Question
      
        The velocity of a particle is $$v = {v_0} + gt + f{t^2}.$$    Its position is $$x=0$$  at $$t=0,$$  then its displacement after unit time ($$t=1$$ )  is-                                
       A.
        $${v_0} + \frac{g}{2} + f$$              
       B.
        $${v_0} + 2g + 3f$$              
       C.
        $${v_0} + \frac{g}{2} + \frac{f}{3}$$                 
              
       D.
        $${v_0} + g + f$$              
            
                Answer :  
        $${v_0} + \frac{g}{2} + \frac{f}{3}$$      
             Solution :
        $$\eqalign{
  & {\text{We know that,}}  \cr 
  & v = \frac{{dx}}{{dt}} \Rightarrow dx = v\,dt  \cr 
  & {\text{Integrating}},\int\limits_0^x {dx}  = \int\limits_0^t {v\,dt}   \cr 
  & {\text{or}},\,x = \int\limits_0^t {\left( {{v_0} + gt + f{t^2}} \right)dt = \left[ {{v_0}t + \frac{{g{t^2}}}{2} + \frac{{f{t^3}}}{3}} \right]_0^t}   \cr 
  & {\text{or}},\,x = {v_0}t + \frac{{g{t^2}}}{2} + \frac{{f{t^3}}}{3}  \cr 
  & {\text{At }}t = 1,\,\,\,x = {v_0} + \frac{g}{2} + \frac{f}{3}. \cr} $$