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:

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