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Homework Help: Super fast average acceleration confusion problem!

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    a car traveling at a constant speed of 20 m/s that is initially traveling due northwast rounds a corner so that after 10s, the car is traveling due northeast. What are the magnitude and direction of the car's average acceleration during this interval of time?

    2. Relevant equations

    a= v2-v2/t

    3. The attempt at a solution
    i divided 20m/s/10s and got 2 m/s and used a motion diagram to get the direction to be northward. This is not right. I'm not sure how to do it because it is a constant speed, so wouldn't the velocity be -20-20/10? and that would definately be 0, and the acceleration is not 0 becasue it is chaning direction.
  2. jcsd
  3. Oct 29, 2009 #2


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    Homework Helper

    The car's changing directions, so you have to find the vector representing the change in velocity. Draw a vector diagram of the car's velocity to see what's going on.
  4. Nov 27, 2010 #3
    ok so for your average acceleration is just 20^2/radius which is going to be your magnitude. in order to find the radius you need to find the length of the arc. And because the car is traveling at a uniform speed you can just use a simple kinematic equation to find the arc distance. d(final)=d(initial) +v(initial)*T+(aT^2)/2 which in this case is shortend to d(final)=v(initial)T. that is 1/4 of your circle so multiply that by four. That is now your circumference. So C=2[tex]\pi[/tex]r. plug in your answer for r into 20^2/r and you have your magnitude. To find the direction is much more simple. If you were to lay that part of a curve on a grid, say from Y at 1 to X at 1 and you draw say 9 total tangent lines. all of their directions will average out to be the same direction the tangent line that is at 45 degrees. So 45 degrees from both the northwest and the northeast leaves you with a direction of north.
    Final answer: 20^2/r to the north
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