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Math Help - physics tourist & bear problem

  1. #1
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    physics tourist & bear problem

    another easy one i think:

    A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 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?

    how many meters?

    thanks alot.
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  2. #2
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    Quote Originally Posted by rcmango View Post
    another easy one i think:

    A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 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?

    how many meters?

    thanks alot.
    The maximum value of d is such that the bear gets to the car at the same time the tourist does.

    So set up a coordinate system such that the bear is at the origin and positive x is in the direction from the bear to the tourist.

    Both are moving at a constant speed. The bear has to cover 27 + d meters in the same time the tourist covers d meters.

    So for the tourist:
    [tex]d = v_t t = 3.5t[tex]

    Thus
    t = \frac{d}{3.5}

    For the bear:
    27 + d = v_b t = 6 \left ( \frac{d}{3.5} \right )

    Now solve for d.

    -Dan
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  3. #3
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    Hello, rcmango!

    Another approach . . .


    A tourist being chased by an angry bear is running in a straight line
    toward his car at a speed of 3.5 m/s. .The car is a distance d meters away.
    The bear is 27 meters behind the tourist and running at 6.0 m/s.
    The tourist reaches the car safely.
    What is the maximum possible value for d?

    The tourist has a 27-meter headstart.

    Relative to the tourist, the bear has a speed of 2.5 m/s.

    To cover 27 meters, it takes the bear: . \frac{27}{2.5} \:=\:10.8 seconds.

    In that time, the tourist can run: . 3.5 \times 10.8 \:=\:37.8 meters.

    [I hope he left a window open so he can dive into the car.]


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    This reminds me of two chased-by-a-bear stories.


    An old-timer was telling about his encounter with an angry grizzly.

    "I was barely keepin' ahead of him and he wuz gainin' on me fast.
    I saw a tree branch about fifteen feet up. .Just as he came up
    'n took a swipe at me, I took this big leap . . ."

    "Well, did you catch it?"

    "Well, not on the way up . . . "



    Two guys are camping when a bear headed towards them.

    One man starting putting on his running shoes.

    "Hey," said his friend, "you can't outrun that bear."

    "I don't have to," was the reply. "I just have to outrun you."

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  4. #4
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    Quote Originally Posted by rcmango View Post
    A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 m away from the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?

    how many meters?

    thanks alot.
    Hello,

    if you change the text of your problem as I've done (see above) then there is a second situation possible: The car is between the bear and the tourist and both are running toward each other. They reach the car (on different sides, I hope for the tourist) after:
    t=\frac{27}{3.5+6}\approx 2.842~s

    During this time the tourist runs d = 9.947 m
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