kean (msk955) – Homework 02 – yao – (56725) 1This print-out should have 25 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 (part 1 of 2) 10.0 pointsExpress the vector~RABCDPRin terms of~A,~B,~C, and~D, the edges of aparallelogram.1.~R =~B +~A correct2.~R =~B +~D3.~R =~A +~D4.~R =~A −~D5.~R =~D −~A6.~R =~B −~A7.~R =~C +~B8.~R =~C +~D9.~R =~A −~B10.~R =~A −~CExplanation:Apply the parallelogram rule of addition:join the tail s of the two vectors~A and~B; theresultant vector is the diagonal of a paral lel-ogram formed with~A and~B as two of itssides.002 (part 2 of 2) 10.0 pointsExpress the vector~P in terms of~A,~B,~C, and~D ,1.~P =~A −~B correct2.~P =~B +~D3.~P =~A +~D4.~P =~C +~B5.~P =~A −~D6.~P =~D −~A7.~P =~C −~A8.~P =~B +~A9.~P =~B −~A10.~P =~C +~DExplanation:By the triangle method of addit ion~B +~P =~A~P =~A −~B .003 (part 1 of 2) 10.0 pointsA person walks 24◦north of east for 3.83 km.Another person walks due north, then dueeast to arrive at the same location.How far due north would this person walk?Correct answer: 1 .5578 km.Explanation:Let : d = 3.83 km andθ = 24◦.dxyθThe north component issin θ =ydy = d sin θ = (3. 83 km) sin 24◦=1.5578 km .004 (part 2 of 2) 10.0 pointsHow far would this person walk due east?kean (msk955) – Homework 02 – yao – (56725) 2Correct answer: 3.49888 km.Explanation:The east component iscos θ =xdx = d cos θ = ( 3.83 km) cos 24◦=3.49888 km .005 (part 1 of 2) 10.0 pointsA person walks the path shown. The tota ltrip consists of four straight-line paths.SNW E119 m437 m54.0◦161 m331 m22.0◦At the end of the walk, what is the mag-nitude of the person’s resultant displacementmeasured from the starting point?Correct answer: 525.643 m.Explanation:Let : d1= 119.0 m ,θ1= 0◦,d2= 437.0 m ,θ2= −90◦,d3= 161.0 m ,θ3= 54.0◦− 180.0◦= −126◦,d4= 331.0 m , andθ4= 180.0◦− 22.0◦= 158◦.dθ119 m437 m−126◦161 m331 m158◦∆x1= (119 m) cos 0◦= 119 m ,∆y2= (437 m) cos(−90◦) = −437 m ,∆x3= (161 m) cos(−126◦)= −94.6334 m ,∆y3= (161 m) sin(−126◦) = −130.252 m ,∆x4= (331 m) cos 158◦= −306.898 m , and∆y4= (331 m) sin 158◦= 123.995 m .∆xtot= 119 m−94.6334 m−306.898 m= −282.531 m and∆ytot= −437 m−130.252 m + 123.995 m= −443.257 m , sod =q(−282.531 m)2+ (−443.257 m)2=525.643 m .006 (part 2 of 2) 10.0 pointsWhat is the direction (measured from duewest, with counterclockwise positive) of theperson’s resultant displacement?Correct answer: 5 7 .4866◦.Explanation:tan θ =∆ytot∆xtotθ = tan−1∆ytot∆xtot= tan−1−443.257 m−282.531 m= 57.4866◦south of west.kean (msk955) – Homework 02 – yao – (56725) 3007 10.0 pointsAn escalator is 19.9 m long. If a person standson the escalator, it takes 49.8 s to ride fromthe bottom to the top.If a person walks up the moving escalatorwith a speed of 0. 598 m/s relative to theescalator, how long does it take the person toget to t he top?Correct answer: 19.9479 s.Explanation:Basic Concepts:~vpg= ~vpe+ ~vegveg=∆x∆tGiven:Let up be positive.∆x = 19.9 m∆t = 49.8 svpe= ±0.598 m/sSolution:Going up:vpg= vpe+ veg= vpe+∆x∆t∆tup=∆xvpg=∆xvpe+∆x∆t·∆t∆t=∆x∆tvpe∆t + ∆x=(19.9 m)(49. 8 s)(0.598 m/s)(49.8 s) + 19.9 m= 19.9479 s008 10.0 pointsA boat crosses a river of width 254 m in whi chthe current has a uniform speed of 2 .36 m/s.The pilot maintains a bearing (i. e. , the direc-tion in which the boat points) perpendicularto the river and a throttle setting to gi ve aconstant speed of 3.1 6 m/s relative to the wa-ter.What is the magnitude of the speed of theboat relative to a statio nary shore observer?Correct answer: 3 .94401 m/s.Explanation:The key to solving probl ems of this type isto first identify the observers (and frames), Sand S′. In this case, it is convenient to takethe shore observer to be S and to considera hypothetical observer S′drifting with theriver. Let the x axes of S and S′are along thestream, and the y axes be perpendicular to it(draw a picture). We will use capital lettersto denote the components of the vectors in S′.The “particle” being observed is the boat. Weare given that the river’s velocity componentsin S areux= 2.36 m/s , uy= 0 ,and the components of the speed of the boatin S′areVx= 0 , Vy= 3.16 m/s .Therefore, taking components of the equation~V =~v −~u ,we havevx= ux+ Vx= 2.36 m/svy= uy+ Vy= 3.94401 m/s ,which give the velocity components a s seen bythe shore observer (S). Then the magnitudeof the vector~v = (vx, vy) isv =qv2x+ v2y=q2.36 m/s2+ 3.16 m/ s2= 3.94401 m/s .009 (part 1 of 2) 10.0 pointsA river flows at a speed vr= 5.89 km/hr withkean (msk955) – Homework 02 – yao – (56725) 4respect to the shoreline. A boat needs to goperpendicular to the shoreline to reach a pieron the river’s other side. To do so, the boatheads upstream at an angle θ = 30◦from thedirection to the boat’s pier.Find the ratio of vbto vr, where vris definedabove and vbis the boat ’s speed wit h respectto the wa ter.1.vbvr= sin2θ2.vbvr=1tan θ3.vbvr= tan θ4.vbvr=1cos2θ5.vbvr=1sin θcorrect6.vbvr=1cos θ7.vbvr=1sin2θ8.vbvr= sin θ9.vbvr= cos2θ10.vbvr= cos θExplanation:Let : vbs=? , boat relat ive to shorevws= vr, water relative to shorevbs= vb, boat r el ative to water .This is a problem about relative velocity.See the figure below.LvbsvbwvwsθThe velocity of the boa t relative to theshore is given by−→vbs=−→vbw+−→vws.It is easy to see from t he figure above thatvws, vbwand vbsform a right triangle if theboat moves northward relative to the earth.Therefore,vws= vbwsin θ=⇒vbwvws=1sin θ.010 (part 2 of 2) 10.0 pointsIf the time taken for the boat to cross the riveris 19 .1 min, determine the width of the river.Correct answer: 3 .24757 k m.Explanation:By similar reasoning, we know relative tothe shore, the velocity of the boat isvbs=vwstan θ= 10.2018 km/hr .Therefore the time taken for the boat to crossthe river is given byt =Lvbs=Lvwstan θ=L tan θvws.=⇒ L =vwsttan θ=(5.89 km/hr) (19. 1 min) (1 hr/60 min)tan 30◦=(5.89 km/hr) (0.3 18334 hr)tan 30◦= 3.24757 km .011 10.0 pointsVertically falling rain makes slanted streakson the side windows of a moving automobile.If the streaks make an angle of 45◦, whatdoes thi s tell you about the r el ative speed ofthe car and the falling rain?kean (msk955) – Homework 02 – yao – (56725) 51.