3. what are the velocity and acceleration function vectors for a particle traveling in the xy-plane with position vector r(t)=(2+t^2, 3t)? a)v(t)=(2t^2, 3t) ; a(t)=(2t, 3) b)v(t)=(2t, 3) ; a(t)=(2, 0) c)v(t)=(2t, 3t) ; a(t)=(2, 3t) d)v(t)=(2+2t, 3) ; a(t)=(4, 0)

Remember::

a) velocity is the first derivative of the position vector

b) acceleration is the second derivative of the position vector or, what is the same, the first derivative of the velocity vector.

Using that, you get:

1) given: r(t)=(2+t^2, 3t)

2) v(t) = d r(t) / dt = (2t, 3)

3) a(t) = d (v(t)) / dt = (2,0)

  option b) v(t)=(2t, 3) ; a(t)=(2, 0)

7.5/0.5 is 15 because there is 2 0.5s in 1 and there are 7 1s in 7 so that's 14 plus the 1 from 2 0.5s so answer is 15