\(x=t^2-4 t+6\)
\(\frac{d x}{d t}=2 t-4\)
At \(t=2 s\), particle is at rest and reverses its position so,
\(\left.\begin{array}{l}\left.x\right|_{t=0}=6 \,m \\ \left.x\right|_{t=2 s }=2 \,m \end{array}\right] 4 \,m\)
\(\left.\begin{array}{l}\left.x\right|_{t=2 s }=2 \,m \\ \left.x\right|_{t=3 s }=3 \,m \end{array}\right] 1 \,m\)
Distance \(=(4+1) \,m =5 \,m\)
Displacement \(=4-1=3 \,m\)