10판/2. 직선운동 70

2-20 할리데이 10판 솔루션 일반물리학

x=25t6.0t3 x = 25t-6.0t^3 v=x˙=dxdt=ddt(25t6.0t3)=2518t2 \begin{aligned} v &= \dot x = \frac{dx}{dt} \\ &= \frac{d}{dt}(25t-6.0t^3) \\ &= 25-18 t^2 \end{aligned} a=v˙=dvdt=ddt(2518t2)=36t \begin{aligned} a &= \dot v = \frac{dv}{dt} \\ &= \frac{d}{dt}(25-18 t^2) \\ &= -36 t \end{aligned} {x=25t6.0t3v=2518t2a=36t(2201) \therefore \begin{cases} x &= 25t-6.0t^3 \\ v &= 25-18 t^2 \\ a &= -36 t \end{cases}\cdots(2-20-1) (a) Ans=tv=0=?\Ans =t_{v=0} = ? (2201), (2-20-1), $$v = 25-18 ..

2-18 할리데이 10판 솔루션 일반물리학

x=12t22t3 x = 12t^2-2t^3 v=x˙= ⁣dx ⁣dt= ⁣d ⁣dt(12t22t3)=24t6t2 \begin{aligned} v &= \dot x = \dxt{x} \\ &= \dt(12t^2-2t^3) \\ &= 24t-6t^2 \end{aligned} a=v˙= ⁣dv ⁣dt=x¨= ⁣d2x ⁣dt2= ⁣d ⁣dt(24t6t2)=2412t \begin{aligned} a &= \dot v = \dxt{v}=\ddot x = \dxtt{x} \\ &= \dt(24t-6t^2) \\ &= 24-12t \end{aligned} {x=12t22t3v=24t6t2a=2412t(2181) \therefore \begin{cases} x &= 12t^2-2t^3 \\ v &= 24t-6t^2 \\ a &= 24-12t \end{cases}\cdots({2-18-1}) (a) Ans=x(3.5)=? \Ans = x(3.5) =? (2181),(2-18-1), $$ \begin{aligned} x(3...

2-17 할리데이 10판 솔루션 일반물리학

x=9.75+1.50t3 x = 9.75 + 1.50t^3 (a) Ans=vˉ=x(t3)x(t2)t3t2=x(3.00)x(2.00)3.002.00=9.75+1.50(3.00)3{9.75+1.50(2.00)3}3.002.00=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} (b) Ans=x˙(2.00)=?\Ans = \dot x(2.00) =? $$ \begin{aligned} \\ \dot x &= \dxt{x} \\ &= \dt(9.75 + 1.50t^3) \\ &= \frac{9}{2}t^2 \cdots..

2-16 할리데이 10판 솔루션 일반물리학

x=4.06.0t2(2161) x = 4.0- 6.0t^2 \cdots({2-16-1}) (a) Ans=tx˙=0=? \Ans = t_{\dot x=0} =? x˙= ⁣dx ⁣dt= ⁣d ⁣dt(4.06.0t2)=12t=0 \begin{aligned} \dot x &= \dxt{x} \\&= \dt(4.0- 6.0t^2) \\&= -12t=0 \end{aligned} t=0[s] t=0\ut{s} (b) Ans=x(0)=?=4.06.0(0)2=4.0[m] \begin{aligned} \Ans &= x(0) =? \\ &= 4.0- 6.0(0)^2 \\ &= 4.0\ut{m} \end{aligned} (c) Ans=tx=0,t0=? \Ans = t_{x=0,t0} =? (2162), (2-16-2), t=23 t = \sqrt{\frac{2}{3}} (e) (f) +20t (그래프참조) (g) 증가시킴 (그래프참조)

2-15 할리데이 10판 솔루션 일반물리학

x=18t+5.0t2x = 18t + 5.0t^2 (a) Ans=x˙(2)=? \Ans = \dot 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} (b) Ans=x(t2)x(t1)t2t1=?=x(3.0)x(2.0)3.02.0={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}

2-14 할리데이 10판 솔루션 일반물리학

x=16tet x = 16te^{-t} Ans=x(tx˙=0)=? \Ans = \abs{x(t_{\dot x=0})} =? x˙= ⁣dx ⁣dt= ⁣d ⁣dt(16tet)=16et16tet=16et(t1)=0 \begin{aligned} \dot x &= \dxt{x} \\&= \dt(16te^{-t}) \\&= 16 e^{-t}-16 te^{-t} \\&= -16 e^{-t} (t-1) = 0 \end{aligned} t=1[s] t=1\ut{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}

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} 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} (b) $$ \beg..

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} (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} ta=dvvst=ntan=tta=tvvsd \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} $$ \begin{aligned} x..

2-11 할리데이 10판 솔루션 일반물리학

{vc=15.00[m/s]t=150.0[s]xc0xa0=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} $$ \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{..