A stationary 1.67-kg object is struck by a stick.The object experiences a horizontal force given by F = at - bt²,where t is the time in milliseconds from the instant the stick first contacts the object.If a = 1500 N/(ms)and b = 20 N/(ms)²,what is the speed of the object just after it comes away from the stick at t = 2.74 ms? A) 3.3 m/s B) 22 m/s C) 3.7 m/s D) 25 m/s

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Explanation:The impulse on the object is:$$J = \int_{0}^{2.74} (1500t - 20t^2)dt = [750t^2 - \frac{20}{3}t^3]_0^{2.74} = 2990 	ext{ N}\cdot 	ext{ms}$$The change in momentum is:$$J = mv_f - mv_i = 2990 	ext{ N}\cdot 	ext{ms}$$Since the object was initially stationary, we have:$$v_f = \frac{J}{m} = \frac{2990 	ext{ N}\cdot 	ext{ms}}{1.67 	ext{ kg}} = 1790 	ext{ m/s}$$

A Stationary 1