Question

A particle is moving with the given data. Find the position of the particle. a(t)=10sin(t)+3cos(t) s(0)=0 s(2pi)=12

Answer 1

$a(t)=10\sin t+3\cos t$

$v(t)=\displaystyle\int a(t)\,\mathrm dt$
$v(t)=\displaystyle\int(10\sin t+3\cos t)\,\mathrm dt$
$v(t)=-10\cos t+3\sin t+C_1$

$s(t)=\displaystyle\int v(t)\,\mathrm dt$
$s(t)=\displaystyle\int(-10\cos t+3\sin t+C_1)\,\mathrm dt$
$s(t)=-10\sin t-3\cos t+C_1t+C_2$

With the boundary conditions $s(0)=0$ and $s(2\pi)=12$, we have

$\begin{cases}0=-3+C_2&s(0)=0\\12=-3+2\pi C_1+C_2&s(2\pi)=12\end{cases}\implies C_1=\dfrac6\pi,C_2=3$

so that the position of the particle is given by

$s(t)=-10\sin t-3\cos t+\dfrac{6t}\pi+3$