The displacement of a particle moving in a straight line is given by $x=16t-2t^2$ (where, $x$ is in meters and $t$ is in second). The distance traveled by the particle in $8$ seconds [starting from $t$ $=$ 0] is

Given: 

$x=16t-2t^2$

Comparing with second equation of motion: 

$s=ut+\frac{1}{2}a{t}^2$

It can be concluded that: 

$u=16m/s,a=4m/s^2$

Now, using first equation of motion:

$v=u+at$

$v=16-4t$
For $v=0$ we get, $t=4s$

Here, direction of velocity will reverse. So total distance travelled will be the sum of displacement in first four seconds and displacement in next four seconds.

 
Now,
Displacement at
$t=0,x=0$
$t=4,x=32$
$t=8,x=0$
$\therefore$ Distance $=$ (distance from $x=0$ to $x=32$) + (distance from $x=32$ to $x=0$) 
$\therefore$ Distance $=32+32=64m$