$(******v(t)=-t^2+5t-6($in $\frac{m}{s}).$ Find the particle's net displacement and total distance traveled over the time interval $-2,6.$

To understand the terms "net displacement" and "total distance travelled," imagine a particle that moves from $x=0$ to $x=2,$ then back to $x=1.$ Its net displacement is $1,$ since it started at position $0$ and ended at position $1.$ Its total distance travelled is $3,$ since it moved $3$ units all together.