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Question-4624




Question Number 4624 by sanusihammed last updated on 14/Feb/16
Answered by Yozzii last updated on 15/Feb/16
(12) The equation of motion of the particle  is given as              (d/ds)((v^2 /2))=−g−kv^2 .  v is a function of s as s and v are   related to each other by time t.  ∴ (d/ds)((v^2 /2))=(1/2)×2v(dv/ds)=v(dv/ds).  ∴ v(dv/ds)=−g−kv^2 =−(g+kv^2 )      (∗)  At the maximum height s=h,v=0 and  at v=u,s=0.  ∴ Separating the variables of (∗) and  integrating between the limits   v=0,v=u and s=h,s=0, we get  ∫_u ^0 (v/(g+kv^2 ))dv=∫_0 ^h −1ds  (1/(2k))∫_u ^0 ((2kv)/(g+kv^2 ))dv=−s∣_0 ^h     (k≠0)  (1/(2k))log_e (g+kv^2 )∣_u ^0 =−h=0  h=((−1)/(2k))(log_e (g)−log_e (g+ku^2 ))  h=(1/(2k))log_e (((g+ku^2 )/g))  (k≠0,g+ku^2 >0⇒k>−g/u^2 )  h=(1/(2k))log_e (1+((ku^2 )/g))       (13)(i) Let a be the acceleration of the   particle in ms^(−2) . a is given as a function  of time t in the form a=t+3.   Since a=(dv/dt)⇒∫_v_0  ^v dv=∫_t_0  ^t adt  ⇒v∣_v_0  ^v =∫_t_0  ^t (t+3)dt  v−v_0 =0.5t^2 +3t∣_t_0  ^t   v=v_0 +0.5t^2 +3t−0.5t_0 ^2 −3t_0   At t_0 =0,v_0 =2.  ∴v=2+3t+0.5t^2 −0−0  v=2+3t+(1/2)t^2     (t≥0).  At t=2, v=2+3×2+(1/2)×4=10ms^(−1)   (ii) Since v=(ds/dt)⇒∫_s_0  ^s ds=∫_t_0  ^t vdt.  At t_0 =0,s_0 =0.  ∴∫_0 ^s ds=∫_0 ^t (2+3t+(1/2)t^2 )dt  s∣_0 ^s =2t+(3/2)t^2 +(1/6)t^3 ∣_0 ^t   s−0=2t+(3/2)t^2 +(1/6)t^3 −0  s=t(2+(3/2)t+(1/6)t^2 )  (t≥0)  s is an increasing function for all t≥0  since (ds/dt)=v=2+3t+0.5t^2 ≥2+0+0=2>0.  The particle moves along a straight   line so that since s≥0, the distance=displacement.  At t=2 s,s(2)=2(2+3+(2/3))=((34)/3)m.  At t=4 s, s(4)=4(2+6+(8/3))=((128)/3) m   ∴ required distance=s(4)−s(2)=((128−34)/3)=((94)/3)m
(12)Theequationofmotionoftheparticleisgivenasdds(v22)=gkv2.visafunctionofsassandvarerelatedtoeachotherbytimet.dds(v22)=12×2vdvds=vdvds.vdvds=gkv2=(g+kv2)()Atthemaximumheights=h,v=0andatv=u,s=0.Separatingthevariablesof()andintegratingbetweenthelimitsv=0,v=uands=h,s=0,wegetu0vg+kv2dv=0h1ds12ku02kvg+kv2dv=s0h(k0)12kloge(g+kv2)u0=h=0h=12k(loge(g)loge(g+ku2))h=12kloge(g+ku2g)(k0,g+ku2>0k>g/u2)h=12kloge(1+ku2g)(13)(i)Letabetheaccelerationoftheparticleinms2.aisgivenasafunctionoftimetintheforma=t+3.Sincea=dvdtv0vdv=t0tadtvv0v=t0t(t+3)dtvv0=0.5t2+3tt0tv=v0+0.5t2+3t0.5t023t0Att0=0,v0=2.v=2+3t+0.5t200v=2+3t+12t2(t0).Att=2,v=2+3×2+12×4=10ms1(ii)Sincev=dsdts0sds=t0tvdt.Att0=0,s0=0.0sds=0t(2+3t+12t2)dts0s=2t+32t2+16t30ts0=2t+32t2+16t30s=t(2+32t+16t2)(t0)sisanincreasingfunctionforallt0sincedsdt=v=2+3t+0.5t22+0+0=2>0.Theparticlemovesalongastraightlinesothatsinces0,thedistance=displacement.Att=2s,s(2)=2(2+3+23)=343m.Att=4s,s(4)=4(2+6+83)=1283mrequireddistance=s(4)s(2)=128343=943m

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