A particle moves along the \(x\)-axis with its position at time \(t\) given by \(x(t)=(t-a)(t-b)\), where \(a\) and \(b\) are constants and \(a\neq b\). For which of the following values of \(t\) is the particle at rest?

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The particle is at rest when the velocity is equal to 0.

$$ x(t)=(t-a)(t-b) $$ $$ x(t)=t^2-at-bt+ab $$ $$ v(t)=\frac{d}{dt}[x(t)]=2t-a-b $$

Set the velocity equation equal to 0:

$$ 0 = 2t-a-b $$ $$ a+b = 2t $$ $$ \boxed{t=\frac{a+b}{2}} $$