Give the acceleration a = d^2s/dt^2, initial velocity, and initial position of a body moving on a coordinate line. Find the body's position at time t. a = -4 sin 2t, v(0)=2, s(0)=-3

Question

anonymous · Answer

$$a(t)=s''(t)$$$$v(t)=s'(t)$$using above, the equation can be written$$s''=-4 \sin 2t$$$$s'(0)=2$$$$s(0)=-3$$solve the second order DE

dumbcow · Answer

in other words...integrate function twice, using the initial values to solve for constants of integration
$$\int\limits -4\sin(2t) dt = 2\cos(2t) +C$$
--> 2cos(0) +C = 2  --> C = 0
$$\int\limits 2 \cos(2t)dt = \sin(2t) +C$$
--> sin(0) +C = -3  ---->  C = -3
$$s(t) = \sin(2t) -3$$