2-22 할리데이 10판 솔루션 일반물리학

$$x= ct^2 - bt^3$$ (a) $$\Ans = c =?$$ $$\put k_n = \text{AnyNumber}$$ $$\begin{aligned} x &= ct^2 - bt^3=k_1\ut{m} \\ &\Rightarrow c(t\ut{s})^2 - b(t\ut{s})^3 = k_1\ut{m} \cdots(2-22-1) \\ &\Rightarrow c(t\ut{s})^2 = k_2\ut{m} \\ &\Rightarrow c(t^2\ut{s^2}) = k_2\ut{m} \\ &\Rightarrow c = \frac{k_2\ut{m}}{t^2\ut{s^2}} \\ &\Rightarrow c = \frac{k_2}{t^2}\ut{m/s^2} \end{aligned}$$ (b) $$\Ans = b =?$$ $$(2-22-1),$$ $$\begin{aligned} \\ &c(t\ut{s})^2 - b(t\ut{s})^3 = k_1\ut{m} \\ &\Rightarrow b(t\ut{s})^3 = k_3\ut{m} \\ &\Rightarrow b(t^3\ut{s^3}) = k_3\ut{m} \\ &\Rightarrow b = \frac{k_3\ut{m}}{t^3\ut{s^3}} \\ &\Rightarrow b = \frac{k_3}{t^3}\ut{m/s^3} \end{aligned}$$ (c)~(m) $$\begin{cases} c &= 3.0 \\ b &= 2.0 \end{cases}$$ $$x = 3.0t^2 - 2.0t^3$$ $$\begin{aligned} v &= \dot x = \dxt{x} \\ &=\dt(3.0t^2 - 2.0t^3) \\ &= 6 t-6 t^2 \end{aligned}$$ $$\begin{aligned} a &= \dot v = \dxt{v} \\ &=\dt(6 t-6 t^2) \\ &=6 - 12 t \end{aligned}$$ $$\therefore \begin{cases} x &= 3.0t^2 - 2.0t^3 \\ v &= 6 t-6 t^2 \\ a &=6 - 12 t \end{cases} \cdots(2-22-2)$$ (c) $$t_{\max x} = ?$$ $$v = 6 t_c-6 t_c^2 = 0$$ $$t_c = 0, 1\cdots(2-22-3)$$ $$a(t_c) = (0,6),(1,-6)$$ $$\therefore \begin{cases} x_{LocalMin} &= x(0) \\ x_{LocalMax} &= x(1) \end{cases}$$ $$\therefore t_{\max x} = 1\ut{s} \ (\because t>0)$$ (d) $$\Ans = \int_0^4 \abs{v}\,\dd t =?$$ $$(2-22-3),$$ $$\abs{v} = \begin{cases} v, &(0 \le t \le1) \\ -v &(1 \le t) \end{cases}$$ $$\begin{aligned} \\ \therefore \Ans &= \int_0^4 \abs{v(t)}\,\dd t \\ &= \int_0^1 v\,\dd t+\int_1^4 -v\,\dd t \\ &= \int_0^1 v\,\dd t-\int_1^4 v\,\dd t \\ &= \[V\]_0^1-\[V\]_1^4 \\ &= \[x\]_0^1-\[x\]_1^4 \\ &= \[x(1)-x(0)\]-\[x(4)-x(1)\] \\ &= 2x(1)-x(0)-x(4) \\&=82\ut{m} \end{aligned}$$ (e) $$\begin{aligned} \Ans &= x(4)-x(0) \\ &=\{3.0(4)^2 - 2.0(4)^3\}-\{3.0(0)^2 - 2.0(0)^3\} \\ &= -8.0\times 10\ut{m} \end{aligned}$$ (f)~(i) $$(2-22-2),$$ $$v = 6 t-6 t^2$$ $$\begin{aligned} \, (f) \, v(1.0) &= 6(1.0)-6(1.0)^2 = 0.0\ut{m/s} \\ \, (g)\, v(2.0) &= 6(2.0)-6(2.0)^2 = -12\ut{m/s} \\ \, (h)\, v(3.0) &= 6(3.0)-6(3.0)^2 = -36\ut{m/s} \\ \, (i)\, v(4.0) &= 6(4.0)-6(4.0)^2 = -72\ut{m/s} \end{aligned}$$ (j)~(l) $$(2-22-2),$$ $$a =6 - 12 t$$ $$\begin{aligned} \, (j) \, a(1.0) &= 6 - 12(1.0) = -6.0\ut{m/s^2} \\ \, (k)\, a(2.0) &=6 - 12(2.0) = -18\ut{m/s^2} \\ \, (l)\, a(3.0) &= 6 - 12(3.0)= -30\ut{m/s^2} \\ \, (m)\, a(4.0) &=6 - 12(4.0)= -42\ut{m/s^2} \end{aligned}$$