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

Submitted by omehgodwin on January 19, 2016 - 5:09pm

The position of a particle is given as S(t)=(2t^2-8t+5). Determine the time when its velocity is zero. Calculate its total distance at t=3seconds.

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## Re: Engineering Mechanics

S(t)=(2t^2-8t+5)

v(t)=(4t-8) = 0

t=2

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