Question

The acceleration of a particle is a(t) = 4 cos (2t). Find the position function for the particle if v(0) = -1 and s(0) = -3

Answer 1

Integrate the function and use the points to find the constant of integration

Answer 2

like @PhysicsGuru said
\[a(t)=\frac{\mathrm{d}v(t)}{\mathrm{d}t}\]
\[\int \mathrm{d}v= \int a(t) \ \mathrm{d}t \]
\[v(t)+v_0=\int a(t) \ \mathrm{d}t\]
\[v(t)=\frac{\mathrm{d}s(t)}{\mathrm{d}t}\]
\[ \int \mathrm{d}s(t) = \int \big(v(t)+v_0\big) \mathrm{d}t\]

also
\[ \int \cos (2t) \mathrm{d} t = \frac{1}{2} \sin (2t) +C\]
\[ \int \sin(2t) = -\frac{1}{2} \cos(2t)+C\]