The distance, in feet, a moving object has traveled after t seconds is given by 2t/(4 + t). find the acceleration of the object after 5 seconds. (round your answer to three decimal places.)
Hi! Let me help you! a = (Vf - Vi)/t ; where distance d = [2(t)]/(4+t), t = 5secs, and Vi = 0 a = [(2t)/(4+t)]/t <---- working equation a = {[2(5)]/9}/5 <---- cancel 5 a = 2/9 ft/s^2 <---- Answer
