Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 65: 31

a) velocity will be zero at $t=3s$ b) $ x = -15 m$ $ a= 18m/s^2$

We have $ x = (2t^3 - 9t^2 + 12) m$ a) $ v_x =\frac{dx}{dt}$ $ v_x = 6t^2 -18t$ now when $v_x =0$ $ 6t^2 - 18t =0$ $ t= 3s$ Hence velocity will be zero at $t=3s$ b) At $ t=3s$ $ x = (2t^3 - 9t^2 + 12) m$ $ x = (2*3^3 - 9*3^2 + 12) m$ $ x = -15 m$ $ a= \frac{dv}{dt} = \frac{d(6t^2 -18t)}{dt}$ $ a = 12t -18$ At $ t=3s$ $ a= 18m/s^2$