Engineering Mechanics: Position of a particle is given as $S(t)=2t^2-8t+5$

Determine the time when its velocity is zero

S(t)=(2t^2-8t+5)
v(t)=(4t-8) = 0
t=2
 

Calculate its total distance at t=3seconds.

S(t) = integral from 0 to 3 of sqrt[1 + (4t-8)^2 ] dt = 10.73
 

Total displacement after t=3seconds
S(t)=(2t^2-8t+5)
S(3)=2(3^2)-8(3)+5 = -1