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Unformatted text preview: alameri (aoa434) – Hwk02 – Stokes – (19102)1This printout should have 17 questions.Multiplechoice questions may continue onthe next column or page – find all choicesbefore answering.This is Online Homework No. 2 (OH02).All answers should be submitted online before0700 on Tuesday 3 Feb (UAE time). Computers and network connections can give trouble,so do not wait until the last minute!00110.0 pointsA car travels along a straight stretch of road.It proceeds for 15.7 mi at 54 mi/h, then26.8 mi at 42 mi/h, and finally 33.7 mi at36.7 mi/h.What is the car’s average velocity duringthe entire trip?Correct answer: 41.254 mi/h.Explanation:The total distance the car traveled isΔd=dA+dB+dC= 76.2 miThe total time the car spent on the road isΔt=dAvA+dBvB+dCvC=15.7 mi54 mi/h+26.8 mi42 mi/h+33.7 mi36.7 mi/h= 1.84709 h.Hence the average velocity isv=ΔdΔt=76.2 mi1.84709 h= 41.254 mi/h.002(part 1 of 4) 10.0 pointsThe position versus time for a certain objectmoving along thexaxis is shown. The object’s initial position is5 m.642246123456789bbbbbbtime (s)position(m)Find the instantaneous velocity at 0.5 s.Correct answer: 10 m/s.Explanation:The instantaneous velocity is the slope ofthe tangent line at that point.v=(5 m)(5 m)(1 s)(0 s)=10 m/s.003(part 2 of 4) 10.0 pointsFind the instantaneous velocity at 6 s.Correct answer:3 m/s.Explanation:v=(4 m)(2 m)(7 s)(5 s)=3 m/s.004(part 3 of 4) 10.0 pointsFind the average velocity between 0 s and 3 s.Correct answer: 2.33333 m/s.Explanation:Average velocity is the change in positionbetween the two points divided by the changein time between the two points.v=(2 m)(5 m)(3 s)(0 s)=2.33333 m/s.005(part 4 of 4) 10.0 pointsFind the average velocity over the whole timeshown.Correct answer: 0.555556 m/s.Explanation:v=(0 m)(5 m)(9 s)(0 s)=.555556 m/s.00610.0 pointsalameri (aoa434) – Hwk02 – Stokes – (19102)2Which of the following describe possible scenarios?A) An object has zero instantaneous velocityand nonzero acceleration.B) An object has negative acceleration andis speeding up.C) An object has positive acceleration andconstant velocity.D) An object has positive velocity and zeroacceleration.E) An object has increasing positive position and negative velocity.F) An object has decreasing positive position and negative acceleration.1.All are possible.2.A, D, and F only.3.B, C, and D only.4.None are possible.5.A, B, D, E, and F only.6.A, B, D, and F only.correct7.A, D, E, and F only.8.D and F only.9.A, C, D, and F only.10.A, B, C, D, and F only.Explanation:Option C is impossible because an objectwhich has a constant velocity MUST have azero acceleration. Acceleration is the rate ofchange in velocity. If the velocity is unchanging then the rate of change is zero....
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This note was uploaded on 02/16/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
 Spring '09
 NAJAFZADEH
 mechanics, Work

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