Question-4624 Tinku Tara June 3, 2023 Others 0 Comments FacebookTweetPin Question Number 4624 by sanusihammed last updated on 14/Feb/16 Answered by Yozzii last updated on 15/Feb/16 (12)Theequationofmotionoftheparticleisgivenasdds(v22)=−g−kv2.visafunctionofsassandvarerelatedtoeachotherbytimet.∴dds(v22)=12×2vdvds=vdvds.∴vdvds=−g−kv2=−(g+kv2)(∗)Atthemaximumheights=h,v=0andatv=u,s=0.∴Separatingthevariablesof(∗)andintegratingbetweenthelimitsv=0,v=uands=h,s=0,weget∫u0vg+kv2dv=∫0h−1ds12k∫u02kvg+kv2dv=−s∣0h(k≠0)12kloge(g+kv2)∣u0=−h=0h=−12k(loge(g)−loge(g+ku2))h=12kloge(g+ku2g)(k≠0,g+ku2>0⇒k>−g/u2)h=12kloge(1+ku2g)(13)(i)Letabetheaccelerationoftheparticleinms−2.aisgivenasafunctionoftimetintheforma=t+3.Sincea=dvdt⇒∫v0vdv=∫t0tadt⇒v∣v0v=∫t0t(t+3)dtv−v0=0.5t2+3t∣t0tv=v0+0.5t2+3t−0.5t02−3t0Att0=0,v0=2.∴v=2+3t+0.5t2−0−0v=2+3t+12t2(t⩾0).Att=2,v=2+3×2+12×4=10ms−1(ii)Sincev=dsdt⇒∫s0sds=∫t0tvdt.Att0=0,s0=0.∴∫0sds=∫0t(2+3t+12t2)dts∣0s=2t+32t2+16t3∣0ts−0=2t+32t2+16t3−0s=t(2+32t+16t2)(t⩾0)sisanincreasingfunctionforallt⩾0sincedsdt=v=2+3t+0.5t2⩾2+0+0=2>0.Theparticlemovesalongastraightlinesothatsinces⩾0,thedistance=displacement.Att=2s,s(2)=2(2+3+23)=343m.Att=4s,s(4)=4(2+6+83)=1283m∴requireddistance=s(4)−s(2)=128−343=943m Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Solve-the-system-of-congruences-2x-1-mod5-3x-2-mod7-4x-1-mod11-Next Next post: Question-70161 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.