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Acceleration to displacement

  1. Aug 31, 2007 #1
    1. two knights on horseback start from rest 88.0 m apart and ride directly toward each other to do battle. Sir George's acceleation has a magnitude of 0.300 m/s^2, Sir Alfred's has a magnitude of 0.200 m/s^2. Relative to Sir George's starting point, where do the knights collide?



    2. a= (v-v0)/(t-t0), v=(x-x0)/(t-t0)



    3. 0.300m/s^2=(v-0)/t, v=0.300m/s^2*t, 0.300m/s^2*t=(x-x0)/t, I don't know where to go from here!
     
    Last edited: Aug 31, 2007
  2. jcsd
  3. Aug 31, 2007 #2

    learningphysics

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    You have a formula for displacement in terms of time and acceleration... try to use that to get the time when they collide...
     
  4. Aug 31, 2007 #3
    x=(0.300m/s^2 *t^2)/2 ?
     
  5. Aug 31, 2007 #4
    nevermind, I got it now!
     
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