# hw2 - al-ameri(aoa434 – Hwk02 – Stokes –(19102)1This...

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Unformatted text preview: al-ameri (aoa434) – Hwk02 – Stokes – (19102)1This print-out should have 17 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This is Online Homework No. 2 (OH-02).All answers should be submitted online before0700 on Tuesday 3 Feb (UAE time). Comput-ers 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 thex-axis is shown. The ob-ject’s initial position is-5 m.-6-4-2246123456789bbbbbbtime (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 pointsal-ameri (aoa434) – Hwk02 – Stokes – (19102)2Which of the following describe possible sce-narios?A) An object has zero instantaneous velocityand non-zero 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 posi-tion and negative velocity.F) An object has decreasing positive posi-tion 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 unchang-ing 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.

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hw2 - al-ameri(aoa434 – Hwk02 – Stokes –(19102)1This...

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