A particle is moving with the given data:a(t) = 3 cos(t)−2 sin(t), s(0) =0, v(0) = 4. Find the position of the particle as a function of time.

Answer 1

We are given the acceleration. To find the position, you need to integrate twice. The first integration will yield the velocity and the 2nd integration will yield the position.

∫3cos t - 2 sin t dt = 3sin t + 2cos t + C

We can determine C, the constant of integration since V(0) = 4

v(0) = 4 = 3sin 0 + 2 cos 0 + c

v(0) = 4 = 0 + 2 + C

Solving for C, C=2 and the velocity equation is:

V(t) = 3 sin t + 2 cos t + 2

Now integrate again

.∫3 sin t + 2 cos t + 2 dt = -3 cos t + 2 sin t + 2t + C

S(0) = 0 as given so

S(0) = 0 = -3 cos 0 + 2 sin 0 + 2(0) + C

S(0) = 0 = -3 + 0 + 0 + C

Solving for C this time, C=3 and the position equations is:

S(t) = -3 cos t + 2 sin t + 2t + 3