Problem
The vertical motion of mass $A$ is defined by the relation $x = 10 \sin 2t + 15 \cos 2t + 100$, where $x$ and $t$ are expressed in $\mathrm{mm}$ and seconds.
Determine:
a) The position, velocity and acceleration of $A$ when $t = 1~\mathrm s$.
b) The maximum velocity and acceleration of $A$.
So far I've solved for $x(t=1), v(t=1), a(t=1)$ and $V_{\mathrm{max}}$.
$ v(t) = \frac{\mathrm dx}{\mathrm dt} \\= 10 \sin 2t + 15 \cos 2t + 100 \\=[(10 \cos 2t)(2)]+[(15 - \sin 2t)(2)]\\=20 \cos 2t - 30 \sin 2t $
$ a(t) = \frac{\mathrm dv}{\mathrm dt} \\= 20 \cos 2t - 30 \sin 2t \\= [(20 - \sin 2t)(2)]-[(20 \cos 2t)(2)] \\=-40 \sin 2t - 60 \cos 2t $
$ v(t=1)=20 \cos 2t - 30 \sin 2t \\=20 \cos 2(1) - 30 \sin 2(1)\\= 18.941 ~\mathrm{mm/sec} $
$ a(t=1)=-40 \sin 2t - 60 \cos 2t \\=-40 \sin 2(1) - 60 \cos 2(1) \\=-61.36 ~\mathrm{mm/sec^2} $
$ x(t=1)= 10 \sin 2t + 15 \cos 2t + 100 \\= 10 \sin 2(1) + 15 \cos 2(1) + 100 \\= 115.34 ~\mathrm{mm} $
$
0=-40 \sin 2t - 60 \cos 2t
$
$
t=-28.155 ~\mathrm{sec}
$
$ V_{\mathrm{MAX}}= 20 \cos 2t - 30 \sin 2t \\= 20 \cos(2(-28.155)) - 30 \sin(2(-28.155)) \\= 36.0.54 ~\mathrm{mm/sec} $
The maximum acceleration happens at a point where the acceleration's derivative (sometimes called the jerk) is zero. Differentiate a third time to get $$\frac{da}{dt}=j(t)=-80\cos2t+120\sin2t=0$$ $$-2\cos 2t+3\sin2t=0\implies t=\frac\pi2+\frac12\tan^{-1}\frac23$$ $$a_{\max}=-40\sin2t-60\cos2t=72.111\dots$$