Chapter 2Page 1CHAPTER 2 - Describing Motion: Kinematics in One Dimension1.We find the time fromaverage speed = d/t;15 km/h = (75 km)/t, which gives t= 5.0 h.2.We find the average speed fromaverage speed = d/t= (280 km)/(3.2 h) = 88 km/h.3.We find the distance traveled fromaverage speed = d/t;(110 km/h)/(3600 s/h) = d/(2.0 s), which gives d= 6.1×10–2km = 61 m.4.We find the average velocity fromæ= (x2– x1)/(t2– t1) = (– 4.2 cm – 3.4 cm)/(6.1 s – 3.0 s) = – 2.5 cm/s (toward – x).5.We find the average velocity fromæ= (x2– x1)/(t2– t1) = (8.5 cm – 3.4 cm)/[4.5 s – (– 2.0 s)] = 0.78 cm/s (toward + x).Because we do not know the total distance traveled, we cannot calculate the average speed.6.(a)We find the elapsed time before the speed change fromspeed = d1/t1;65 mi/h = (130 mi)/t1, which gives t1= 2.0 h.Thus the time at the lower speed ist2= T– t1= 3.33 h – 2.0 h = 1.33 h.We find the distance traveled at the lower speed fromspeed = d2/t2;55 mi/h = d2/(1.33 h), which gives d2= 73 mi.The total distance traveled isD= d1+ d2= 130 mi + 73 mi = 203 mi.(b)We find the average speed fromaverage speed = d/t= (203 mi)/(3.33 h) = 61 mi/h.Note that the average speed is not!(65 mi/h + 55 mi/h). The two speeds were not maintained forequal times.7.Because there is no elapsed time when the light arrives, the sound travels one mile in 5 seconds.We find the speed of sound fromspeed = d/t= (1 mi)(1610 m/1 mi)/(5 s) ≈300 m/s.8.(a)We find the average speed fromaverage speed = d/t= 8(0.25 mi)(1.61×103m/mi)/(12.5 min)(60 s/min) = 4.29 m/s.(b)Because the person finishes at the starting point, there is no displacement;thus the average velocity isæ= ∆x/∆t= 0.9.(a)We find the average speed fromaverage speed = d/t= (160 m + 80 m)/(17.0 s + 6.8 s) = 10.1 m/s.(b)The displacement away from the trainer is 160 m – 80 m = 80 m; thus the average velocity isæ= ∆x/∆t= (80 m)/(17.0 s + 6.8 s) = + 3.4 m/s, away from trainer.10.Because the two locomotives are traveling with equal speeds in opposite directions, each locomotive will travel halfthe distance, 4.25 km. We find the elapsed time fromspeed1= d1/t;(95 km/h)/(60 min/h) = (4.25 km)/t, which gives t= 2.7 min.
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