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Question Number 24025 by ajfour last updated on 11/Nov/17

Commented by ajfour last updated on 11/Nov/17

Find the closest distance of approach of B and R (if it is yet to come) and time after which they are closest. Situation has been shown for t=0 . (Assume constant velocities) B continues along its road and so does R . B is the blue vehicle, and R the red one, both roads are inclined at ∠ α to each other.

FindtheclosestdistanceofapproachofBandR(ifitisyettocome)andtimeafterwhichtheyareclosest.Situationhasbeenshownfort=0.(Assumeconstantvelocities)BcontinuesalongitsroadandsodoesR.Bisthebluevehicle,andRtheredone,bothroadsareinclinedat∠αtoeachother.

Commented by ajfour last updated on 12/Nov/17

Commented by ajfour last updated on 12/Nov/17

Let shortest distance of approach be d_s . let initial distance be L_0 . L_0 ^2 =(a+bcos α)^2 +b^2 sin^2 α d_s =L_0 sin (β+θ) v_(net) ^2 =(u−vcos α)^2 +v^2 sin^2 α t_s =((L_0 cos (β+θ))/v_(net) ) tan β =((bsin α)/(a+bcos α)) tan θ = ((vsin α)/(u−vcos α)) d_s =L_0 sin (β+θ) =L_0 (sin βcos θ+cos βsin θ) d_s =bsin α(((u−vcos α)/v_(net) ))+ +(a+bcos α)(((vsin α)/v_(net) )) t_s = L_0 cos (β+θ)/v_(net) =L_0 (cos βcos θ−sin βsin θ) =(a+bcos α)(((u−vcos α)/v_(net) ^2 ))− (bsin α)(((vsin α)/v_(net) ^2 )) . where, v_(net) =(√((u−vcos α)^2 +v^2 sin^2 α)) .

Letshortestdistanceofapproachbeds.letinitialdistancebeL0.L02=(a+bcosα)2+b2sin2αds=L0sin(β+θ)vnet2=(u−vcosα)2+v2sin2αts=L0cos(β+θ)vnettanβ=bsinαa+bcosαtanθ=vsinαu−vcosαds=L0sin(β+θ)=L0(sinβcosθ+cosβsinθ)ds=bsinα(u−vcosαvnet)++(a+bcosα)(vsinαvnet)ts=L0cos(β+θ)/vnet=L0(cosβcosθ−sinβsinθ)=(a+bcosα)(u−vcosαvnet2)−(bsinα)(vsinαvnet2).where,vnet=(u−vcosα)2+v2sin2α.

Commented by Physics lover last updated on 12/Nov/17

thats exactly how i thought.

thatsexactlyhowithought.

Answered by mrW1 last updated on 12/Nov/17

at t=0, the distance between B and R: L_0 =(√((a+bcos α)^2 +(bsin α)^2 )) =(√(a^2 +b^2 +2abcos α)) at t: L=(√([a−ut+(b+vt)cos α]^2 +[(b+vt)sin α]^2 )) let D=L^2 D=[a−ut+(b+vt)cos α]^2 +[(b+vt)sin α]^2 =[a+bcos α−(1−cos α)ut]^2 +[bsin α+sin αvt]^2 =(a+bcos α)^2 −2(a+bcos α)(1−cos α)ut+(1−cos α)^2 u^2 t^2 +b^2 sin^2 α+2bsin^2 αvt+sin^2 α v^2 t^2 =[(a+bcos α)^2 +b^2 sin^2 α]−2[u(a+bcos α)(1−cos α)+vbsin^2 α]t+[u^2 (1−cos α)^2 +v^2 sin^2 α]t^2 =[u^2 (1−cos α)^2 +v^2 sin^2 α]t^2 −2[u(a+bcos α)(1−cos α)+vbsin^2 α]t+[a^2 +b^2 +2abcos α] =pt^2 −2qt+r =p(t−(q/p))^2 +(r−(q^2 /p)) since p>0, there is a minimum for D at t=t_m =(q/p)=((u(a+bcos α)(1−cos α)+vbsin^2 α)/(u^2 (1−cos α)^2 +v^2 sin^2 α)) D_(min) =r−(q^2 /p) =[a^2 +b^2 +2abcos α]−(([u(a+bcos α)(1−cos α)+vbsin^2 α]^2 )/(u^2 (1−cos α)^2 +v^2 sin^2 α)) L_(min) =(√([a^2 +b^2 +2abcos α]−(([u(a+bcos α)(1−cos α)+vbsin^2 α]^2 )/(u^2 (1−cos α)^2 +v^2 sin^2 α))))

att=0,thedistancebetweenBandR:L0=(a+bcosα)2+(bsinα)2=a2+b2+2abcosαatt:L=[a−ut+(b+vt)cosα]2+[(b+vt)sinα]2letD=L2D=[a−ut+(b+vt)cosα]2+[(b+vt)sinα]2=[a+bcosα−(1−cosα)ut]2+[bsinα+sinαvt]2=(a+bcosα)2−2(a+bcosα)(1−cosα)ut+(1−cosα)2u2t2+b2sin2α+2bsin2αvt+sin2αv2t2=[(a+bcosα)2+b2sin2α]−2[u(a+bcosα)(1−cosα)+vbsin2α]t+[u2(1−cosα)2+v2sin2α]t2=[u2(1−cosα)2+v2sin2α]t2−2[u(a+bcosα)(1−cosα)+vbsin2α]t+[a2+b2+2abcosα]=pt2−2qt+r=p(t−qp)2+(r−q2p)sincep>0,thereisaminimumforDatt=tm=qp=u(a+bcosα)(1−cosα)+vbsin2αu2(1−cosα)2+v2sin2αDmin=r−q2p=[a2+b2+2abcosα]−[u(a+bcosα)(1−cosα)+vbsin2α]2u2(1−cosα)2+v2sin2αLmin=[a2+b2+2abcosα]−[u(a+bcosα)(1−cosα)+vbsin2α]2u2(1−cosα)2+v2sin2α

Commented by ajfour last updated on 12/Nov/17

thank you Sir, please view my solution sir ..

thankyouSir,pleaseviewmysolutionsir..

Commented by mrW1 last updated on 14/Nov/17

your solution is more simple and direct.

yoursolutionismoresimpleanddirect.

Commented by ajfour last updated on 15/Nov/17

thanks for the appreciation, Sir!

thanksfortheappreciation,Sir!

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