A particle moves along the \(x\)-axis so that its position at time \(t\) is given by \(x(t)=t^2-8t+7\). For what value of \(t\) is the velocity of the particle zero?

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Start by taking the derivative with respect to time. This will give us the velocity equation.

$$ x(t)=t^2-8t+7 $$ $$ v(t)=2t-8 $$

Set it equal to 0:

$$ 0 = 2t-8 $$ $$ t=\boxed{4} $$