Physics 8.01 Assignment 11.1Figure 1: sunsetWe treat the sun as an object that is rotating about the earth withuniform angular velocity, and that rises and sets vertically (this isassured by being on the equator on the equinox). We first need toknow the angle sun must rotate through during the sunset (α). Ifwe can then get the angular speed of the sun as we see it we can getthe time it takes for the sunset.α is small ⇒ α tan α =DR1D =13.92 × 108mR =1.50 × 1011mα =9.28 × 10−3radiansThe angular velocity follows from fact sun will rotate 2π radiansduring 24 hours:ωsun=2π24×60×60=7.27 × 10−5radianssThe time is the angular size of the sun divided by the angularvelocity, analogous to the case with distance and linear velocity.Tsunset=αωsun=9.28×10−37.27×10−5= 128s ≈ 2min.21.2a)Dimensional consistency ofv :msx : ma :ms2RHS:m2s2LHS:m2s2+ m ×ms2∝m2s2√b)Both LHS and RHS have the same dimensions. These are nowmultiplied by mass with units kg.kg ×m2s2: Joules31.3a)Figure 2: Problem 34b)The Person who spends less time walking and more time run-ning will win. Initially they walk along together, but Tim startsrunning sooner because he has only to walk 1/2 the time it will takehim, not 1/2 the distance which makes Rick take longer to reachD as the graph shows (note the specific times designated on thegraph).c)Rickt1=D/2Vw(t1/ : the time that Rick reaches D/2)tR− t1=D/2VrtR=D2VwVrVr+Vwd)VR=DtR=2VrVwVr+Vwe)TimtT2Vr+tT2Vw= DtT=D(Vr+Vw)/251.4I:a =vf−vi∆tDefinition of average accelerationvi= 0 (starts at rest)∆t =6svf=90m/sII:Distance ∼ area under v(t) curveIII:amax=45⇒ Dminarchived by a straight line from a point on t axisto vf=90m/s at tf=6. ∆t =90/45 = 2 ⇒ starts moving at 4.Any line before 4 to 90 m/s has a lower slope so we wouldn’t violatethe amaxcondition. To get twice the distance I look for a line thathas twice the area under it. The easiest choice is the line from 2sto 90 m/s. The triangle under it has twice the base and the samealtitude, hence twice the area.Figure 3: Problem 461.5a = −10m/s2∆t = 2.0g: hitting the ground r: reflection from the groundf =0.71These are the three basic kinematic equations written with arbitrarytime for the initial conditions (instead of the conventional t=0).Eq. 1: x(t)=x(t0)+v(t)(t − t0)+12a(t − t0)2Eq. 2: v(t)=v(t0)+a(t − t0)Eq. 3: v2(t) − v2(t0)=+2a[x(t) − x(t0)]We should pay attention that t0in these equations is the initial time,not necessarily zero. So for each region as you see in the picture thereare different initial conditions for v0and t0. We should first writethe appropriate y(t) for each region. Setting the two y(t)’s for thetwo balls gives us the time of the collision time tc.Forthefirstball :Eq.1 ⇒ tg1=√2=1.41sEq.3 ⇒ vg1=√200 = 14.1m/svr1= fvg1=10m/svg2= −vr1∧ Eq.2 ⇒ tg2− tg1=2×1010= 2.0Because the time it takes that second ball hit the ground is again1.41 and because 2.0 +1.41 = 2.0+ 1.41! The second hits the firstball when the first ball hits the ground itself.However, if you change ∆t say 1.8 you would get a different answer.I also write the systematic way of solving this problem:Between the first hit and the second one ball #1 is at y1(t) and ball32 is at y2(t)y1(t)=0+14.1f(t − 1.41) − 5(t − 1.41)2y2(t)=10− 5(t − 2)2Collision time: tc7y2(tc)=y1(tc)Solving this with will give you 3.41 for˜t as expected! (But you haveto be careful that your solution is in the region of validity of bothy1and y2.Figure 4:
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