GIVEN
The function of the object's velocity is given as follows:
[tex]v(t)=-2\sin t[/tex]
Also given:
[tex]s(0)=0[/tex]
SOLUTION
To get the position's function (s(t)), the velocity function needs to be integrated:
[tex]s(t)=\int v(t)dt[/tex]
Therefore:
[tex]\begin{gathered} s(t)=\int(-2\sin t)dt \\ \mathrm{Take\:the\:constant\:out}: \\ s(t)=-2\cdot\int\sin\left(t\right)dt \\ \mathrm{Use\:the\:common\:integral}:\quad \int \sin \left(t\right)dt=-\cos \left(t\right) \\ s(t)=-2\left(-\cos\left(t\right)\right) \\ \mathrm{Simplify}\text{ and add a constant to the solution} \\ s(t)=2\cos\left(t\right)+C \end{gathered}[/tex]
Recall that s(0) = 0. Therefore:
[tex]\begin{gathered} s(0)=2\cos(0)+C=0 \\ \therefore \\ C=-2 \end{gathered}[/tex]
Hence, the position function is:
[tex]s(t)=2\cos t-2[/tex]
The THIRD OPTION is correct.
