Phn I: Chuyn ng c hc

Phn I: Chuyn ng c hcI.Chuyn ng ca mt hay nhiu vt trn mt ng thng

1/.lc 6 gi, mt ngi i xe p xut pht t A i v B vi vn tc v1=12km/h.Sau 2 gi mt ngi i b t B v A vi vn tc v2=4km/h. Bit AB=48km/h.

a/. Hai ngi gp nhau lc my gi?ni gp nhau cch A bao nhiu km?

B/. Nu ngi i xe p ,sau khi i c 2km ri ngi ngh 1 gi th 2 ngi gp nhau lc my gi?ni gp nhau cch A bao nhiu km?

c. v th chuyn ng ca 2 xe trn cng mt h trc ta

d. v th vn tc -thi gian ca hai xe trn cung mt h trc ta .

2/.Mt ngi i xe p t A n B vi d nh mt t=4h. do na qung ng sau ngi tng vn tc thm 3km/h nn n sm hn d nh 20 pht.

A/. Tnh vn tc d nh v qung ng AB.

B/. Nu sau khi i c 1h, do c vic ngi y phi gh li mt 30 ph.hi on ng cn li ngi phi i v vn tc bao nhiu n ni nh d nh ?

3/. Mt ngi i b khi hnh t C n B vi vn tc v1=5km/h. sau khi i c 2h, ngi ngi ngh 30 ph ri i tip v B.Mt ngi khc i xe p khi hnh t A (AC >CBv C nm gia AB)cng i v B vi vn tc v2=15km/h nhng khi hnh sau ngi i b 1h.

a. Tnh qung ng AC v AB ,Bit c 2 ng n B cng lc v khi ngi i b bt u ngi ngh th ngi i xe p i c 3/4 qung ng AC.

b*.V th v tr v th vn tc ca 2 ngi trn cng mt h trc ta

c. gp ngi i b ti ch ngi ngh,ngi i xe p phi i vi vn tc bao nhiu?

4/. Mt thuyn nh c chuyn ng ngc dng nc lm rt mt ci phao.Do khng pht hin kp,thuyn tip tc chuyn ng thm 30 ph na th mi quay li v gp phao ti ni cch ch lm rt 5km. Tm vn tc ca dng nc,bit vn tc ca thuyn i vi nc l khng i.

5/. Lc 6h20ph hai bn ch nhau i hc vi vn tc v1=12km/h.sau khi i c 10 ph mt bn cht nh mnh b qun bt nh nn quay li v ui theo vi vn tc nh c.Trong lc bn th 2 tip tc i b n trng vi vn tc v2=6km/h v hai bn gp nhau ti trng.

A/. Hai bn n trng lc my gi ? ng gi hay tr hc?

B/. Tnh qung ng t nh n trng.

C/. n ni ng gi vo hc ,bn quay v bng xe p phi i vi vn tc bng bao nhiu?Hai bn gp nhau lc my gi?Ni gp nhau cch trng bao xa?

6/. Hng ngy t 1 xut pht t A lc 6h i v B, t th 2 xut pht t B v A lc 7h v 2 xe gp nhau lc 9h.Mt hm, t th 1 xut pht t A lc 8h, cn t th 2 vn khi hnh lc 7h nn 2 xe gp nhau lc 9h48ph.Hi hng ngy t 1n B v t 2 n B lc my gi.Cho vn tc ca mi xe khng i.

7/. Hai ngi i xe my cng khi hnh t A i v B.Sau 20ph 2 xe cch nhau 5km.

A/. Tnh vn tc ca mi xe bit xe th 1 i ht qung ng mt 3h,cn xe th 2 mt 2h

B/.Nu xe 1 khi hnh trc xe 2 30ph th 2 xe gp nhau bao lu sau khi xe th 1 khi hnh?Ni gp nhau cach A bao nhiu km?C/.xe no n B trc?Khi xe n B th xe kia cn cch B bao nhiu km?

8*/Vo lc 6h ,mt xe ti i t A v C,n 6h 30ph mt xe ti khc i t B v C vi cng vn tc ca xe ti 1.Lc 7h, mt t i t A v C, t gp xe ti th 1lc 9h, gp xe ti 2 lc 9h 30ph.Tm vn tc ca xe ti v t. Bit AB =30km

9/ Hai a im A v B cch nhau 72km.cng lc,mt t i t A v mt ngi i xe p t B ngc chiu nhau v gp nhau sau 1h12ph. Sau t tip tc v B ri quay li vi vn tc c v gp li ngi i xe p sau 48ph k t ln gp trc

a/. Tnh vn tc ca t v xe p.

b/. Nu t tip tc i v A ri quay li th s gp ngi i xe p sau bao lu( k t ln gp th hai)

c*/. V th chuyn ng , th vn tc ca ngi v xe ( cu b) trn cng mt h trc ta .

10/ Mt ngi i t A n B.Trn qung ng u ngi i vi vn tc v1,na thi gian cn li i vi vn tc v2 ,na qung ng cn li i vi vn tc v1 v on cui cng i vi vn tc v2 .tnh vn tc trung bnh ca ngi trn c qung ng

11/. Cho th chuyn ng ca 2 xe nh hnh v. x(km)

a. Nu c im ca mi chuyn ng. Tnh thi 80

im v v tr hai xe gp nhau.

b. xe 2 gp xe 1 bt u khi hnh sau khi ngh

th vn tc ca xe 2 l bao nhiu? Vn tc xe 2 l 40

bao nhiu th n gp xe 1 hai ln.

c. Tnh vn tc trung bnh ca xe 1 trn c qung 20

ng i v v.

Gi phng php gii

1. lp phng trnh ng i ca 2 xe:

a/. S1 =v1t; S2= v2(t-2) ( S1+S2=AB ( v1t+v2(t-2)=AB, gii p/t ( t ( s1,,S2 ( thi im v v tr 2 xe gp nhau.

b/. gi t l thi gian tnh t lc ngi i xe xut pht n lc 2 ngi gp nhau ta c p/t

S1= v1 (t-1); S2= v2 (t-2) ; S1 + S2 = AB ( v1 (t-1)+ v2 (t-2)=48 ( t=4,25h=4h 15ph

(thi im gp nhau T=10h 15 ph

nigp nhau cch A: xn=S1=12(4,25-1)=39km.

2 a/.lp p/t: (1); AB=4v (2)

gii 2 p/t (1)v (2) (v=15km/h; AB=60km/h

b/. lp p/t AB=4.1+(t-1-0,5)v2 (v2=18km/h A E C D B

3 a .. . . . .

khi ngi i b bt u ngi ngh D th ngi i xe p

i mt t2 =2h-1h=1h .

Qung ng ngi i trong 1h l :

AE=V2t2=1.15=15km.

Do AE=3/4.AC (AC=...20km

V ngi i b khi hnh trc ngi i xe 1hnhng li ngi ng 0,5h nn tng thi gian ni ib i nhiu hn ngi i xe l 1h-0,5h = 0,5h.Ta c p/t

EMBED Equation.3 (AB-AC)/v1-AB/v2=0,5 ((AB-20)/5-AB/15=0,5 (AB=33,75km

b.chn mc thi gian l lc ngi i b khi hnh t C (V tr ca ngi i b i vi A:

Ti thi im 0h :X0=20km

Ti thi im 2h: X01=X0+2V1=20+2. 5=30km

Ti thi im 2,5h: X01=30km

Sau 2,5 h X1= X01+(t-2,5)v1.

V tr ca ngi i xe i vi A: X2=v2(t-1).

T0122,53

X12025303032,5

X2001522,530

Ta c bng bin thin:

Biu din cc cp gi tr tng ng ca x, t len h trc ta cc vung gc vi trc tung biu din v tr, trc honh biu din thi gian chuyn ng ta c th nh hnh v

Bng bin thin vn tc ca 2 xe theo thi gian

T gi0122,53

V1km/h555-00-55

V2km/h00-15151515

Ta c th nh hnh v bn

c./ gp ngi i b ti v tr D cch A 30km th thi gian ng i xe p n D phi tha mn iu kin: 2

5 a. qung ng 2 bn cng i trong 10 ph tc 1/6h l AB= v1/6=2km

khi bn i xe v n nha ( mt 10 ph )th bn i b n D :BD=v2/6=6/6=1km

k/c gia 2 bnkhi bn i xe bt u ui theo : AD=AB+BD=3km

thi gian tlc bn i xe ui theon lc gp ngi i bb trng l:

t=AD/(v1-v2)= 3/6=1/2h=30ph

tng thi gian i hc:T=30ph+2.10ph=50ph (tr hc 10 ph.

A B C D

b. qung ng t nh n trng: AC= t. v1=1/2.12=6km c.* gi vn tc ca xe p phi i saukhi pht hin b qunl v1*

ta c: qung ng xe p phi i: S=AB+AC=8km

8/12-8/v1*=7h10ph-7h (v1*=16km/h

* thi gian bn i xe quay v n nh: t1=.....AB/v1*=2/16=0,125h=7,5ph. khi bn i bb n D1 cch A l AD1= AB+ v2 .0,125=2,75km.

*Thi gian ngi i xe dui kpngi i b: t2=AD1/(v1*-v2)=....0,275h=16,5ph

Thi im gp nhau: 6h20ph+ 7,5ph + 16,5ph + 6h 54ph

* v tr gp nhau cch A: X= v1*t2=16.0,125=4,4km (cch trng 6-4,4=1,6km.

6.gi v1 ,v2 l vn tc cae 1 v xe 2 ta c:

thng ngy khi gp nhau, xe1 i c t1-9-6=3h, xe 2 i c t2= 9-7=2h (p/t

v1 t1+ v2t2=AB hay 3 v1+2v2=AB (1)

hm sau,khigp nhau, xe 1 i mt t01=..1,8h,xe 2 i mt t02=...2,8h. (p/t

v1t01+ v2t02=AB hay 1,8v1+2,8v2=AB (2)

t (1) v (2)( 3v1= 2v2.(3)

t (3) v (1) ( t1=6h, t1=4h (thi im n ni T1=6+6=12h, T2= 7+4=11h

7 gi v1 , v2 ln lt l vn tc ca 2 xe.khi i ht qung ng AB, xe 1 i mt t1=3h, xe 2 i mt t2=2h . ta c p/t v1t1=v2t2=AB (v1/v2=t2/t1=2/3 (1)

mt khc ( v1-v2=5:1/3=15 (2)

t (1) v (2) ( v1=30km/h,v2=45km/h

b qung ng 2 xe i trong thi gian t tnh t lc xe 1 bt u xut pht

S1= v1t=30t, S2=v2(t-0,5)=45t-22,5

Khi 2 xe gp nhau: S1=S2= ( t=1,5h x

Ni gp nhau cch A l x=s1=30.1,5=45km

c. p s 15km.

8 gi vn tc t l a, vn tc xe ti l b.

Khi t gp xe ti 1 (xe ti 1 i mt 3h, xe t i

mt 2h. v qung ng i bng nhau nn: 3.a=2.b (1) t

Khi t gp xe ti 2 th xe ti 2 i mt 3h,cn t i mt 2,5 h. v t i nhiu hn xe ti mt on AB=30km nn : 2,5b-3a=30 (2)

t (1) v (2) ( a=40km/h, b=60km/h.9 A D C BT khi xut pht n ln gp nhau th nht : (tv1+v2) =AB/t1=72:1,2=60km/h (1)

T ln gp nhau th nht C n ln gp nhau th 2 D t i c qung ng di hn xe dp l (v1-v2). 0,8=2.CB ((v1-v2).0,8=2.v2.1,2 (v1=4v2 (2)

T 1 v 2 ( v1=48km/h, v2=12km/h

b. khi gp nhau ln th 3 tng qung ng hai xe i l 3.AB (p/t:( v1+v2)t=3.AB (t=...

c. bng bin thin v tr ca 2 xe i vi A theo thi gian t tnh t luc khi hnh

T01,534,5

X1072072

X272543618

Dng th nh hnh v trn

**Bng bin thin vn tc ca 2 xe theo thi gian tnh t lckhi hnh

T(h)011,534,55

V1km/h484848 --48-48

4848

-48-48

V1km/h121212121212

chuyn ng(Bi tp b xung)I.Vn tc trung bnh

1.1.1.Mt ngi i trn qung ng S chia thnh n chng khng u nhau, chiu di cc chng ln lt l S1, S2, S3,......Sn. Thi gian ngi i trn cc chng ng tng ng l t1, t2 t3....tn . Tnh vn tc trung bnh ca ngi trn ton b qung ng S. Chng minh rng:vn trung bnh ln hn vn tc b nht v nh hn vn tc ln nht.

Gii: Vn tc trung bnh ca ngi trn qung ng S l: Vtb=

Gi V1, V2 , V3 ....Vn l vn tc trn cc chng ng tng ng ta c:

.......

gi s Vkln nht v Vi l b nht ( n ( k >i ( 1)ta phi chng minh Vk > Vtb > Vi.Tht vy:

Vtb= = .Do ; ... >1 nn

t1+ t2.+.. tn> t1 +t2+....tn ( Vi< Vtb (1)

Tng t ta c Vtb= = .Do ; ... Vtb (2) PCM

2. Hp 2 vn tc cng phng1.2.1 Cc nh th thao chy thnh hng di l, vi vn tc v nh nhau. Hun luyn vin chy ngc chiu vi h vi vn tc u