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Black Mamba Physics Problem

  1. Nov 1, 2005 #1
    I have a physics midterm tomorrow and this is the only problem that I cannot figure out. Thanks for the help!
    The black mamba is one of the world's most poisonous snakes, and with a maximum speed of 18.0 km/h, it is also the fastest. Suppose a mamba waiting in a hide-out sees prey and begins slithering toward it with a velocity of
    +18.0 km/h. After 2.50 s, the mamba realizes that its prey can move faster than it can. The snake then turns around and slowly returns to its hide'out in 12.0 s. Calculate
    a. the mamba's average velocity during its return to the hide-out.
    b. the mamba's average velocity for the complete trip.
    c. the mamba's average speed for the complete trip.
  2. jcsd
  3. Nov 1, 2005 #2


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    Gold Member

    Please, no triple posting.

    Also, please show us our work or what you have done so far.
  4. Nov 1, 2005 #3
    sorry for the triple posting. I didn't know it was sending. I noticed that my answers were half of what they should be. PART A: For the average velocity on the way to they prey, I used (0 km/h + 18km/h) / 2. When I looked at the answer, it was showing that d = Average Velocity x t = 18km/h. I'm wondering how they got the distance to equal a velocity. Basically, I was saying the Vavg = 9 while the book said Vavg = 18.
  5. Nov 1, 2005 #4


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    Homework Helper

    The wording of the question does not mention
    ANY time spent at slow speed, once you began to time it.
    It moved at constant speed 18 km/hr for 2.5 sec.
  6. Nov 1, 2005 #5
    ahhhh, thank you very much, so the initial velocity and the final velocity are the same.
  7. Nov 1, 2005 #6


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    Homework Helper

    So all your distances are now twice as large as you had thought.
  8. Nov 2, 2005 #7


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    Homework Helper

    Have you solved this one?
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