kean (msk955) – Homework 01 – yao – (56725) 1This print-out should have 25 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsConvert the volume 12.9 in.3to m3, recal lingthat 1 i n. = 2.54 cm and 100 cm = 1 m.Correct answer: 0.00021139 3 m3.Explanation:V =12.9 in.32.54 cm1 in31 m100 cm3=0.00021139 3 m3.002 10.0 pointsIn May 1998, forest fires in southern Mexicoand Guatemala spread smoke all the way toAustin. T hose fires consumed forest l a nd at arate o f 21800 acres/week.On the average, how many square metersof forest are burned down every minute? Anacre is an area equivalent to that of a rectangle60.5 yd wide and 80 yd long, there are 36inches in one yard, and t here are 2.54 cm inone inch.Correct answer: 8752.13 m2/min.Explanation:Let : R = 21800 acres/wk ,w = 60.5 yd , andℓ = 80 yd .R =21800 acreswk1 wk7 d1 d24 h×1 h60 min(60.5 yd) (80 yd)acre×36 in1 yd22.54 cm1 in21 m100 cm2=8752.13 m2/min .003 10.0 pointsA creature moves at a speed of2.25 furlongs/fortnight (not a very commonunit of speed).If 1.0 furlong = 220 yards, 1 fortnight =14 days, 1 ya r d = 0. 9144 meter, and 1 day= 86400 sec, find the speed of the creature(which is probably a snail).1. 13529 m/s2. 1.51534 × 10−6m/s3. 0.0003741 96 m/s correct4. 0.0733425 m/s5. 6.54803 × 108m/s6. 3.34083 × 106m/s7. 11312 m/sExplanation:v = (2.25 furlongs/fortnight) ×220 ydfurlong×0.9144 myd×1 fortni ght14 d×1 d86400 s= 0.000374196 m/s .004 10.0 pointsChuck drove 3 5 mi from Austin to San Mar-cos in 40 min, stopped 30 min for a ham-burger, and then drove 45 mi to San Antonioin 50 min. What was Chuck’s average speed?1. 56 mph2. 43 mph3. 30 mph4. 50 mphkean (msk955) – Homework 01 – yao – (56725) 25. 53 mph6. 40 mph correctExplanation:v =total distancetime=80 mi120 min60 min1 h=40 mph .005 10.0 pointsA student t hrows a baseball at a large gong44 m away and hears the sound of the gong1.35556 s later. The speed of sound in air is330 m/s.What was the average speed of the base-ball on its way to the gong? (For simplici ty,assume its trajectory to be a straight line.)Correct answer: 36 m/s.Explanation:Let : d = 44 m,t = 1.35556 s, andv = 330 m/s .It takests=dvs=44 m330 m/s= 0.133333 sfor the sound to travel back from the gong, sothe flight time of the baseball was onlytb= ttot− ts= 1.35556 s − 0.133333 s = 1.22222 sand the baseball’s average speed wasvb=∆d∆t=44 m1.22222 s=36 m/s .006 (part 1 of 2) 10.0 pointsA runner is jogging at a st eady 7.4 km/hr.When the runner is 6.5 k m from the finishline, a bird begi ns flying from the runner tothe finish line at 22.2 km/hr (3 times as fastas t he runner). When the bird reaches thefinish line, it turns around and flies back tothe runner.LvbvrfinishlineHow far does the bird t r avel? Even thoughthe bird is a dodo, assume that it occupiesonly one point in space (a “zero” length bird)and that it can turn without loss of speed.Correct answer: 9.75 km.Explanation:Let : vr= 7.4 km/hr ,L = 6.5 km , andvb= 3 vr.finishlineL1dr1db1The runner travels a distance x until theencounter with the bird. In that time, the birdhas traveled a dist a nce L +(L − x) = 2 L − x .The bird travel 3 times as fast as the runnerduring this time frame, sodb= 3 dr2 L − x = 3 x2 L = 4 xx =12Land the bird flies a distance ofdb= 2 L −12L =32L =32(6.5 km)=9.75 k m .kean (msk955) – Homework 01 – yao – (56725) 3007 (part 2 of 2) 10.0 pointsAfter this first encounter, the bird then turnsaround and flies from the runner back to thefinish li ne, turns around again and flies backto the runner. The bird repeats the back andforth trips until the runner reaches the finishline.How far does the bird travel from the be-ginning (including the distance traveled to thefirst encounter)?Correct answer: 19.5 k m.Explanation:The distance r emaining for the runner afterthe first encounter is L1=12L , and again thebird will fly 3 times as far as the runner unt ilthe next encounter. This pattern repeats overthe entire original distance, sodb= 3 L = 3 (6.5 km) =19.5 km .008 (part 1 of 2) 10.0 pointsConsider a moving object whose positi on xis pl otted a s a function of the time t. Theobject moved in different ways during thetime intervals denoted I, II and III on thefigure.2 4 6246txI II IIIDuring these three intervals, when was theobject’s speed highest? Do not confuse thespeed with the velocity.1. During interval II2. Same speed during each of the three in-tervals.3. During interval III correct4. Same speed during intervals II and III5. During interval IExplanation:The velocity v is the slope of the x(t) curve;the magnitude v = |v| of this slo pe is thespeed. The curve is steepest (in absolutemagnitude) during the interval III a nd that iswhen the object had the highest speed.009 (part 2 of 2) 10.0 pointsDuring which interval(s) did the object’s ve-locity remain constant?1. During interval II only2. During interval I only3. During interval III only4. During each of the three intervals cor-rect5. During none of the three intervalsExplanation:For each of the three intervals I, II or III, thex(t) curve i s linear, so its slop e (the velocityv) is constant. Between the intervals, thevelocity changed in an abrupt manner, but itdid remain constant during each interval.010 10.0 pointsThe plot shows x(t) for a particle.xtWhich statement is correct about the mo-tion?1. The particle at first speeds up, then slowsdown, then speeds up again.2. The particle moves at constant speed, butnot at constant velocity.kean (msk955) – Homework 01 – yao – (56725) 43. The particle slows down, comes momen-tarily to a rest, then returns with increasingspeed to its starting point. correct4. The particle follows the path of a projec-tile, like a thrown baseball.5. The particle moves along a portion of acircle, as the sketch shows directl y.Explanation:Since vxis the slope of x(t), the particl eslows down, comes to momentary rest, andthen reverses direction, increasing speed as itreturns to its starting point.011 10.0 pointsA car is parked in i ts dri veway facing awayfrom the street. Its owner backs the car outonto the street. As the car starts from restand gains speed up to 5 mph backing out ofthe driveway, its1. velocity points toward the rear of the car,and its accelerati on points toward its rear.correct2. velocity points toward the rear of the car,while its