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(D) Given the distance function $s = a \sin t + b \cos 2t$. The velocity $v$ is the first derivative of distance with respect to time $t$: $v = \frac{

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$-4b$

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(D) Given the distance function $s = a \sin t + b \cos 2t$. The velocity $v$ is the first derivative of distance with respect to time $t$: $v = \frac{ds}{dt} = \frac{d}{dt}(a \sin t + b \cos 2t) = a \cos t - 2b \sin 2t$. The acceleration $a_{acc}$ is the second derivative of distance with respect to time $t$: $a_{acc} = \frac{d^2s}{dt^2} = \frac{d}{dt}(a \cos t - 2b \sin 2t) = -a \sin t - 4b \cos 2t$. At $t = 0$,the acceleration is: $a_{acc} = -a \sin(0) - 4b \cos(0) = -a(0) - 4b(1) = -4b$. Thus,the acceleration at $t = 0$ is $-4b$.

If the distance $s$ traveled by a particle in time $t$ is $s = a \sin t + b \cos 2t$,then the acceleration at $t = 0$ is

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