11판/2. 직선운동 66

2-56 할리데이 11판 솔루션 일반물리학

{R:Red CarG:Green Cart0:Car Startt1:Car MeetA:Case AB:Case B \begin{cases} R&:\text{Red Car}\\ G&:\text{Green Car}\\ t_0&:\text{Car Start}\\ t_1&:\text{Car Meet}\\ A&:\text{Case A}\\ B&:\text{Case B}\\ \end{cases} {xR0=0xG0=230[m]vRA=20.0[km/h]=509[m/s]xA1=44.5[m]vRB=40.0[km/h]=1009[m/s]xB1=76.6[m] \begin{cases} x_{R0}&=0\\ x_{G0}&=230\ut{m}\\ v_{RA}&=20.0\ut{km/h}=\frac{50}{9}\ut{m/s}\\ x_{A1}&=44.5\ut{m}\\ v_{RB}&=40.0\ut{km/h}=\frac{100}{9}\ut{m/s}\\ x_{B1}&=76.6\ut{m}\\ \end{cases} [Red Car]\title{Red Car} $$ \begin{aligned} \De..

2-55 할리데이 11판 솔루션 일반물리학

{t=120[ms]=0.12[s]v=3400[km/h]=85009[m/s] \begin{cases} t&=120\ut{ms}=0.12\ut{s}\\ v&=3400\ut{km/h}=\frac{8500}{9}\ut{m/s} \end{cases} S=vt=0.1285009=3403[m]113.3333333333333[m]1.1×102[m] \begin{aligned} S&=vt\\ &=0.12\cdot\frac{8500}{9}\\ &=\frac{340}{3}\ut{m}\\ &\approx 113.3333333333333\ut{m}\\ &\approx 1.1\times10^2\ut{m}\\ \end{aligned}

2-54 할리데이 11판 솔루션 일반물리학

{a={2(t[s])+8}[m/s2]v(2.0)=11.0[m/s] \begin{cases} a&=\bra{-2(t\ut{s})+8}\ut{m/s^2}\\ v(-2.0)&=11.0\ut{m/s} \end{cases} v=a ⁣dt=(2t+8) ⁣dt=t2+8t+C \begin{aligned} v&=\int a\dd t\\ &=\int (-2t+8)\dd t\\ &=-t^2+8t+C\\ \end{aligned} v(2)=(2)2+8(2)+C=11C=31 \begin{aligned} v(-2)&=-(-2)^2+8(-2)+C=11\\ C&=31\\ \end{aligned} v(t)=t2+8t+31\therefore v(t)=-t^2+8t+31 v(6.0)=(6)2+8(6)+31=43[m/s] \begin{aligned} v(6.0)&=-(6)^2+8(6)+31\\ &=43\ut{m/s}\\ \end{aligned}

2-53 할리데이 11판 솔루션 일반물리학

{v=169.1[km/h]=169136[m/s]S=18.4[m] \begin{cases} v&=169.1\ut{km/h}=\frac{1691}{36}\ut{m/s}\\ S&=18.4\ut{m} \end{cases} t=Sv=18.4169136[s]=33128455[s]0.3917208752217622[s]0.392[s]392[ms] \begin{aligned} t&=\frac{S}{v}\\ &=\frac{18.4}{\frac{1691}{36}}\ut{s}\\ &=\frac{3312}{8455}\ut{s}\\ &\approx0.3917208752217622\ut{s}\\ &\approx0.392\ut{s}\\ &\approx392\ut{ms}\\ \end{aligned}

2-52 할리데이 11판 솔루션 일반물리학

{t0:Startt1:Max Speedt2:End \begin{cases} t_0&:\text{Start}\\ t_1&:\text{Max Speed}\\ t_2&:\text{End}\\ \end{cases} {v1=11.00[m/s]S01=12.0[m]S02=100[m] \begin{cases} v_{1}&=11.00\ut{m/s}\\ S_{0\rarr1}&=12.0\ut{m}\\ S_{0\rarr2}&=100\ut{m} \end{cases} (a)\ab{a} S=12(v+v0)t, S=\frac{1}{2}(v+v_0)t, S01=12(v1+v0)t01=12(11+0)t01t01=2S0111 \begin{aligned} S_{0\rarr1}&=\frac{1}{2}(v_1+v_0)t_{0\rarr1}\\ &=\frac{1}{2}(11+0)t_{0\rarr1}\\ t_{0\rarr1}&=\frac{2S_{0\rarr1}}{11}\\ \end{aligned} $$ \be..

2-51 할리데이 11판 솔루션 일반물리학

{R:Red TrainG:Green Train \begin{cases} R&:\text{Red Train}\\ G&:\text{Green Train}\\ \end{cases} {vR=72[km/h]=20[m/s]vG=160[km/h]=4009[m/s]S=950[m]a=1.0[m/s2] \begin{cases} v_R&=72\ut{km/h}=20\ut{m/s}\\ v_G&=160\ut{km/h}=\frac{400}{9}\ut{m/s}\\ S&=950\ut{m}\\ a&=-1.0\ut{m/s^2}\\ \end{cases} 2aS=v2v02, 2aS=v^2-{v_0}^2, 2(1)SStop=(0)2v02SStop=v022 \begin{aligned} 2(-1)S_{\text{Stop}}&=(0)^2-{v_0}^2\\ S_{\text{Stop}}&=\frac{{v_0}^2}{2}\\ \end{aligned} $$ \begin{cases} S_{R\text{stop}}&=\cfrac{{v_{R0..

2-50 할리데이 11판 솔루션 일반물리학

S=vt12at2, S=vt-\frac{1}{2}at^2, {S1.5=v1.5t1.512a1.5t1.52S1=v1t112a1t12 \begin{cases} S_{1.5}=v_{1.5}t_{1.5}-\frac{1}{2}a_{1.5}{t_{1.5}}^2\\ S_1=v_1t_1-\frac{1}{2}a_1{t_1}^2 \end{cases} {H1.5=(0)t1.512(g)t1.52H1=(0)t112(g)t12 \begin{cases} H_{1.5}=(0)t_{1.5}-\frac{1}{2}(-g){t_{1.5}}^2\\ H_1=(0)t_1-\frac{1}{2}(-g){t_1}^2 \end{cases} {H1.5=12gt1.52H1=12gt12 \begin{cases} H_{1.5}=\frac{1}{2}g{t_{1.5}}^2\\ H_1=\frac{1}{2}g{t_1}^2 \end{cases} $$ \begin{aligned} \frac{H_{1.5}}{H_1}&=\frac{..

2-49 할리데이 11판 솔루션 일반물리학

{H=45.0[m]t=2.00[s]h1=11.8[m]g=9.80665[m/s2] \begin{cases} H&=45.0\ut{m}\\ t&=2.00\ut{s}\\ h_1&=11.8\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} S=vt12at2, S=vt-\frac{1}{2}at^2, h1H=v(2)12(g)(2)2(11.8)(45)=v(2)12(g)(2)2 \begin{aligned} h_1-H&=v(2)-\frac{1}{2}(-g)(2)^2\\ (11.8)-(45)&=v(2)-\frac{1}{2}(-g)(2)^2\\ \end{aligned} v=(835+g)=52813320000[m/s]=26.40665[m/s]26.4[m/s] \begin{aligned} \abs{v}&=\abs{-\(\frac{83}{5}+g\)}\\ &=\frac{528133}{20000}\ut{m/s}\\ &=26.40665\ut{m/s}\\ &\approx 26.4\ut{m/s}\\ \end{aligned}

2-48 할리데이 11판 솔루션 일반물리학

{a=50.0[m/s2]v=80.0[km/h] \begin{cases} a&=50.0\ut{m/s^2}\\ v&=80.0\ut{km/h} \end{cases} 80[km/h]=80[km/h]1000[m]1[km]1[h]3600[s]=2009[m/s] \begin{aligned} 80\ut{km/h}&=80\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}}\\ &=\frac{200}{9}\ut{m/s} \end{aligned} v=v0+at, v=v_0+at, $$ \begin{aligned} \frac{200}{9}\ut{m/s}&=(0)+(50\ut{m/s^2})t\\ t&=\frac{4}{9}\ut{s}\\ &\approx 0.4444444444444444\ut{s}\\ &\approx 0.444\ut{s}\\ &\approx 444\ut{ms..

2-47 할리데이 11판 솔루션 일반물리학

{v1=65[mi/h]v2=70[mi/h]S=700[km] \begin{cases} v_1&=65\ut{mi/h}\\ v_2&=70\ut{mi/h}\\ S &= 700\ut{km}\\ \end{cases} {1[mile]=5280[ft]1[ft]=12[in]1[in]=2.54[cm]1[mile]=1.609344[km] \begin{cases} 1\ut{mile}&=5280\ut{ft}\\ 1\ut{ft}&=12\ut{in}\\ 1\ut{in}&=2.54\ut{cm}\\ \therefore 1\ut{mile}&=1.609344\ut{km} \end{cases} $$ \begin{aligned} \Delta t &= t_1-t_2\\ &=\frac{S}{v_1}-\frac{S}{v_2}\\ &=S\(\frac{1}{v_1}-\frac{1}{v_2}\)\\ &=(700\ut{km})\(\frac{1}{65\ut{mi/h}}-\frac{1}{70\ut{mi/h..