Board Problems

BOAR D PROB LEMS  TRANSPOR  TRANSPORT TATION ENGINEERING Submitted by: Jan Roald B. Triumfante

em: 1.Probl   I ti sest i mat edt hatbet weent hehour sofnoonand7: 00P. M.t he speedofahi ghwayt r afficflowi ngpastt hei nt er sec t i onofEDSAandOr t i gas  Av enuei sappr oxi mat el y 3 2 S=t –9t  +15t+45kph  Wher e“ t ”i st henumberofhour spas tnoon

Requi r ed:   atwhatt i mebet weennoonand7: 00P. M.i st het r afficmovi ngt he f as t e s t ? 3 2 Gi ven:S=t –9t   +15t+45kph

For mul a:d/dx( un)=nun-1 3 2 S=t -9t +15t+45

Sol ut i on:

 Appl yt hef or mul ad/dx( un) =nun-1 2 dS/dt=3t   –18t+15=0

di vi det hewhol eequat i onby3 2 t   –6t+5=0

( t-1) ( t-5)=0 t=5hr s t=1hr  whent=1 3 2 S=( 1)   –9( 1)   +15( 1)+45

S=52kph  Whent=5 S=( 5)3  –9( 5)+15( 5)+45 S=

20kph

00 P. M.  Ther ef or et het r afficmov esf as t es ti sat1: em: 2.Probl   Thepeakhourf act orf orat r afficdur i ngrushhouri sequalt o0. 60  wi t hahi ghes t5mi nv ol umeof250v ehi cl est hespacemeanspeedoft het r affic i s90kph Requi r ed:   Comput et heflow oft r affici nve hi cl es /hour Gi ven:PHF=0. 60;t r afficv ol ume=250vehi cl es ;t i me=5mi n For mul a:PHF=flow oft r affic/t r affic( 60/5) Sol ut i on: PHF=flow oft r affic/t r affic( 60/5) 0. 60=flow oft r affic/250( 60/5) q=1800vehi cl es/hour obl em: 3.Pr  At2: 00P. M.anai r pl anet akesoffat340mphonanai r cr af tcar r i er .  Theai r cr af tcar r i ermov esduesout hat25kphi nt hesamedi r ect i onast he pl ane.At4: 05P. M.t hecommuni cat i onr angei nmi l esbet weent hepl aneand t hec ar r i e rwasl o s t . Requi r ed:   Det er mi net hecommuni cat i onr angei nmi l esbet wee nt hepl aneand t hec ar r i e r Gi ven:t i me=af t er2: 00P. M. ;at4: 05P. M. ,t=2hr sand5mi n=2+5/60

Sol uti on : =340t–25t=315t  x 25mi l es  x=315( 2+5/60)=656. obl em: 4.Pr  Acarwast r av el l i ngataspeedof50mph.Thedr i v ersaw ar oadbl oc k 80m aheadandst eppedont hebrakecausi ngt hecart odece l er at euni f orml yat 2 10m/s. Requi r ed:   Fi ndt hedi st ancef r om t her oadbl ockt ot hepoi ntwher et hecar st opped.Assumeper cept i onr eact i ont i me2sec. Formul a:S1 =vot 1 2 2 2  v =v +2aS o  Gi ven:d=80m;vo  =50mi /hr=22. 36m/s;S1  =22. 36( 2)=44. 72m 2 2 2 Sol ut i on : =vo    v +2aS 02  =22. 362  +2( 10) S2 S2  =25m 28m d=80–S1 –S2  =80–44. 72–25=10. em: 5.Probl   I ti sest i mat edt hatbet weent hehour sofnoonand7: 00P. M.t he speedofahi ghwayt r afficflowi ngpastt hei nt er sec t i onofEDSAandOr t i gas  Av enuei sappr oxi mat el y 3 2 S=t –9t  +15t+45kph  Wher e“ t ”i st henumberofhour spas tnoon

Requi r ed:   Atwhatt i mebet wee nnoonand7: 00P. M.i st het r afficmovi ngt he s l o we s t ? 3 2 Gi ven:S=t –9t   +15t+45kph

For mul a:d/dx( un)=nun-1 Sol ut i on:

3 2 S=t -9t +15t+45

 Appl yt hef or mul ad/dx( un) =nun-1 2 dS/dt=3t   –18t+15=0

di vi det hewhol eequat i onby3 2 t   –6t+5=0

( t-1) ( t-5)=0 t=5hr s t=1hr  whent=1 3 2 S=( 1)   –9( 1)   +15( 1)+45

S=52kph  Whent=5 S=( 5)3  –9( 5)+15( 5)+45 S=20kph

00P. M.  Ther ef or et het r afficmov essl owes ti sat5: em: 6.Probl   I ti sest i mat edt hatbet weent hehour sofnoonand7: 00P. M.t he speedofahi ghwayt r afficflowi ngpastt hei nt er sec t i onofEDSAandOr t i gas  Av enuei sappr oxi mat el y

3 2 S=t –9t   +15t+45kph

 Wher e“ t ”i st henumberofhour spas tnoon

Requi r ed : ati  Wh st hesl owes tspeedi ti smovi ngatt hi st i me? 3 2 Gi ven:S=t –9t   +15t+45kph

For mul a:d/dx( un)=nun-1 Sol ut i on:

3 2 S=t -9t +15t+45

 Appl yt hef or mul ad/dx( un) =nun-1 2 dS/dt=3t   –18t+15=0

di vi det hewhol eequat i onby3 2 t   –6t+5=0

( t-1) ( t-5)=0 t=5hr s t=1hr  whent=1 3 2 S=( 1)   –9( 1)   +15( 1)+45

S=52kph  Whent=5 S=( 5)3  –9( 5)+15( 5)+45

S=20kph em: 7.Probl   Thepeakhourf act orf orat r afficdur i ngrushhouri sequalt o0. 60  wi t hahi ghes t5mi nv ol umeof250v ehi cl est hespacemeanspeedoft het r affic i s90kph Requi r ed:   Comput et hedensi t yoft r affici nvehi cl es/km Gi ven:   PHF=0. 60;t r afficv ol ume=250ve hi cl es ;t i me=5mi n;us =90 Formul a:q=usK Sol ut i on: q=usK 1800=90K K=20 vehi cl es/hour em: 8.Probl   Thepeakhourf act orf orat r afficdur i ngrushhouri sequalt o0. 60  wi t hahi ghes t5mi nv ol umeof250v ehi cl est hespacemeanspeedoft het r affic i s90kph Requi r ed:   Comput et hedensi t yoft r affici nvehi cl es/km Gi ven:   PHF=0. 60;t r afficv ol ume=250ve hi cl es ;t i me=5mi n;us =90 Formul a:spaci ng=1000/K Sol ut i on: spaci ng=1000/20 Spaci ng=50m centert ocent er em: 9.Probl   Thepeakhourf act orf orat r afficdur i ngrushhouri sequalt o0. 60  wi t hahi ghes t10mi nv ol umeof500v ehi cl est hespacemeanspeedoft het r affic i s100kph Requi r ed:   Comput et heflow oft r affici nve hi cl es /hour

Gi ven:PHF=0. 60;t r afficv ol ume=500ve hi cl es ;t i me=10mi n Formul a:PHF=flowoft r affic/t r affic( 60/10) Sol ut i on: PHF=flow oft r affic/t r affic( 60/10) 0. 60=flow oft r affic/500( 60/10) cl es/hr q=1800vehi Probl em: 10.   Thepeakhourf act orf orat r afficdur i ngrushhouri sequalt o0. 60  wi t hahi ghes t10mi nv ol umeof500v ehi cl est hespacemeanspeedoft het r affic i s100kph Requi r ed:   Comput et heflow oft r affici nve hi cl es /hour Gi ven:PHF=0. 60;t r afficv ol ume=500ve hi cl es ;t i me=10mi n;us  =100 For mul a:q=usK Sol ut i on:

q=usK

1800=100K

K =18vehi cl es/hour