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Positions and vectors

  1. Sep 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A motorist drives south at 20.0 m/2 for 3.00 min, then turns west and travels at 25.0 m/s for 2 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00 min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity.

    2. Relevant equations

    3. The attempt at a solution
    Well I tried to convert all that to distance, so multiplying velocity and time. What I don't understand is the north east part. If I were to represent the first part. it would be 3600 m to the south, so -3600j and the second one would be -3000i to and what about the north east part?
  2. jcsd
  3. Sep 17, 2008 #2


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    Homework Helper

    The problem says North West.

    The length of the vector will be 1800 but the component vectors would be multiplied by (1/2)*sqrt(2).
  4. Sep 18, 2008 #3
    where did you get (1/2)*sqrt(2) from?
  5. Sep 18, 2008 #4


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    That's the value of sine and cosine of a 45 degree angle. A right triangle of unit sides has a hypotenuse of sqrt(2). So finding the sine and cosine is simply 1/sqrt(2) or it can be written as (1/2)*sqrt(2).

    You are going Northwest so the sine and cosine components of the displacement vector at that angle can be given as that.
  6. Sep 18, 2008 #5
    so if it says north west then you're assuming the angle is 45 degress with the x axis
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