A  particle moves along $x-$axis with constant acceleration and its  $x-$position depend on time $\text{'}t\text{'}$  as shown in the following graph (parabola) ; then in interval 0 to 4 sec
 

Since the graph is like a parabola 
$\therefore$   let  $x\left(t\right)=At+B{t}^2+C$
From graph  $x\left(0\right)=0\Rightarrow C=0$
  $x\left(t\right)=B{t}^2+At$
 $x\left(4\right)=0\Rightarrow 16B+4A=0$   ……………. (i)
${\left(\frac{dx}{dt}\right)}_0=1\Rightarrow {(A+2Bt)}_{t=0}=1$  
Put in (i), we get   $B=-\frac{1}{4}$
 $\therefore x=t-\frac{t^2}{4}$
Max.  $x$ coordinate = 1 (from max. and min)
$\to$  Since motion is a straight line motion 
$\to$  total distance traveled  $=2\times 1=2m$
Average speed   $=\frac{2}{4}=0.5m/\sec$