contestada

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

Respuesta :

LammettHash
[tex]a(t)=10\sin t+3\cos t[/tex]

[tex]v(t)=\displaystyle\int a(t)\,\mathrm dt[/tex]
[tex]v(t)=\displaystyle\int(10\sin t+3\cos t)\,\mathrm dt[/tex]
[tex]v(t)=-10\cos t+3\sin t+C_1[/tex]

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

With the boundary conditions [tex]s(0)=0[/tex] and [tex]s(2\pi)=12[/tex], we have

[tex]\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[/tex]

so that the position of the particle is given by

[tex]s(t)=-10\sin t-3\cos t+\dfrac{6t}\pi+3[/tex]