Consider a particle moving along a straight line, whose position as...

Given $s(t) = t^2 - 6t + 5$

Velocity $v(t) = \dfrac{ds}{dt} = 2t - 6$

The particle comes to rest when $v(t) = 0$, which gives $t = 3$ s.

Position at $t = 0$ s is $s(0) = 5$ m.

Position at $t = 3$ s is $s(3) = 3^2 - 6(3) + 5 = -4$ m.

Position at $t = 6$ s is $s(6) = 6^2 - 6(6) + 5 = 5$ m.

Distance travelled from $t = 0$ to $t = 3$ s is $d_1 = |-4 - 5| = 9$ m.

Distance travelled from $t = 3$ to $t = 6$ s is $d_2 = |5 - (-4)| = 9$ m.

Total distance travelled is $D = d_1 + d_2 = 9 + 9 = 18$ m.

Average speed $= \dfrac{18}{6} = 3$ ms$^{-1}$

Answer: $3$