a)=o'Acos(o1=Io'Acos(@)=0A (d)Using r((=[Acos(ot)]i, so when cos(ot=0, we have r=Acos(ot)=0 Thus ot ==+ki k=0.2. and t k=0,1,2, 20 Then the time is Imin 2o(k=0)during (2 0s at which the position vector first attain a magnitude of o m At this time, the magnitude of the velocity vector is Asin(@t) OA The magnitude of the acceleration vector is a()=o'Acos(o =Fo'Acos(o 7)=0m/s 2. One model for the motion of a particle moving in a resistive medium suggests that the spee decrease exponentially according to the expression v(o=voe, where vo is the speed of th particle when t=0 s and B is a positive constant (a)How long will it take the particle to reach half its initial speed? (b) What distance does the particle traverse during the time interval calculated in part(a)? (c) Through what distance does the particle move before it is brought to rest? Solution (a)v(=voe In 2 =2→-B=-m2 (b) The distance that the particle traverse is v()dt voe 1) (c) According to the expression l()=vne-", When t→∞,v()→0 The distance the particle moves before it is brought to rest is S= v(dt=Lvoe"dr =-0(0-1)=o Ba t A t A A 2 2 2 ( ) cos( ) cos( ) ω ω π = −ω ω = −ω ω = v (d) Using r t A t i ˆ ( ) =[ cos(ω )] r , so when cos(ωt) = 0 , we have r = Acos(ωt) = 0 . Thus 0,1,2,... 2 t = ± kπ k = π ω and 0,1,2,... 2 = ± k = k t ω π ω π Then the time is ( 0) 2 tmin = k = ω π during t ≥ 0s at which the position vector first attain a magnitude of 0 m. At this time, the magnitude of the velocity vector is ) m/s 2 v(t) Asin( t) Asin( ωA ω π = −ω ω = −ω ω = v . The magnitude of the acceleration vector is 2 2 2 ) 0 m/s 2 ( ) = − cos( ) = − cos( = ω π a t ω A ωt ω A ω v 2. One model for the motion of a particle moving in a resistive medium suggests that the speed decrease exponentially according to the expression t v t v e−β = 0 ( ) , where 0 v is the speed of the particle when t = 0 s and β is a positive constant. (a) How long will it take the particle to reach half its initial speed? (b) What distance does the particle traverse during the time interval calculated in part (a)? (c) Through what distance does the particle move before it is brought to rest? Solution: (a) β β β β ln 2 ln 2 2 1 2 ( ) 0 = 0 = ⇒ = ⇒ − = − ⇒ = − − e t t v v t v e t t (b) The distance that the particle traverse is β β β β β β β β 2 ( )d d ( 1) 0 ln 2 0 ln2 0 0 ln2 0 0 ln2 0 v e v e v S v t t v e t t t = = = − = − − = − − − ∫ ∫ (c) According to the expression t v t v e−β = 0 ( ) , When t → ∞ , v(t) → 0 . The distance the particle moves before it is brought to rest is β β β β β 0 0 0 0 0 0 0 ( )d d (0 1) v v e v S v t t v e t - t - t = = = − = − − = ∞ ∞ ∞ ∫ ∫