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How to find velocity as a function of time?

  1. Sep 4, 2008 #1
    1. The problem statement, all variables and given/known data


    A ship with an initial velocity of 3 m/s moves toward a port 4km away. The ship's acceleration is a constant -0.1 cm/s^2.
    What is the ship's velocity after 1 minute?
    Will the ship stop before crahing into the port?

    2. Relevant equations

    v final = velocity initial +a delta t
     
  2. jcsd
  3. Sep 4, 2008 #2

    Hootenanny

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    Welcome to PF veronicak5678,

    You missed out one vital section:
     
  4. Sep 4, 2008 #3
    attempt:

    v final = 300 cm/s + -.01 cm/s^2

    v final = 299.99 cm/s
     
  5. Sep 4, 2008 #4

    Hootenanny

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    Notice that the acceleration is given in centimetres per second2, but you are asked for the velocity after one minute.
     
  6. Sep 4, 2008 #5
    so the final velocity will be 299.4?
     
  7. Sep 4, 2008 #6

    Hootenanny

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    Your method is correct, but there seems to be a typo in you're previous post:
     
  8. Sep 5, 2008 #7
    Oops! I was rushing around yesterday. So I used v final = 300 cm/s + -.1 cm/s^2 (60 s) to get 294 cm/s. Is this correct for velocity after 1 minute? Seems to make sense...

    I used (v final )^2 - (v initial)^2 = 2a(delta x), giving me 0-300 = 2 (-.1) delta x

    so delta x = 1500. My units don't seem right, but i think this means it will stop at 1500 cm?
     
    Last edited: Sep 5, 2008
  9. Sep 5, 2008 #8

    Kurdt

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    Yes that answer is correct.

    Are you familiar with other kinematic equations?

    https://www.physicsforums.com/showpost.php?p=905663&postcount=2

    Think about what info you have, what you need and what equation will help you find it.
     
  10. Sep 5, 2008 #9
    OK, using (v final )^2 = (v initial)^2 + 2a delta x, I get

    300 + 2 (-.1) 400000 = v final ^2, so v final = 100. I think this means that at 400000 cm, the ship will still have a velocity of 100 cm/s, and will crash. Is that correct?
     
  11. Sep 5, 2008 #10

    Kurdt

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    Yes that looks fine.
     
  12. Sep 5, 2008 #11
    Cool. Thanks a lot for all your help!
     
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