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Linear motion

  1. Feb 12, 2008 #1
    [SOLVED] linear motion

    1. The problem statement, all variables and given/known data
    A car enters a turn at 85 km/h, slows to 55 km/h, and emerges 28 s later at 35 degrees to its original motion, still moving at 55 km/h. What is the magnitude an direction of the average acceleration measure with respect to the car's original direction? Book gives the answer as 0.5 m/s^2 and 142 degrees.


    2. Relevant equations
    a=(v2-v1)/delta t


    3. The attempt at a solution

    V2x= 55 cos 35 degrees=55(.819)=.45 km/h

    V1x=80 km/h

    a= 35 km/h (1000m)/28s(3600s)= 0.35 m/s^2

    I have no idea how to arrive at the direction of the acceleration vector.
     
  2. jcsd
  3. Feb 12, 2008 #2
    Consider the two speeds you know to be two sides of a triangle, with the third side unknown but opposite an angle of 35 degrees.

    From there you can use the law of cosines to work out the third side length, which will be the total change in speed. From that you can also work out the acceleration. You also have enough information to determine the angle of the acceleration relative to the cars initial heading.

    Law of cosines:
    [tex]c^{2} = a^{2} + b^{2} - 2*a*b*Cos(C)[/tex]

    Oh, and with questions like these, it really helps to draw a diagram.
     
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