For \(t\geq 0\), the position of a particle moving along the \(x\)-axis is given by \(x(t)=\sin t-\cos t\). What is the acceleration of the particle at the point where the velocity is first equal to \(0\) ?

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Obtaining the velocity and acceleration functions:

$$ x(t)=\sin t-\cos t $$ $$ v(t)= \cos t + \sin t $$ $$ a(t)= -\sin t + \cos t $$

Finding the value of \(t\) where velocity is \(0\):

$$ 0 = \cos t + \sin t $$ $$ -\sin t = \cos t $$ $$ -\frac{\sin t}{\cos t}=1 $$ $$ \tan t = -1 $$ $$ t= \frac{3\pi}{4} $$

Calculating the acceleration at this time:

$$ a\left(\frac{3\pi}{4}\right)= -\sin \left(\frac{3\pi}{4}\right) + \cos \left(\frac{3\pi}{4}\right) $$ $$ = -\frac{\sqrt{2}}{2} + \left(-\frac{\sqrt{2}}{2}\right) $$ $$ = -\frac{2\sqrt{2}}{2} $$ $$ = \boxed{-\sqrt{2}} $$