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posted by  sem29 on 5/21/2008 12:10:45 AM  |  status: Live  

stuck!

Course Textbook Chapter Problem
Calculus Based Physics Physics (7th) by Cutnell, Johnson 2 30P
Question Details:
two knights head toward each other from 88m away.  one accelerates .3m/s^2 the other accelerates .2 m/s^2.  How far away are they from the first knight when they collide.  I understand up until I divide the equation for the first night by the equation for the second night. Why would I divide them?
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posted by Justin_08 on 5/21/2008 12:31:44 AM  |  status: Live
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sem29's comment:
"Thanx!"
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           the distance between the two knights  are d = s1 +s2 = 88 m
                             accelaration of the knight   is  a1 =  3m/s^2
                     accelaration of the second knight is a2 = 2 m/s^2.
----------------------------------------------------------------------
 
                   
 
         Let they collide after time  ' t ' at point C then 
 
                      the accelaration of the first body is  a1 = s1 / t^2    
                                                                             t^2 = s1 / a1 ---------------(1)
                  the accelaration of the second body is a2 = s2 / t^2
                                                                             t^2 = s2 / a2  ----------------(2)                 
    
            from (1) & (2) we get  s1 / a1  = s2 / a2 
                              and         s1 +s2 = 88 m
                                     
                                solving these equations we get     s2 = 35.2 m     
                           35.2 m      away are they from the second knight when they collide
                                     solving these equations we get     s1 = 52.8 m     
                           52.8 m      away are they from the first  knight when they collide
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posted by Pradeep on 5/21/2008 12:43:13 AM  |  status: Live
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sem29's comment:
"thank you, i just wasnt sure why the equations were divided! haha"
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