a. If s(t) is the displacement at time t, then the net displacement over the interval [0, 2] is
∆s = s(2) - s(0) = (2² - 3•2 + 2) - (0² - 3•0 + 2) = -2
The average velocity over this interval is
v [ave] = ∆s/∆t = -2/(2 - 0) = -1
b. Differentiate s with respect to t to get the velocity at time t :
v(t) = ds/dt = 2t - 3
At t = 0, the velocity is v(0) = 2•0 - 3 = -3, so the speed here is 3.
At t= 2, the velocity is v(2) = 2•2 - 3 = 1.
Differentiate again to get the acceleration:
a(t) = dv/dt = 2
so that a(0) = a(2) = 2.
c. The body changes direction when its velocity changes sign. In other words, its moves forward relative to its starting position if v(t) > 0, and backward if v(t) < 0.
The moment of changing direction happens when v(t) = 0. Solve for t :
2t - 3 = 0 ⇒ t = 3/2
