Answer
$t= 1s$
Work Step by Step
Given that $x=3t^2-2t^3$, we take the derivative to find:
$v=\frac{dx}{dt}=6t-6t^2$
For maximum positive $x$ position, $v$ will be zero:
$6t-6t^2=0$
$t-t^2=0$
$t(1-t)=0$
$t=0, 1$
We plug in these values into $x_{(t)}=3t^2-2t^3$ to find what the position of the particle is at these points:
For $t=0s$
$x_{(0)}=3(0)^2-2(0)^3=0m$
for $t=1s$
$x_{(1)}=3(1)^2-2(1)^3=1m$
As $x_{(1)}>x_{(0)}$, so $x$ reaches its maximum value at time $t= 1s$
