2-20 할리데이 10판 솔루션 일반물리학 x=25t−6.0t3 x = 25t-6.0t^3 x=25t−6.0t3 v=x˙=dxdt=ddt(25t−6.0t3)=25−18t2 \begin{aligned} v &= \dot x = \frac{dx}{dt} \\ &= \frac{d}{dt}(25t-6.0t^3) \\ &= 25-18 t^2 \end{aligned} v=x˙=dtdx=dtd(25t−6.0t3)=25−18t2 a=v˙=dvdt=ddt(25−18t2)=−36t \begin{aligned} a &= \dot v = \frac{dv}{dt} \\ &= \frac{d}{dt}(25-18 t^2) \\ &= -36 t \end{aligned} a=v˙=dtdv=dtd(25−18t2)=−36t ∴{x=25t−6.0t3v=25−18t2a=−36t⋯(2−20−1) \therefore \begin{cases} x &= 25t-6.0t^3 \\ v &= 25-18 t^2 \\ a &= -36 t \end{cases}\cdots(2-20-1) ∴⎩⎨⎧xva=25t−6.0t3=25−18t2=−36t⋯(2−20−1) (a) Ans=tv=0=?\Ans =t_{v=0} = ? Ans=tv=0=? (2−20−1), (2-20-1), (2−20−1), $$v = 25-18 .. 10판/2. 직선운동 2019.07.24
2-19 할리데이 10판 솔루션 일반물리학 Ans=aˉ=v(2.4)−v(0)t2.4−t0=−30[m/s]−18[m/s]2.4[s]=−20[m/s2] \begin{aligned} \Ans &= \bar a = \frac{v(2.4) - v(0)}{t_{2.4}-t_0} \\ &=\frac{-30\ut{m/s} - 18\ut{m/s}}{2.4\ut{s}} \\ &=-20\ut{m/s^2} \end{aligned} Ans=aˉ=t2.4−t0v(2.4)−v(0)=2.4[s]−30[m/s]−18[m/s]=−20[m/s2] 10판/2. 직선운동 2019.07.24
2-18 할리데이 10판 솔루션 일반물리학 x=12t2−2t3 x = 12t^2-2t^3 x=12t2−2t3 v=x˙= dx dt= d dt(12t2−2t3)=24t−6t2 \begin{aligned} v &= \dot x = \dxt{x} \\ &= \dt(12t^2-2t^3) \\ &= 24t-6t^2 \end{aligned} v=x˙=dtdx=dtd(12t2−2t3)=24t−6t2 a=v˙= dv dt=x¨= d2x dt2= d dt(24t−6t2)=24−12t \begin{aligned} a &= \dot v = \dxt{v}=\ddot x = \dxtt{x} \\ &= \dt(24t-6t^2) \\ &= 24-12t \end{aligned} a=v˙=dtdv=x¨=dt2d2x=dtd(24t−6t2)=24−12t ∴{x=12t2−2t3v=24t−6t2a=24−12t⋯(2−18−1) \therefore \begin{cases} x &= 12t^2-2t^3 \\ v &= 24t-6t^2 \\ a &= 24-12t \end{cases}\cdots({2-18-1}) ∴⎩⎨⎧xva=12t2−2t3=24t−6t2=24−12t⋯(2−18−1) (a) Ans=x(3.5)=? \Ans = x(3.5) =? Ans=x(3.5)=? (2−18−1),(2-18-1), (2−18−1), $$ \begin{aligned} x(3... 10판/2. 직선운동 2019.07.24
2-17 할리데이 10판 솔루션 일반물리학 x=9.75+1.50t3 x = 9.75 + 1.50t^3 x=9.75+1.50t3 (a) Ans=vˉ=x(t3)−x(t2)t3−t2=x(3.00)−x(2.00)3.00−2.00=9.75+1.50(3.00)3−{9.75+1.50(2.00)3}3.00−2.00=572[cm/s]=28.5[cm/s] \begin{aligned} \Ans &= \bar v = \frac{x(t_3)-x(t_2)}{t_3-t_2} \\ &= \frac{x(3.00)-x(2.00)}{3.00-2.00} \\ &= \frac{9.75 + 1.50(3.00)^3-\bra{9.75 + 1.50(2.00)^3}}{3.00-2.00} \\ &= \frac{57}{2}\ut{cm/s} \\ &= 28.5\ut{cm/s} \\ \end{aligned} Ans=vˉ=t3−t2x(t3)−x(t2)=3.00−2.00x(3.00)−x(2.00)=3.00−2.009.75+1.50(3.00)3−{9.75+1.50(2.00)3}=257[cm/s]=28.5[cm/s] (b) Ans=x˙(2.00)=?\Ans = \dot x(2.00) =? Ans=x˙(2.00)=? $$ \begin{aligned} \\ \dot x &= \dxt{x} \\ &= \dt(9.75 + 1.50t^3) \\ &= \frac{9}{2}t^2 \cdots.. 10판/2. 직선운동 2019.07.24
2-16 할리데이 10판 솔루션 일반물리학 x=4.0−6.0t2⋯(2−16−1) x = 4.0- 6.0t^2 \cdots({2-16-1})x=4.0−6.0t2⋯(2−16−1) (a) Ans=tx˙=0=? \Ans = t_{\dot x=0} =? Ans=tx˙=0=? x˙= dx dt= d dt(4.0−6.0t2)=−12t=0 \begin{aligned} \dot x &= \dxt{x} \\&= \dt(4.0- 6.0t^2) \\&= -12t=0 \end{aligned} x˙=dtdx=dtd(4.0−6.0t2)=−12t=0 t=0[s] t=0\ut{s} t=0[s] (b) Ans=x(0)=?=4.0−6.0(0)2=4.0[m] \begin{aligned} \Ans &= x(0) =? \\ &= 4.0- 6.0(0)^2 \\ &= 4.0\ut{m} \end{aligned} Ans=x(0)=?=4.0−6.0(0)2=4.0[m] (c) Ans=tx=0,t0=? \Ans = t_{x=0,t0} =? Ans=tx=0,t0=? (2−16−2), (2-16-2), (2−16−2), t=23 t = \sqrt{\frac{2}{3}} t=32 (e) (f) +20t (그래프참조) (g) 증가시킴 (그래프참조) 10판/2. 직선운동 2019.07.23
2-15 할리데이 10판 솔루션 일반물리학 x=18t+5.0t2x = 18t + 5.0t^2 x=18t+5.0t2 (a) Ans=x˙(2)=? \Ans = \dot x(2) =? Ans=x˙(2)=? x˙= dx dt= d dt(18t+5.0t2)=18+10tx˙(2)=18+10(2)=38[m/s] \begin{aligned} \dot x &= \dxt{x} \\&= \dt(18t + 5.0t^2) \\&= 18 + 10 t \\ \dot x(2)&=18 + 10 (2) \\ &= 38\ut{m/s} \end{aligned} x˙x˙(2)=dtdx=dtd(18t+5.0t2)=18+10t=18+10(2)=38[m/s] (b) Ans=x(t2)−x(t1)t2−t1=?=x(3.0)−x(2.0)3.0−2.0={18(3.0)+5.0(3.0)2}−{18(2.0)+5.0(2.0)2}=43[m/s] \begin{aligned} \Ans &= \frac{x(t_2)-x(t_1)}{t_2-t_1} =? \\ &= \frac{x(3.0)-x(2.0)}{3.0-2.0} \\ &= \bra{18(3.0) + 5.0(3.0)^2}-\bra{18(2.0) + 5.0(2.0)^2} \\ &= 43\ut{m/s} \end{aligned} Ans=t2−t1x(t2)−x(t1)=?=3.0−2.0x(3.0)−x(2.0)={18(3.0)+5.0(3.0)2}−{18(2.0)+5.0(2.0)2}=43[m/s] 10판/2. 직선운동 2019.07.23
2-14 할리데이 10판 솔루션 일반물리학 x=16te−t x = 16te^{-t} x=16te−t Ans=∣x(tx˙=0)∣=? \Ans = \abs{x(t_{\dot x=0})} =? Ans=∣x(tx˙=0)∣=? x˙= dx dt= d dt(16te−t)=16e−t−16te−t=−16e−t(t−1)=0 \begin{aligned} \dot x &= \dxt{x} \\&= \dt(16te^{-t}) \\&= 16 e^{-t}-16 te^{-t} \\&= -16 e^{-t} (t-1) = 0 \end{aligned} x˙=dtdx=dtd(16te−t)=16e−t−16te−t=−16e−t(t−1)=0 t=1[s] t=1\ut{s} t=1[s] ∣x(1)∣=∣16(1)e−(1)∣=16e[m]≈5.886071058743077[m] \begin{aligned} \\\abs{x(1)} &= \abs{16(1)e^{-(1)}} \\ &= \frac{16}{e}\ut{m} \\&\approx 5.886071058743077\ut{m} \end{aligned} ∣x(1)∣=16(1)e−(1)=e16[m]≈5.886071058743077[m] 10판/2. 직선운동 2019.07.23
2-13 할리데이 10판 솔루션 일반물리학 (a) {v1=55[km/h]v2=90[km/h]t=t1=t2 \begin{cases} v_1 &= 55\ut{km/h} \\ v_2 &= 90\ut{km/h} \\ t &= t_1 = t_2 \end{cases} ⎩⎨⎧v1v2t=55[km/h]=90[km/h]=t1=t2 Ans=vˉa=ΣSΣt=S1+S2t1+t2=S1+S22t=12(S1t+S2t)=v1+v22=55[km/h]+90[km/h]2=72.5[km/h]≈73[km/h] \begin{aligned} \Ans &= \bar v_a = \frac{\Sigma S}{\Sigma t} =\frac{S_1+S_2}{t_1+t_2} \\ &=\frac{S_1+S_2}{2t} \\ &=\frac{1}{2}\left(\frac{S_1}{t}+\frac{S_2}{t}\right) \\ &=\frac{v_1+v_2}{2} \\ &=\frac{55\ut{km/h}+90\ut{km/h}}{2} \\ &=72.5\ut{km/h} \\ &\approx 73\ut{km/h} \end{aligned} Ans=vˉa=ΣtΣS=t1+t2S1+S2=2tS1+S2=21(tS1+tS2)=2v1+v2=255[km/h]+90[km/h]=72.5[km/h]≈73[km/h] (b) $$ \beg.. 10판/2. 직선운동 2019.07.23
2-12 할리데이 10판 솔루션 일반물리학 {vs=5.00[m/s]v=25.0[m/s]L=12.0[m]\begin{cases} v_s &= 5.00\ut{m/s} \\ v &= 25.0\ut{m/s} \\ L &= 12.0\ut{m} \end{cases} ⎩⎨⎧vsvL=5.00[m/s]=25.0[m/s]=12.0[m] (a) put {x=Wave End PointLa=Added Wave Lengthn=Number Of Added Carsta=Time Of Add 1 Car \put \begin{cases} x &= \text{Wave End Point} \\ L_a &= \text{Added Wave Length} \\ n &= \text{Number Of Added Cars} \\ t_a &= \text{Time Of Add 1 Car} \end{cases} put ⎩⎨⎧xLanta=Wave End Point=Added Wave Length=Number Of Added Cars=Time Of Add 1 Car ta=dv−vst=ntan=tta=tv−vsd \begin{aligned} t_a &= \frac{d}{v-v_s} \\ t &= nt_a \\ n &= \frac{t}{t_a} = t\frac{v-v_s}{d} \end{aligned} tatn=v−vsd=nta=tat=tdv−vs $$ \begin{aligned} x.. 10판/2. 직선운동 2019.07.23
2-11 할리데이 10판 솔루션 일반물리학 {vc=15.00[m/s]t=150.0[s]xc0−xa0=1.500[km] \begin{cases} v_c &= 15.00\ut{m/s} \\ t &= 150.0\ut{s} \\ x_{c0}-x_{a0} &= 1.500\ut{km} \end{cases} ⎩⎨⎧vctxc0−xa0=15.00[m/s]=150.0[s]=1.500[km] $$ \begin{aligned} \Ans &= v_a = ? \\ x_{c0}+\Delta x_c &= x_{a0}+\Delta x_a \\ x_{c0}-x_{a0} &= \Delta x_a-\Delta x_c \\ &= v_at-v_ct \\ &= (v_a-v_c)t \\ v_a-v_c&= \frac{x_{c0}-x_{a0}}{t} \\ v_a&= \frac{x_{c0}-x_{a0}}{t}+v_c \\ &= \frac{1.500\ut{km}}{150.0\ut{s}}\cdot\frac{1000\ut{m}}{1\ut{.. 10판/2. 직선운동 2019.07.22