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The displacement of a particle after time 't' is given by x=(k)/(b^(2))(1-e^(-bt)) , where b is a constant. What is the acceleration of the particle |
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Answer» `ke^(-bt)` `F_("NET")=d_(2)vg-d_(1)vg` `a=((d_(2)-d_(1))vg)/(d_(1)v)` `implies a=((d_(2)-d_(1))/(d_(1)))g` `implies v_(F)=sqrt(2gh)-at` `O=sqrt(2gh)-((d_(2)-d_(1))/(d_(1)))g timplies t=(sqrt(2gh))/(g)(d_(1))/((d_(2)-d_(1)))=sqrt((2H)/(g))((d_(1))/(d_(2)-d_(1)))` 
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