## Fundamentals of Physics Extended (10th Edition)

$t= 1s$
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 $t= 1s$