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Displacement and Velocity

  1. Sep 7, 2004 #1
    This problem makes absolutely no sense to me what-so-ever. I haven't got the faintest idea where to even get started figuring this out. It's probably childs play to most here, but I am really beginng to get extremely stressed.

    Here's the problem:

    A tourist being chased by an angry bear is running in a straight line toward his car as a speed of 3.9 m/s. The car is distance d away. The bear is 20 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?


    So, like I said I don't even know where to begin. I'm not looking for an easy answer here, just a little guidance to start me on my way. It would be greatly appreciated.
     
  2. jcsd
  3. Sep 7, 2004 #2
    the maximum distance is just when the bear catches up to the tourist... How long would that be?
     
  4. Sep 7, 2004 #3
    Blackmamba, write an equation describing the distance between the bear and tourist, and make it relative to time:

    distance between bear and car = 20 m -(bear speed * time - runner speed * time)....work from there to find the time.

    With time, and runner speed, you can find distance.

    Remember what needhelp said...

    work from there...
     
    Last edited: Sep 7, 2004
  5. Sep 8, 2004 #4
    Thank you both for your input. However, my main issue here is finding time. I don't understand how to find time. There isn't a single example in my book, and my professor never went over it in class. Most normal problems at least give a time, whereas this one does not.
     
  6. Sep 8, 2004 #5
    Define the car as the origin. The velocity of the tourist and the bear are given as [itex]\dot{x}_T = -3.9[/itex] and [itex]\dot{x}_B = -6.0[/itex]. The positions of the tourist and the bear as functions of time are thus [itex]x_T = -3.9t + d[/itex] and [itex]x_B = -6.0t + d+20[/itex]. For tourist to arrive safely, [itex]x_B \geq x_T[/itex]. Solving the equation gives [itex]t\leq 200/21[/itex]. Plug that into the equation for [itex]x_T[/itex] to give [itex]d\leq 260/7[/itex]
     
    Last edited: Sep 8, 2004
  7. Sep 8, 2004 #6
    Can you explain further how you solved for: [itex]x_T = -3.9t + d[/itex] and [itex]x_B = -6.0t + d+20[/itex] ?

    Further more I don't understand this: [itex]x_B \geq x_T[/itex]

    Thank you for your help.
     
  8. Sep 8, 2004 #7

    HallsofIvy

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    distance= velocity times time. We know the bear has velocity 6 m/s so the distance moved in t seconds is 6.0 t. Why negative? xB is "distance to the car" and that is becoming smaller. We are told that, when t= 0 (at the start of the "race") the distance from the bear to the car, xB is 20 m so xB is 20 minus the distance the bear has run.

    If the man doesn't stay closer to the car than the bear, he's in big trouble!

    To really be realistic you really should take into account the time required to open and close the car door! (I hope that door isn't locked!)

    Oh, and many grizzly bears are quite capable of ripping a car door off.
     
  9. Sep 8, 2004 #8
    Wow, I really am having an off morning. There's nothing like a physics problem to make someone feel really stupid. I don't understand why I can't just grasp what is going on.

    HallsofIvy ~ Thanks for your explanation. That makes things somewhat clearer. So am I to understand that this answer ---> [itex]t\leq 200/21[/itex] is the result of this equation ---> [itex]x_B = -6.0t + d+20[/itex] ? If so I don't understand how that was achieved.
     
  10. Sep 9, 2004 #9
    The equation you use is [itex]x_B \geq x_T[/itex]. So [itex]-6.0t + d+20\geq -3.9t + d[/itex]. Now it's just a matter of solving the equation for [itex]t[/itex]. This will give you the maximum time necessary to arrive at the car safely. Now to find the corresponding value of [itex]d[/itex], go back to the equations for the position of A or B and let the position equal zero, that is the position of the car.
     
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