,t=-1. t =0.

    This means that the cars do the following:

    in the first second: car1 has traveled v1*1 and car2 has traveled v2*1.
    in the second second: car1 has traveled v1*2 and car2 has traveled v2*2
    etc.

    RobertH
     
  7. Sep 23, 2004 #6
    Yes, that is correct. However, that is the reason why I said you have to think of t as negative time -t.

    Start at t=-4, for example. At that point in time you have to think: the point of intersection is v*t away (t is now positive since we started at t=-4, think about it!). Sit down and make a table with t=-4, t=-3, t=-2,t=-1. t =0.

    This means that the cars do the following:

    in the first second: car1 has traveled v1*1 and car2 has traveled v2*1.
    in the second second: car1 has traveled v1*2 and car2 has traveled v2*2
    etc.

    I understand your conceptual problem, but please make the table...
    RobertH
     
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