A2Physics物理出國英語VectorsPart3Velocityvectors

1.Aboatismovingwithavelocityof10m/sduenorthrelativetothewater.Thewaterisflowingwithavelocityof5m/sdueeast.Whatistheresultantvelocityoftheboat?

First,weusethePythagoreantheoremforvectoraddition.Thetwovelocityvectorsareperpendiculartoeachother.Letthevelocityoftheboatrelativetothewater(v_{b/w}=10m/s)(north)andthevelocityofthewater(v_w=5m/s)(east).

Themagnitudeoftheresultantvelocity(v)isgivenby(v=sqrt{v_{b/w}^2+v_w^2}).

Substitutingthevalues,wehave(v=sqrt{10^{2}+5^{2}}=sqrt{100+25}=sqrt{125}=5sqrt{5}approx11.2m/s).

Tofindthedirection,weusethetangentfunction.Let(theta)betheanglenorthofeast.Then(tantheta=frac{v_{b/w}}{v_w}=frac{10}{5}=2),so(theta=arctan(2)approx63.4^{circ})northofeast.

2.Anairplaneisflyingwithavelocityof200m/sinadirectionof30°northofwest.Whatarethecomponentsofthevelocityvector?

Ifweconsiderthewesteastdirectionasthexaxis(negativexforwest)andthenorthsouthdirectionastheyaxis.

Thexcomponent(v_x=vcos30^{circ}),where(v=200m/s).So(v_x=200timesfrac{sqrt{3}}{2}=100sqrt{3}approx173.2m/s).

Theycomponent(v_y=vsin30^{circ}=200timesfrac{1}{2}=100m/s).

3.Aswimmercanswimataspeedof2m/sinstillwater.Theswimmerwantstocrossariverthatisflowingataspeedof1m/s.Iftheswimmerswimsperpendiculartotheflowoftheriver,whatistheresultantvelocityoftheswimmer?

Thevelocityoftheswimmerinstillwater(v_s=2m/s)(perpendiculartotheriverflow)andthevelocityoftheriver(v_r=1m/s).

UsingthePythagoreantheorem,themagnitudeoftheresultantvelocity(v=sqrt{v_s^2+v_r^2}).

Substitutingthevalues,(v=sqrt{2^{2}+1^{2}}=sqrt{4+1}=sqrt{5}approx2.24m/s).

Thedirection(theta)oftheresultantvelocitywithrespecttothedirectionperpendiculartotheriverflowisgivenby(tantheta=frac{v_r}{v_s}=frac{1}{2}=0.5),so(theta=arctan(0.5)approx26.6^{circ})downstream.

4.Acyclistismovingwithavelocityof15m/sintheeastdirection.Awindisblowingwithavelocityof5m/sinthenorthdirection.Whatistheresultantvelocityofthecyclist?

Themagnitudeoftheresultantvelocity(v=sqrt{15^{2}+5^{2}}=sqrt{225+25}=sqrt{250}=5sqrt{10}approx15.8m/s).

Thedirection(theta)northofeastisgivenby(tantheta=frac{5}{15}=frac{1}{3}),so(theta=arctan(frac{1}{3})approx18.4^{circ})northofeast.

5.Acarismovingwithavelocityof30m/sinadirectionof45°southofwest.Findthecomponentsofthevelocityvector.

Forthexcomponent(westeast),(v_x=vcos45^{circ}),where(v=30m/s).So(v_x=30timesfrac{sqrt{2}}{2}=15sqrt{2}approx21.2m/s).

Fortheycomponent(northsouth),(v_y=vsin45^{circ}=30timesfrac{sqrt{2}}{2}=15sqrt{2}approx21.2m/s).

6.Aboatismovingwithavelocityof8m/sinthesouthdirectionrelativetothewater.Thewaterisflowingwithavelocityof3m/sinthewestdirection.Whatistheresultantvelocityoftheboat?

Themagnitudeoftheresultantvelocity(v=sqrt{8^{2}+3^{2}}=sqrt{64+9}=sqrt{73}approx8.54m/s).

Thedirection(theta)westofsouthisgivenby(tantheta=frac{3}{8}=0.375),so(theta=arctan(0.375)approx20.6^{circ})westofsouth.

7.Anobjectismovingwithavelocityvector(vec{v}=(3hat{i}+4hat{j})m/s).Whatisthemagnitudeanddirectionofthevelocity?

Themagnitudeofthevelocity(v=sqrt{3^{2}+4^{2}}=sqrt{9+16}=5m/s).

Thedirection(theta)withrespecttothexaxisisgivenby(tantheta=frac{4}{3}),so(theta=arctan(frac{4}{3})approx53.1^{circ})abovethexaxis.

8.Aplaneisflyingwithavelocityof180m/sinadirectionof60°northofeast.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vcos60^{circ}=180timesfrac{1}{2}=90m/s).

Theycomponent(v_y=vsin60^{circ}=180timesfrac{sqrt{3}}{2}=90sqrt{3}approx155.9m/s).

9.Aswimmeraimstoswimdirectlyacrossariver.Theswimmercanswimataspeedof1.5m/sinstillwaterandtheriverisflowingataspeedof0.8m/s.Atwhatangleshouldtheswimmerswimrelativetothedirectionperpendiculartotheriverflow?

Let(theta)betheangleupstreamfromtheperpendiculardirection.

Weknowthat(tantheta=frac{v_r}{v_s}),where(v_r=0.8m/s)istherivervelocityand(v_s=1.5m/s)istheswimmer'svelocityinstillwater.

So(theta=arctan(frac{0.8}{1.5})approx28.1^{circ})upstreamfromtheperpendiculardirection.

10.Aparticlehasavelocityvector(vec{v}=(2hat{i}+5hat{j})m/s).Whatisthemagnitudeanddirectionofthevelocity?

Themagnitudeofthevelocity(v=sqrt{(2)^{2}+5^{2}}=sqrt{4+25}=sqrt{29}approx5.39m/s).

Thedirection(theta)withrespecttothenegativexaxisisgivenby(tantheta=frac{5}{2}=2.5),so(theta=arctan(2.5)approx68.2^{circ})abovethenegativexaxis.

11.Acarismovingwithavelocityof25m/sinadirectionof30°eastofsouth.Findthecomponentsofthevelocityvector.

Thexcomponent(v_x=vsin30^{circ}=25timesfrac{1}{2}=12.5m/s)(east).

Theycomponent(v_y=vcos30^{circ}=25timesfrac{sqrt{3}}{2}approx21.65m/s)(south).

12.Aboatismovingwithavelocityof6m/sinthenorthdirectionrelativetothewater.Thewaterisflowingwithavelocityof2m/sintheeastdirection.Whatistheresultantvelocityoftheboat?

Themagnitudeoftheresultantvelocity(v=sqrt{6^{2}+2^{2}}=sqrt{36+4}=sqrt{40}=2sqrt{10}approx6.32m/s).

Thedirection(theta)eastofnorthisgivenby(tantheta=frac{2}{6}=frac{1}{3}),so(theta=arctan(frac{1}{3})approx18.4^{circ})eastofnorth.

13.Anairplanehasavelocityvector(vec{v}=(120hat{i}50hat{j})m/s).Whatisthemagnitudeanddirectionofthevelocity?

Themagnitudeofthevelocity(v=sqrt{120^{2}+(50)^{2}}=sqrt{14400+2500}=sqrt{16900}=130m/s).

Thedirection(theta)belowthexaxisisgivenby(tantheta=frac{50}{120}=frac{5}{12}),so(theta=arctan(frac{5}{12})approx22.6^{circ})belowthexaxis.

14.Acyclistismovingwithavelocityof12m/sinthewestdirection.Awindisblowingwithavelocityof4m/sinthesouthdirection.Whatistheresultantvelocityofthecyclist?

Themagnitudeoftheresultantvelocity(v=sqrt{12^{2}+4^{2}}=sqrt{144+16}=sqrt{160}=4sqrt{10}approx12.65m/s).

Thedirection(theta)southofwestisgivenby(tantheta=frac{4}{12}=frac{1}{3}),so(theta=arctan(frac{1}{3})approx18.4^{circ})southofwest.

15.Aswimmercanswimataspeedof2.2m/sinstillwater.Theswimmerwantstocrossariverthatisflowingataspeedof1.1m/s.Iftheswimmerswimsatanangleof60°upstreamfromtheperpendiculardirectiontotheriverflow,whatistheresultantvelocityoftheswimmer?

First,findthecomponentsoftheswimmer'svelocityrelativetothewater.Thecomponentperpendiculartotheriverflow(v_{sperp}=2.2cos60^{circ}=1.1m/s)andthecomponentagainsttheriverflow(v_{sparallel}=2.2sin60^{circ}=2.2timesfrac{sqrt{3}}{2}approx1.91m/s).

Thenetcomponentalongtheriverflow(v_{netparallel}=v_{sparallel}1.1=1.911.1=0.81m/s).

Theresultantvelocity(v=sqrt{v_{sperp}^2+v_{netparallel}^2}=sqrt{1.1^{2}+0.81^{2}}=sqrt{1.21+0.6561}=sqrt{1.8661}approx1.36m/s).

16.Anobjecthasavelocityvector(vec{v}=(4hat{i}3hat{j})m/s).Whatisthemagnitudeanddirectionofthevelocity?

Themagnitudeofthevelocity(v=sqrt{(4)^{2}+(3)^{2}}=sqrt{16+9}=5m/s).

Thedirection(theta)withrespecttothenegativexaxisisgivenby(tantheta=frac{3}{4}),so(theta=arctan(frac{3}{4})approx36.9^{circ})belowthenegativexaxis.

17.Aplaneisflyingwithavelocityof220m/sinadirectionof40°southofwest.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vcos40^{circ}=220times0.766approx168.5m/s).

Theycomponent(v_y=vsin40^{circ}=220times0.643approx141.5m/s).

18.Aboatismovingwithavelocityof7m/sintheeastdirectionrelativetothewater.Thewaterisflowingwithavelocityof3m/sinthenorthdirection.Whatistheresultantvelocityoftheboat?

Themagnitudeoftheresultantvelocity(v=sqrt{7^{2}+3^{2}}=sqrt{49+9}=sqrt{58}approx7.62m/s).

Thedirection(theta)northofeastisgivenby(tantheta=frac{3}{7}approx0.429),so(theta=arctan(0.429)approx23.2^{circ})northofeast.

19.Aparticlehasavelocityvector(vec{v}=(5hat{i}12hat{j})m/s).Whatisthemagnitudeanddirectionofthevelocity?

Themagnitudeofthevelocity(v=sqrt{5^{2}+(12)^{2}}=sqrt{25+144}=13m/s).

Thedirection(theta)belowthexaxisisgivenby(tantheta=frac{12}{5}=2.4),so(theta=arctan(2.4)approx67.4^{circ})belowthexaxis.

20.Acarismovingwithavelocityof35m/sinadirectionof70°northofeast.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vcos70^{circ}=35times0.342approx12m/s).

Theycomponent(v_y=vsin70^{circ}=35times0.94approx32.9m/s).

21.Aswimmercanswimataspeedof1.8m/sinstillwater.Theriverisflowingataspeedof0.6m/s.Iftheswimmerwantstoreachthepointdirectlyoppositeontheothersideoftheriver,atwhatangleshouldtheswimmerswimrelativetotheperpendiculardirectiontotheriverflow?

Let(theta)betheangleupstreamfromtheperpendiculardirection.

Weknowthat(sintheta=frac{v_r}{v_s}),where(v_r=0.6m/s)and(v_s=1.8m/s).

So(theta=arcsin(frac{0.6}{1.8})approx19.5^{circ})upstreamfromtheperpendiculardirection.

22.Aplaneisflyingwithavelocityof200m/sinadirectionof35°westofnorth.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vsin35^{circ}=200times0.574approx114.8m/s).

Theycomponent(v_y=vcos35^{circ}=200times0.819approx163.8m/s).

23.Aboatismovingwithavelocityof9m/sinthesouthdirectionrelativetothewater.Thewaterisflowingwithavelocityof4m/sinthewestdirection.Whatistheresultantvelocityoftheboat?

Themagnitudeoftheresultantvelocity(v=sqrt{9^{2}+4^{2}}=sqrt{81+16}=sqrt{97}approx9.85m/s).

Thedirection(theta)westofsouthisgivenby(tantheta=frac{4}{9}approx0.444),so(theta=arctan(0.444)approx23.9^{circ})westofsouth.

24.Anobjecthasavelocityvector(vec{v}=(3hat{i}4hat{j})m/s).Whatisthemagnitudeanddirectionofthevelocity?

Themagnitudeofthevelocity(v=sqrt{3^{2}+(4)^{2}}=sqrt{9+16}=5m/s).

Thedirection(theta)belowthexaxisisgivenby(tantheta=frac{4}{3}),so(theta=arctan(frac{4}{3})approx53.1^{circ})belowthexaxis.

25.Acyclistismovingwithavelocityof16m/sintheeastdirection.Awindisblowingwithavelocityof6m/sinthenorthdirection.Whatistheresultantvelocityofthecyclist?

Themagnitudeoftheresultantvelocity(v=sqrt{16^{2}+6^{2}}=sqrt{256+36}=sqrt{292}=2sqrt{73}approx17.1m/s).

Thedirection(theta)northofeastisgivenby(tantheta=frac{6}{16}=frac{3}{8}=0.375),so(theta=arctan(0.375)approx20.6^{circ})northofeast.

26.Aswimmercanswimataspeedof2.5m/sinstillwater.Theswimmerswimsatanangleof45°downstreamfromtheperpendiculardirectiontotheriverflowinariverflowingat1.2m/s.Whatistheresultantvelocityoftheswimmer?

Thecomponentperpendiculartotheriverflow(v_{sperp}=2.5cos45^{circ}=frac{2.5}{sqrt{2}}approx1.77m/s)andthecomponentalongtheriverflow(v_{sparallel}=2.5sin45^{circ}+1.2=frac{2.5}{sqrt{2}}+1.2approx1.77+1.2=2.97m/s).

Theresultantvelocity(v=sqrt{v_{sperp}^2+v_{sparallel}^2}=sqrt{1.77^{2}+2.97^{2}}=sqrt{3.1329+8.8209}=sqrt{11.9538}approx3.46m/s).

27.Aplaneisflyingwithavelocityof250m/sinadirectionof50°southofwest.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vcos50^{circ}=250times0.643approx160.8m/s).

Theycomponent(v_y=vsin50^{circ}=250times0.766approx191.5m/s).

28.Aboatismovingwithavelocityof7.5m/sinthenorthdirectionrelativetothewater.Thewaterisflowingwithavelocityof3m/sintheeastdirection.Whatistheresultantvelocityoftheboat?

Themagnitudeoftheresultantvelocity(v=sqrt{7.5^{2}+3^{2}}=sqrt{56.25+9}=sqrt{65.25}approx8.08m/s).

Thedirection(theta)eastofnorthisgivenby(tantheta=frac{3}{7.5}=0.4),so(theta=arctan(0.4)approx21.8^{circ})eastofnorth.

29.Aparticlehasavelocityvector(vec{v}=(6hat{i}+8hat{j})m/s).Whatisthemagnitudeanddirectionofthevelocity?

Themagnitudeofthevelocity(v=sqrt{(6)^{2}+8^{2}}=sqrt{36+64}=10m/s).

Thedirection(theta)abovethenegativexaxisisgivenby(tantheta=frac{8}{6}=frac{4}{3}),so(theta=arctan(frac{4}{3})approx53.1^{circ})abovethenegativexaxis.

30.Acarismovingwithavelocityof40m/sinadirectionof20°eastofsouth.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vsin20^{circ}=40times0.342approx13.7m/s)(east).

Theycomponent(v_y=vcos20^{circ}=40times0.94approx37.6m/s)(south).

31.Aswimmercanswimataspeedof1.6m/sinstillwater.Theriverisflowingataspeedof0.4m/s.Iftheswimmerswimsatanangleof30°upstreamfromtheperpendiculardirectiontotheriverflow,whatistheresultantvelocityoftheswimmer?

Thecomponentperpendiculartotheriverflow(v_{sperp}=1.6cos30^{circ}=1.6timesfrac{sqrt{3}}{2}approx1.39m/s)andthecomponentagainsttheriverflow(v_{sparallel}=1.6sin30^{circ}=0.8m/s).

Thenetcomponentalongtheriverflow(v_{netparallel}=0.80.4=0.4m/s).

Theresultantvelocity(v=sqrt{v_{sperp}^2+v_{netparallel}^2}=sqrt{1.39^{2}+0.4^{2}}=sqrt{1.9321+0.16}=sqrt{2.0921}approx1.45m/s).

32.Anairplaneisflyingwithavelocityof300m/sinadirectionof45°northofwest.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vcos45^{circ}=300timesfrac{sqrt{2}}{2}=150sqrt{2}approx212.1m/s).

Theycomponent(v_y=vsin45^{circ}=300timesfrac{sqrt{2}}{2}=150sqrt{2}approx212.1m/s).

33.Aboatismovingwithavelocityof6.5m/sinthesouthdirectionrelativetothewater.Thewaterisflowingwithavelocityof2.5m/sinthewestdirection.Whatistheresultantvelocityoftheboat?

Themagnitudeoftheresultantvelocity(v=sqrt{6.5^{2}+2.5^{2}}=sqrt{42.25+6.25}=sqrt{48.5}approx6.96m/s).

Thedirection(theta)westofsouthisgivenby(tantheta=frac{2.5}{6.5}=frac{5}{13}approx0.385),so(theta=arctan(0.385)approx21.1^{circ})westofsouth.

34.Aparticlehasavelocityvector(vec{v}=(8hat{i}15hat{j})m/s).Whatisthemagnitudeanddirectionofthevelocity?

Themagnitudeofthevelocity(v=sqrt{8^{2}+(15)^{2}}=sqrt{64+225}=17m/s).

Thedirection(theta)belowthexaxisisgivenby(tantheta=frac{15}{8}=1.875),so(theta=arctan(1.875)approx61.9^{circ})belowthexaxis.

35.Acarismovingwithavelocityof45m/sinadirectionof80°northofeast.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vcos80^{circ}=45times0.174approx7.83m/s).

Theycomponent(v_y=vsin80^{circ}=45times0.985approx44.3m/s).

36.Aswimmercanswimataspeedof2m/sinstillwater.Theriverisflowingataspeedof0.7m/s.Iftheswimmerwantstoswimacrosstheriverintheshortesttime,inwhatdirectionshouldtheswimmerswim?

Toswimacrosstheriverintheshortesttime,theswimmershouldswimperpendiculartotheriverflow.Thetimetaken(t=fraclfpdfjp{v_s}),where(d)isthewidthoftheriverand(v_s)istheswimmer'sspeedinstillwater.Theresultantvelocitywillhaveacomponentalongtheriverflowduetotheriver'svelocity,butthetimeisindependentoftheriver'svelocitywhenswimmingperpendiculartotheflow.

37.Aplaneisflyingwithavelocityof280m/sinadirectionof55°westofnorth.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vsin55^{circ}=280times0.819approx229.3m/s).

Theycomponent(v_y=vcos55^{circ}=280times0.574approx160.7m/s).

38.Aboatismovingwithavelocityof8.2m/sinthenorthdirectionrelativetothewater.Thewaterisflowingwithavelocityof3.8m/sintheeastdirection.Whatistheresultantvelocityoftheboat?

Themagnitudeoftheresultantvelocity(v=sqrt{8.2^{2}+3.8^{2}}=sqrt{67.24+14.44}=sqrt{81.68}approx9.04m/s).

Thedirection(theta)eastofnorthisgivenby(tantheta=frac{3.8}{8.2}approx0.463),so(theta=arctan(0.463)approx24.9^{circ})eastofnorth.

39.Aparticlehasavelocityvector(vec{v}=(7hat{i}24hat{j})m/s).Whatisthemagnitudeanddirectionofthevelocity?

Themagnitudeofthevelocity(v=sqrt{(7)^{2}+(24)^{2}}=sqrt{49+576}=25m/s).

Thedirection(theta)belowthenegativexaxisisgivenby(tantheta=frac{24}{7}approx3.43),so(theta=arctan(3.43)approx73.7^{circ})belowthenegativexaxis.

40.Acarismovingwithavelocityof50m/sinadirectionof30°southofeast.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vcos30^{circ}=50timesfrac{sqrt{3}}{2}=25sqrt{3}approx43.3m/s)(east).

Theycomponent(v_y=vsin30^{circ}=50timesfrac{1}{2}=25m/s)(south).

41.Aswimmercanswimataspeedof1.9m/sinstillwater.Theriverisflowingataspeedof0.9m/s.Iftheswimmerswimsatanangleof40°downstreamfromtheperpendiculardirectiontotheriverflow,whatistheresultantvelocityoftheswimmer?

Thecomponentperpendiculartotheriverflow(v_{sperp}=1.9cos40^{circ}=1.9times0.766approx1.46m/s)andthecomponentalongtheriverflow(v_{sparallel}=1.9sin40^{circ}+0.9=1.9times0.643+0.9approx1.22+0.9=2.12m/s).

Theresultantvelocity(v=sqrt{v_{sperp}^2+v_{sparallel}^2}=sqrt{1.46^{2}+2.12^{2}}=sqrt{2.1316+4.4944}=sqrt{6.626}approx2.57m/s).

42.Anairplaneisflyingwithavelocityof320m/sinadirectionof60°northofwest.Whatarethecomponentsofthevelocityvector?

Thexcomponent(v_x=vcos60^{circ}=320timesfrac{1}{2}=160m/s).

Theycomponent(v_y=vsin60^{circ}=320timesfrac{sqrt{3}}{2}=160sqrt{3}approx277.1m/s).

43.Aboatismovingwithavelocityof7.8m/sinthesouthdirectionrelativetothewater.Thewaterisflowingwithavelocityof2.2m/sinthewestdirection.Whatistheresultantvelocityoftheboat?

Themagnitudeoftheresultantvelocity(v=sqrt{7.8^{2}+2.2^{2}}=sqrt{60.84+4.84}=sqrt{65.68}approx8.1m/s).

Thedirection(theta)westofsouthisgivenby(tantheta=frac{2.2}{7.8}approx0.282),so(theta=arctan(0.282)approx15.7^{circ})westofsouth.

44.Aparticlehasavelocityvector(vec{v}=(9hat{i}40hat{j})m/s).Whatisthemagnitudeanddirectionofthevelocity?

Themagnitudeofthevelocity(v=sqrt{9^{2}+(40)^{2}}=sqrt{81+1600}=41m/s).

Thedirection(theta)belowthexaxisisgivenby(tantheta=frac{40}{9}approx4.44),so(theta=arctan(4.44)approx77.3^{circ})belowthexaxis.

45.Acarismovingwithavelocityof55m/sinadirectio