Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

$x = (2t^3-3t^2)~m$ $v = \frac{dx}{dt} = (6t^2-6t)~m/s$ $a = \frac{dv}{dt} = (12~t-6)~m/s^2$ (a) At t = 0: $a = [(12)(0) - 6]~m/s^2$ $a = -6~m/s^2$ We can use the acceleration to find the net horizontal force $F$ $F = ma$ $F = (2.0~kg)(-6~m/s^2)$ $F = -12~N$ (b) At t = 1 s: $a = [(12)(1) - 6]~m/s^2$ $a = 6~m/s^2$ We can use the acceleration to find the net horizontal force $F$ $F = ma$ $F = (2.0~kg)(6~m/s^2)$ $F = 12~N$