Question

If a particle moves in the xy-plane so that at time t > 0 its position vector is (sin t, cos 2t), then at time t = 1, its acceleration vector is?

Answer 1

For this case we can do the following:

the velocity can be founded with the first derivate and we got:

v(t)= (cos t, -2sin t)

And the acceleration is the derivate of the velocity so we got:

a(t)= (-sin t, -2 cos t)

And if we replace t=1 we got:

a(1)= (-sin(1), -2 cos(1))