Question

Chapter 02, Problem 25 A jogger accelerates from rest to 3.11 m/s in 2.16 s. A car accelerates from 19.4 to 31.4 m/s also in 2.16 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 2.16 s? (a) Number Units (b) Number Units (c) Number Units

Answer 1

(Given V= 3.11 m/s t2-16 sec using equation of molion Va votat a = V-Volt Vo 20 , t= 2.16 sec a = 3•1/2-16 - 1-439 m/s² a = 1-439 m/s2 6 Vo = 19.4 m/s to a uspoed of V = 31•4 m/s in remo interval t = 2.16 sec Acceleration Aca V-VO t ac = 31.4-1904/2016 = 5.555 m/s © using equation x= vot / cat2 for jogger J 2 aj = VO + 24 at ² Pj- 0+ (5-555)(-16) for cal * j = 12.96 m He = vott hat? c=(19.4X2016) Ź (5-555)(2-16) 2 xe = + ac = 41.904 + 12.96 1- 54.864 m cal travels X-Mj - 54-864-12-96 ©1*-*; 41-9040)