10. To solve the problem, we note that acceleration is the second time derivative of the position function, and the net force is related to the acceleration via Newton’s second law. Thus, differentiating x(t ) = −13.00 + 2.00t + 4.00t 2 − 3.00t 3 twice with respect to t, we get dx = 2.00 + 8.00t − 9.00t 2 , dt
d 2x = 8.00 − 18.0t dt 2
The net force acting on the particle at t = 3.40 s is G d 2x F = m 2 ˆi = (0.150) [8.00 − 18.0(3.40) ] ˆi = (−7.98 N)iˆ dt